What is Probability? & What is Randomness?

Chapter 7: The Mathematics of Maybe — Introduction to Probability

Page 1 of 7: What is Probability? & What is Randomness?

Concept Introduction

Have you ever watched the toss before a cricket match? The captain flips a coin, calls "Heads" or "Tails", and the outcome decides which team bats first. It seems simple, but have you ever wondered why we do this? We do it because the outcome is uncertain. Neither captain can know for sure if it will be heads or tails. This uncertainty, this element of chance, is at the heart of daily life, from weather forecasts predicting a 70% chance of rain to a doctor estimating the success rate of a surgery.

Probability is the branch of mathematics that allows us to measure this uncertainty. It gives us a precise number, from 0 to 1, to describe how likely an event is to happen. It's the mathematics of "maybe," "perhaps," and "probably." In this chapter, we will learn how to calculate these chances, moving from simple guesses to logical predictions.

{{FORMULA: expr=P(E) = (Number of favourable outcomes) / (Total number of outcomes) | symbols=P(E):Probability of Event E}}

Definitions & Formulas

Before we dive into calculations, let's get our vocabulary right. These are the building blocks of probability.

The Logic Behind the Formula

How do we arrive at the fundamental formula for probability? It's not magic; it's a logical way of comparing what we want to happen with what can happen. Let's build it step-by-step.

1. Identify the Whole Picture: First, consider any experiment, for example, rolling a standard six-sided die. We must list every single possible outcome. This collection of all outcomes is the Sample Space.

   S = {1, 2, 3, 4, 5, 6}
   

   The total number of possible outcomes is 6. This forms the denominator of our probability fraction – it represents the entire universe of possibilities.

2. Isolate the Desired Part: Next, we define the specific event (E) we are interested in. Let's say we want to find the probability of rolling a number greater than 4.

   The outcomes that satisfy this condition are {5, 6}. These are our "favourable" outcomes.

3. Count the Favourable Outcomes: We simply count how many outcomes belong to our event E.

   Number of favourable outcomes = 2
   

4. Form a Ratio: Probability is fundamentally a comparison. We compare the count of our desired outcomes to the count of all possible outcomes. This comparison is best represented as a fraction.

   Likelihood = (Number of ways our event can happen) / (Total number of things that can happen)
   

5. Formalize the Formula: This logic leads us directly to the classical definition of probability for equally likely outcomes.

   P(E) = (Number of outcomes favourable to E) / (Total number of possible outcomes)
   

   For our example, the probability of rolling a number greater than 4 is:

   P(Number > 4) = 2 / 6 = 1/3
   

{{KEY: type=concept | title=The Range of Probability | text=The value of probability, P(E), will always be a number between 0 and 1, inclusive. P(E) = 0 means the event is impossible (like rolling a 7 on a standard die). P(E) = 1 means the event is certain (like getting a number less than 7 on a standard die).}}

What is Randomness?

A key idea in probability is randomness. An experiment is called random if it has more than one possible outcome, and we cannot predict with certainty which outcome will occur in advance.

• When you toss a fair coin, you know the possibilities are Heads or Tails, but you don't know which one it will be. This is randomness.
• When you roll a die, you know the outcome will be from 1 to 6, but you can't be sure which number will come up. This is also randomness.

Crucially, "random" does not mean "without any rules." In fact, it's the opposite! Randomness has a predictable long-term pattern. While you can't predict the next coin toss, if you toss it 1000 times, you can be very confident that you'll get very close to 500 Heads and 500 Tails. Probability is the tool that helps us understand and predict these long-term patterns within random events.

Solved Examples

Example 1: The Classic Coin Toss (Easy)

Given: A fair coin is tossed once.

To Find: The probability of getting a Tail.

Solution:

1. First, list all possible outcomes. This is our sample space, S.

   S = {Head, Tail}
   

2. Count the total number of possible outcomes.

   Total number of outcomes = 2
   

3. Next, identify the event (E) we are interested in, which is "getting a Tail". List the favourable outcomes for this event.

   Favourable outcomes = {Tail}
   

4. Count the number of favourable outcomes.

   Number of favourable outcomes = 1
   

5. Apply the probability formula: P(E) = (Number of favourable outcomes) / (Total number of outcomes).

   P(Tail) = 1 / 2
   

Final Answer: The probability of getting a Tail is 1/2 or 0.5.

Example 2: Rolling a Standard Die (Medium)

Given: A standard six-sided die is rolled once.

To Find: The probability of getting a prime number.

Solution:

1. List the sample space (all possible outcomes) for rolling a die.

   S = {1, 2, 3, 4, 5, 6}
   

2. Count the total number of outcomes.

   Total number of outcomes = 6
   

3. Identify the event (E) as "getting a prime number". A prime number is a number greater than 1 that has only two factors: 1 and itself. Let's list the prime numbers from our sample space.

   Favourable outcomes (prime numbers) = {2, 3, 5}
   

   Note: 1 is not a prime number.

4. Count the number of favourable outcomes.

   Number of favourable outcomes = 3
   

5. Use the probability formula.

   P(Prime Number) = 3 / 6
   

6. Simplify the fraction.

   P(Prime Number) = 1/2
   

Final Answer: The probability of getting a prime number is 1/2.

Example 3: Marbles in a Bag (Hard)

Given: A bag contains 4 red marbles, 8 blue marbles, and 3 green marbles. One marble is drawn at random from the bag.

To Find: The probability that the marble drawn is blue.

Solution:

1. First, calculate the total number of marbles in the bag. This will be the total number of possible outcomes.

   Total marbles = 4 (Red) + 8 (Blue) + 3 (Green) = 15
   

2. So, the total number of possible outcomes is 15.

3. Identify the event (E) as "drawing a blue marble".

4. Count the number of outcomes favourable to this event. This is simply the number of blue marbles.

   Number of favourable outcomes (blue marbles) = 8
   

5. Apply the probability formula.

   P(Blue) = (Number of blue marbles) / (Total number of marbles)
   

   P(Blue) = 8 / 15
   

Final Answer: The probability of drawing a blue marble is 8/15.

Example 4: The 'Not' Question (Tricky)

Given: A standard deck of 52 playing cards. One card is drawn at random.

To Find: The probability that the card drawn is not a King.

Solution:

1. The total number of possible outcomes is the total number of cards in the deck.

   Total number of outcomes = 52
   

2. We can solve this in two ways. Let's try the direct method first. The event is "not a King". We need to count all the cards that are not Kings.

   A standard deck has 4 Kings. So, the number of cards that are not Kings is:

   Number of non-Kings = 52 - 4 = 48
   

3. This is our number of favourable outcomes. Now, apply the formula.

   P(Not a King) = 48 / 52
   

4. Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 4.

   P(Not a King) = 12 / 13
   

5. Alternative Method (using the complement rule): Let's first find the probability of the opposite event: drawing a King.

   The number of Kings is 4.

   P(King) = 4 / 52 = 1 / 13
   

6. The probability of an event not happening is 1 minus the probability of it happening. P(not E) = 1 - P(E).

   P(Not a King) = 1 - P(King)
   

   P(Not a King) = 1 - (1/13) = 13/13 - 1/13 = 12/13
   

   Both methods give the same result.

Final Answer: The probability of not drawing a King is 12/13.

Tips & Tricks

Here are a few shortcuts and checks to keep in mind when solving probability problems.

Common Mistakes

Many students make similar small errors. Watch out for these common traps!

Brain-Teaser Questions

1. A letter is chosen at random from the word "PROBABILITY". What is the probability that the chosen letter is the letter 'B'?

   💡 Answer: Total letters = 11. The letter 'B' appears 2 times. So, P(B) = (Number of B's) / (Total letters) = 2/11.

2. You roll a standard six-sided die. Is it more likely that you will roll an even number, or that you will roll a number less than 4?

   💡 Answer: The even numbers are {2, 4, 6}. So, P(Even) = 3/6 = 1/2. The numbers less than 4 are {1, 2, 3}. So, P(<4) = 3/6 = 1/2. Both events are equally likely.

3. A weather forecast says there is a 70% chance of rain tomorrow. What is the probability that it will not rain tomorrow? Express your answer as a fraction.

   💡 Answer: The total probability of all outcomes must be 1 (or 100%). P(Rain) = 70% = 70/100 = 7/10. P(Not Rain) = 1 - P(Rain) = 1 - 7/10 = 3/10.

Mini Cheatsheet

Here's a small summary you can screenshot for quick revision of today's concepts.

The Probability Scale

Page 2: The Probability Scale

Concept Introduction

Have you ever checked the weather forecast and seen a "40% chance of rain"? What does that number actually mean? It doesn't guarantee rain, but it also doesn't promise a sunny day. Instead, it gives you a measure of likelihood. This single number helps you decide whether to carry an umbrella. It's a way of quantifying uncertainty.

This idea is the very heart of probability. We use numbers to describe how likely an event is to happen. To keep things consistent, mathematicians created a standard scale for this purpose: The Probability Scale. It's a simple number line that starts at 0 and ends at 1. Every possible event, from the absolutely impossible to the absolutely certain, can be placed somewhere on this scale. It's our universal ruler for measuring the "chance" of something happening.

Definitions & Formulas

In probability, we use specific terms and a foundational principle to understand the scale. The most important idea is that probability is a ratio comparing favorable outcomes to all possible outcomes.

{{KEY: type=concept | title=The Golden Rule of Probability | text=The probability of any event E must always be a number between 0 and 1, inclusive. This can be written as 0 ≤ P(E) ≤ 1. A probability can never be negative, nor can it be greater than 1.}}

Logic Behind the 0-to-1 Scale

Why is probability measured on a scale from 0 to 1 and not, say, 0 to 100 or -1 to 1? The logic comes directly from its definition as a ratio of a part to a whole.

1. Probability is a Fraction: At its core, probability is a fraction. The numerator (the top part) is the number of ways your desired event can happen, which we call favorable outcomes. The denominator (the bottom part) is the total number of things that could possibly happen, known as the total possible outcomes.

   Probability = (Part) / (Whole)
   

2. The Lowest Possible Value (The "Floor"): Let's consider the smallest possible number of favorable outcomes. Can you have a negative number of outcomes? No. The absolute minimum is zero. This happens when the event you're interested in simply cannot occur. Example: Rolling a '7' on a standard six-sided die. The number of favorable outcomes is 0.

   P(rolling a 7) = 0 / 6 = 0
   

   This sets the minimum value of our scale at 0.

3. The Highest Possible Value (The "Ceiling"): Now, what's the maximum number of favorable outcomes? The most favorable outcomes you can have is when every single possible outcome is a favorable one. In this case, the number of favorable outcomes is equal to the total number of possible outcomes. Example: Getting a number less than 7 on a standard die roll. All six outcomes (1, 2, 3, 4, 5, 6) are favorable.

   P(number < 7) = 6 / 6 = 1
   

   This sets the maximum value of our scale at 1.

4. Everything in Between: For any other event, the number of favorable outcomes will be more than 0 but less than the total number of outcomes. This means the probability will be a fraction that is strictly between 0 and 1. Example: Rolling a '3' on a die. There is 1 favorable outcome out of 6 total outcomes.

   P(rolling a 3) = 1/6 ≈ 0.167
   

   This value, 1/6, is clearly between 0 and 1.

Therefore, this logical framework naturally confines all probabilities to the range from 0 (impossible) to 1 (certain).

{{VISUAL: diagram: A number line from 0 to 1. Labels at 0 for "Impossible", 0.5 for "Even Chance", and 1 for "Certain". An arrow points from 0 towards 0.5 labeled "Less Likely", and another arrow points from 0.5 towards 1 labeled "More Likely".}}

Solved Examples

Let's apply this understanding to classify different events.

Example 1: The Certainty of Calendars (Easy)

Given: The event is "The day after Sunday will be Monday."

To Find: Place this event on the probability scale and classify it.

Solution:

1. Analyze the event. According to the standard calendar system used worldwide, the day that follows Sunday is always Monday.

2. There are no other possible outcomes. This is a fact that is guaranteed to happen.

3. An event that is guaranteed to happen is a certain event.

4. The probability value for a certain event is 1.

   P(Monday after Sunday) = 1
   

Final Answer: The event is Certain, with a probability of 1.

Example 2: The Classic Coin Toss (Medium)

Given: A fair coin is tossed once. The event is getting "Heads".

To Find: Calculate the probability and classify the event on the scale.

Solution:

1. Identify the total possible outcomes. When a coin is tossed, there are only two possibilities: Heads (H) or Tails (T).

   Total Possible Outcomes = 2
   

2. Identify the number of favorable outcomes. We are interested in the event "getting Heads". There is only one face on the coin that is Heads.

   Number of Favorable Outcomes = 1
   

3. Calculate the probability using the formula P(E) = Favorable / Total.

   P(Heads) = 1 / 2 = 0.5
   

4. A probability of 0.5 corresponds to an even chance on the scale. It's exactly in the middle, meaning getting heads is just as likely as not getting heads.

Final Answer: The probability of getting heads is 0.5, which is classified as an Even Chance.

Example 3: The Student Council Election (Hard)

Given: In a class of 30 students, 18 are girls and 12 are boys. One student is randomly selected to be the class representative. The event is "a girl is selected".

To Find: Calculate the probability of selecting a girl and classify this event.

Solution:

1. Determine the total number of possible outcomes. Since any student can be selected, the total number of outcomes is the total number of students.

   Total Possible Outcomes = 30
   

2. Determine the number of favorable outcomes. We want to select a girl. The number of girls in the class is the number of favorable outcomes.

   Number of Favorable Outcomes = 18
   

3. Calculate the probability.

   P(Girl is selected) = 18 / 30
   

4. Simplify the fraction to understand its position on the scale. Both 18 and 30 are divisible by 6.

   P(Girl is selected) = 18 ÷ 6 / 30 ÷ 6 = 3/5
   

5. Convert the fraction to a decimal to easily place it on the scale.

   3 / 5 = 0.6
   

6. Compare the result to the benchmark of 0.5. Since 0.6 is greater than 0.5, the event is more likely to happen than not.

Final Answer: The probability of selecting a girl is 0.6, which is classified as More Likely.

Example 4: Vowels in "PROBABILITY" (Tricky)

Given: All the letters of the word "PROBABILITY" are written on separate cards and shuffled in a bag. One card is drawn at random.

To Find: Calculate the probability of drawing a vowel and classify the event.

Solution:

1. First, identify the total number of possible outcomes. This is the total number of letters in the word "PROBABILITY".

   P, R, O, B, A, B, I, L, I, T, Y → Total letters = 11
   

   So, Total Possible Outcomes = 11.

2. Next, identify the number of favorable outcomes, which are the vowels in the word. The vowels in English are A, E, I, O, U. Let's find them in "PROBABILITY".

   O, A, I, I → Total vowels = 4
   

   So, Number of Favorable Outcomes = 4.

3. Calculate the probability using the formula.

   P(Vowel) = 4 / 11
   

4. To classify this, let's convert the fraction to a decimal.

   4 ÷ 11 ≈ 0.3636...
   

5. Now, compare this value to the 0.5 benchmark. Since 0.36 is less than 0.5, the event is less likely to happen than not.

Final Answer: The probability of drawing a vowel is 4/11 (approximately 0.36), which is classified as Less Likely.

Tips & Tricks

Use these mental shortcuts to master the probability scale.

Common Mistakes

Many students stumble on the same conceptual points. Here's how to stay on the right track.

Brain-Teaser Questions

1. A box contains 91 red marbles and 9 black marbles. Is it "certain" that you will pick a red marble on your first try? Why or why not?

   💡 Answer: No, it is not certain. It is more likely (in fact, very likely, with P(Red) = 91/100 = 0.91), but it is not certain. Certainty requires a probability of exactly 1, which would only happen if all 100 marbles were red. As long as there is even one black marble, there is a non-zero chance of picking it, so the event is not certain.

2. There are two spinners. Spinner A is split into 2 equal sections (Red, Blue). Spinner B is split into 100 equal sections (50 Red, 50 Blue). Is it more likely to land on Red with Spinner A, Spinner B, or are they the same?

   💡 Answer: The likelihood is exactly the same. For Spinner A, P(Red) = 1/2 = 0.5. For Spinner B, P(Red) = 50/100 = 1/2 = 0.5. Both represent an "Even Chance". The number of sections doesn't matter as long as the ratio of favorable outcomes to total outcomes is the same.

3. An event has a probability P(E) = x. If this event is "less likely" to occur than not, what must be true about the value of x?

   💡 Answer: For an event to be "less likely" to occur than not, its probability must be less than the benchmark for an even chance, which is 0.5. Therefore, it must be true that 0 ≤ x < 0.5. It cannot be 0.5 (that's an even chance), and it cannot be 0 because that would make it "impossible," not just "less likely."

Mini Cheatsheet

Measuring Probability Objectively & Experimental Probability

Page 3: Measuring Probability Objectively & Experimental Probability

Introduction: From Guesses to Evidence

When you flip a coin, you might say, "There's a 50% chance of getting heads." But how do you know that? You could flip the coin 100 times and record how many times heads appears. If you get 52 heads out of 100 flips, the experimental probability would be 52/100 = 0.52 or 52%.

This approach — repeating an experiment and counting outcomes — is called measuring probability objectively through experimentation. Unlike subjective probability (which relies on personal belief), experimental probability is rooted in actual data and repeated trials.

It's widely used in quality control (testing 1000 light bulbs to estimate defect rates), medicine (clinical trials testing drug effectiveness), and sports analytics (calculating a cricket player's strike rate from past matches). The more trials you perform, the more reliable your probability estimate becomes.

{{FORMULA: expr=P(Event) = Number of times event occurred / Total number of trials | symbols=P(Event):Experimental probability of the event, Number of times event occurred:Frequency of favorable outcome, Total number of trials:Total repetitions of experiment}}

Core Definitions

The Logic Behind Experimental Probability

Step 1: Define the experiment and the event you want to study.

For example, rolling a standard 6-sided die and studying the event "getting a 4."

Step 2: Perform the experiment multiple times (the more trials, the better).

Roll the die 50 times, 100 times, or even 1000 times.

Step 3: Count how many times the event of interest actually occurred.

Suppose out of 50 rolls, the number 4 appeared 8 times.

Step 4: Calculate the relative frequency.

Relative Frequency = Number of favorable outcomes / Total trials

Relative Frequency = 8/50 = 0.16

Step 5: Express this as experimental probability.

P(getting a 4) = 0.16 or 16%

Step 6: Understand that as the number of trials increases, experimental probability tends to approach the theoretical probability (which for a fair die would be 1/6 ≈ 0.167).

This convergence is known as the Law of Large Numbers.

{{KEY: type=concept | title=Law of Large Numbers | text=As the number of trials in an experiment increases, the experimental probability gets closer and closer to the theoretical probability. This is why casinos and insurance companies rely on millions of data points — randomness evens out over large samples.}}

Solved Examples

Example 1: Coin Toss Experiment (Easy)

Given: A coin is tossed 40 times. Heads appears 23 times.

To Find: Experimental probability of getting heads.

Solution:

1. Identify the number of favorable outcomes (heads).

Number of heads = 23

2. Identify the total number of trials.

Total tosses = 40

3. Apply the experimental probability formula.

P(Heads) = 23/40

4. Convert to decimal.

P(Heads) = 0.575 or 57.5%

Final Answer: 0.575 or 57.5%

Example 2: Drawing Cards from a Deck (Medium)

Given: A card is drawn from a standard 52-card deck, replaced, and the deck is shuffled. This is repeated 200 times. A heart appears 48 times.

To Find: Experimental probability of drawing a heart.

Solution:

1. Count the number of times a heart was drawn.

Number of hearts drawn = 48

2. Count the total number of trials.

Total draws = 200

3. Calculate experimental probability.

P(Heart) = 48/200

4. Simplify the fraction.

P(Heart) = 12/50 = 6/25

5. Convert to decimal.

P(Heart) = 0.24 or 24%

Final Answer: 0.24 or 24%

Example 3: Manufacturing Defect Analysis (Hard)

Given: A factory produces 5000 light bulbs in a day. A quality inspector randomly tests 250 bulbs and finds 5 defective ones.

To Find: (a) Experimental probability of a bulb being defective. (b) Estimate the total number of defective bulbs in the entire day's production.

Solution:

Part (a):

1. Identify favorable outcomes (defective bulbs in sample).

Defective bulbs = 5

2. Identify total trials (bulbs tested).

Bulbs tested = 250

3. Calculate experimental probability.

P(Defective) = 5/250

4. Simplify.

P(Defective) = 1/50 = 0.02 or 2%

Part (b):

5. Apply this probability to the entire production.

Estimated defective bulbs = Total production × P(Defective)

Estimated defective = 5000 × 0.02 = 100 bulbs

Final Answer: (a) 0.02 or 2%, (b) Approximately 100 defective bulbs

Example 4: Cricket Strike Rate (Tricky)

Given: A cricketer has faced 180 balls in a tournament and scored runs off 72 of those balls (either 1, 2, 3, 4, or 6 runs).

To Find: (a) Experimental probability that the batsman scores a run off any ball. (b) If the batsman faces 300 balls in the next match, estimate how many balls he will score runs from.

Solution:

Part (a):

1. Identify balls on which runs were scored.

Balls with runs = 72

2. Identify total balls faced.

Total balls = 180

3. Calculate experimental probability.

P(Scoring run) = 72/180

4. Simplify the fraction.

P(Scoring run) = 2/5 = 0.4 or 40%

Part (b):

5. The tricky part: we assume past performance predicts future performance (a common statistical assumption in sports analytics).

Expected scoring balls = Total balls × P(Scoring run)

Expected scoring balls = 300 × 0.4 = 120 balls

6. Note: This is an estimate — actual performance may vary due to form, opposition, pitch conditions, etc.

Final Answer: (a) 0.4 or 40%, (b) Approximately 120 balls

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Q1: A die is rolled 300 times. The number 6 appears 55 times. What is the experimental probability of NOT getting a 6?

💡 Answer: First find how many times 6 did NOT appear: 300 - 55 = 245 times. Then P(Not 6) = 245/300 = 49/60 ≈ 0.817 or 81.7%. Alternatively, P(Not 6) = 1 - P(6) = 1 - (55/300) = 1 - 0.183 = 0.817.

Q2: In a bag, you randomly pick a marble, note its color, and replace it. After 80 trials, you get: Red = 32 times, Blue = 28 times, Green = 20 times. If you perform 400 more trials, approximately how many times would you expect to pick a blue marble?

💡 Answer: P(Blue) from experiment = 28/80 = 7/20 = 0.35 or 35%. In 400 trials: Expected blue = 400 × 0.35 = 140 times.

Q3: A student claims their coin is "lucky" because it landed heads 60 times out of 100 tosses. Is this strong evidence the coin is unfair? Explain using probability.

💡 Answer: Experimental P(Heads) = 60/100 = 0.6 or 60%. Theoretical for fair coin = 0.5 or 50%. A 10% difference in 100 tosses is not unusual due to random variation. We'd need many more trials (1000+) to confidently claim the coin is biased. In statistics, this is related to "margin of error" — small samples have larger margins.

Mini Cheatsheet

{{KEY: type=exam-tip | title=CBSE Pattern Alert | text=CBSE often asks: "A die is rolled n times, outcome x appears k times. Find experimental probability." Then they may ask you to estimate outcomes for a larger number of trials. Master the two-step process: (1) Find P from sample, (2) Multiply P by new total to estimate.}}

Theoretical Probability & Analysing Statistical Data Using Probability — Part 1

Page 4: Theoretical Probability & Analysing Statistical Data Using Probability — Part 1

Concept Introduction

Imagine you're at a cricket match, and the umpire is about to toss a coin to decide which team bats first. Before the coin lands, what is the chance it will show heads? You might intuitively say "half" or "50%". This intuition is the foundation of theoretical probability — predicting outcomes based on logic and mathematical reasoning, not by actually flipping the coin hundreds of times.

Theoretical probability assumes all outcomes are equally likely in a perfectly fair situation. A fair coin, a balanced die, a well-shuffled deck of cards — these are ideal scenarios where we can calculate probabilities without conducting any experiments. In the real world, we also use statistical data from samples to estimate probabilities for entire populations, like predicting election results from opinion polls or forecasting product sales based on customer surveys. This page will help you master both concepts.

{{FORMULA: expr=P(Event) = (Number of favourable outcomes) / (Number of all possible outcomes) | symbols=P(Event):Probability of the event, favourable outcomes:outcomes matching our interest, possible outcomes:all outcomes in the sample space}}

Definitions & Formulas

Core Formula for Theoretical Probability

For any event A in a sample space with equally likely outcomes:

P(A) = n(A) / n(S)

where n(A) is the number of favourable outcomes and n(S) is the total number of possible outcomes.

Derivation & Logic: Why Does This Formula Work?

Let us understand the reasoning behind the theoretical probability formula through a step-by-step approach.

1. Start with a simple experiment: Consider rolling a fair 6-sided die. The sample space is:

S = {1, 2, 3, 4, 5, 6}

2. Count total possible outcomes: The die has 6 faces, so there are 6 equally likely outcomes.

n(S) = 6

3. Define your event: Suppose we want to find the probability of rolling an even number. The event E consists of:

E = {2, 4, 6}

4. Count favourable outcomes: There are 3 even numbers among the 6 possible outcomes.

n(E) = 3

5. Apply the fairness assumption: Since the die is fair, each outcome has the same chance. If we divide the number of favourable outcomes by the total outcomes, we get the proportion of "success":

P(E) = n(E) / n(S) = 3 / 6 = 1/2

6. Generalize the principle: This logic works for ANY situation with equally likely outcomes. The probability is simply the ratio of how many outcomes we want to the total number of outcomes possible. This is the foundation of classical or theoretical probability.

{{KEY: type=concept | title=The Equally Likely Assumption | text=Theoretical probability ONLY works when all outcomes have the same chance of occurring. A biased die or a weighted coin violates this assumption, and we would need experimental probability instead.}}

Solved Examples

Example 1: Single Die Roll (Easy)

Given: A fair 6-sided die is rolled once.

To Find: The probability of getting a number greater than 4.

Solution:

1. Identify the sample space and total outcomes.

S = {1, 2, 3, 4, 5, 6}
n(S) = 6

2. Define the event: numbers greater than 4.

A = {5, 6}
n(A) = 2

3. Apply the probability formula.

P(A) = n(A) / n(S) = 2 / 6 = 1/3

Final Answer: 1/3 or approximately 0.333 or 33.3%

Example 2: Letter Selection (Medium)

Given: A letter is chosen at random from the word 'PROBABILITY'.

To Find: The probability of selecting the letter 'I'.

Solution:

1. Count total letters in 'PROBABILITY'.

Letters: P-R-O-B-A-B-I-L-I-T-Y
n(S) = 11

2. Count how many times 'I' appears.

'I' appears 2 times
n(I) = 2

3. Calculate the probability.

P(I) = n(I) / n(S) = 2 / 11

4. Convert to decimal for clarity.

P(I) ≈ 0.182 or 18.2%

Final Answer: 2/11 ≈ 0.182 or 18.2%

Example 3: Card Selection (Medium-Hard)

Given: A standard deck of 52 playing cards contains 4 suits (hearts, diamonds, clubs, spades), each with 13 cards.

To Find: The probability of drawing a King or a Heart from a well-shuffled deck.

Solution:

1. Identify total possible outcomes.

n(S) = 52

2. Count Kings in the deck (one per suit).

Number of Kings = 4

3. Count Hearts in the deck.

Number of Hearts = 13

4. Check for overlap: the King of Hearts is counted in both groups, so we must subtract 1 to avoid double-counting.

Favourable outcomes = 4 + 13 - 1 = 16

5. Apply the probability formula.

P(King or Heart) = 16 / 52 = 4 / 13

Final Answer: 4/13 ≈ 0.308 or 30.8%

Example 4: Statistical Probability from Sample Data (Hard)

Given: A survey of 80 randomly selected students in a school reveals: 32 prefer cricket, 28 prefer football, 12 prefer basketball, 8 prefer hockey.

To Find: a) The probability that a randomly chosen student from this sample prefers cricket. b) If the school has 1200 students, estimate how many would prefer football.

Solution:

Part a):

1. Total students in the sample.

n(Sample) = 80

2. Number of students who prefer cricket.

n(Cricket) = 32

3. Calculate probability using sample data.

P(Cricket) = 32 / 80 = 2/5 = 0.4

Part b):

4. Use the sample probability to estimate for the whole population.

Population size = 1200

5. Estimated number who prefer football.

P(Football) = 28 / 80 = 0.35

6. Apply to population.

Estimated students preferring football = 0.35 × 1200 = 420

Final Answer:

a) 2/5 or 0.4 or 40%

b) Approximately 420 students

Tips & Tricks

Common Mistakes

{{KEY: type=warning | title=The Gambler's Fallacy | text=Past outcomes do NOT affect future independent events. If you flip a coin and get 5 heads in a row, the probability of heads on the next flip is STILL 1/2, not lower. The coin has no memory!}}

Brain-Teaser Questions

1. A bag contains 5 red balls, 3 green balls, and 2 yellow balls. If you pick one ball at random, what is the probability it is NOT green?

💡 Answer: Total balls = 5 + 3 + 2 = 10 Green balls = 3, so Not green = 10 - 3 = 7 P(Not green) = 7/10 = 0.7 or 70%

2. In a class of 40 students, 15 like only pizza, 10 like only burgers, 8 like both, and 7 like neither. If one student is chosen randomly, what is the probability they like at least one of pizza or burgers?

💡 Answer: Students who like at least one = Total - Neither = 40 - 7 = 33 P(at least one) = 33/40 = 0.825 or 82.5%

3. A survey of 200 people shows 120 prefer tea and 80 prefer coffee. If a town has 5000 people, estimate how many prefer tea. What assumption are you making?

💡 Answer: Sample probability for tea = 120/200 = 0.6 Estimated tea lovers in town = 0.6 × 5000 = 3000 Assumption: The sample is representative of the entire town population (no bias in selection).

Mini Cheatsheet

Remember: Theoretical probability is your mathematical prediction in a perfect world. Statistical probability uses real-world data to estimate what might happen. Both are powerful tools — one relies on logic, the other on evidence. Master both, and you'll see probability everywhere in daily life!

Analysing Statistical Data Using Probability — Part 2

Page 5 of 7: Analysing Statistical Data Using Probability — Part 2

From Data to Decisions: Experimental Probability

So far, we've explored the basics of probability. Now, let's see how probability acts in the real world, not just in theory. Imagine a company launching a new flavour of chips. It's too expensive and slow to ask everyone in the country if they like it. Instead, they give out free samples at a mall to 500 people and record their feedback. If 300 people love it, they've gathered statistical data.

This data helps them calculate the experimental probability that a random person will like the new flavour (300/500 = 0.6 or 60%). Based on this, they can predict how many packets to produce for the entire country. This is the power of using small, real-world experiments (samples) to make big decisions about a larger group (population). It's using math to make an educated guess.

{{FORMULA: expr=P(E) = (Number of trials where an event occurred) / (Total number of trials) | symbols=P(E):Experimental Probability of Event E}}

Definitions & Formulas

Understanding the language of statistical probability is key. These terms form the foundation for everything we will discuss.

Logic: The Law of Large Numbers

How do we connect a real-world experiment to a perfect theory? The bridge is the Law of Large Numbers. It explains why experimental probability can seem 'off' at first but becomes more reliable as we perform more trials.

1. Start with Theory. In a perfect world, a fair die has six equally likely outcomes. The theoretical probability of rolling a '4' is fixed.

   P(rolling a 4) = 1/6 ≈ 0.167 or 16.7%
   

2. Conduct a Small Experiment. Now, let's roll a real die 12 times. You might get a '4' three times. The experimental probability is:

   Experimental P(rolling a 4) = 3/12 = 1/4 = 0.25 or 25%
   

   Notice that 0.25 is not very close to the theoretical value of 0.167.

3. Recognize Short-Term Randomness. In a small number of trials, results can be very random and stray far from the theoretical value. You might even roll the die 12 times and get no '4's at all! (Experimental Probability = 0/12 = 0). This doesn't mean the die is broken; it's just short-term chance at play.

4. Increase the Number of Trials. Now, imagine you have a lot of time and roll the die 12,000 times. You will find that the number of '4's you get will be very close to 2,000. Let's say you get 2,015 fours.

5. Calculate the New Experimental Probability. The new experimental probability is:

   Experimental P(rolling a 4) = 2015 / 12000 ≈ 0.1679
   

6. The Convergence. This new value, 0.1679, is extremely close to the theoretical probability of 0.167. The Law of Large Numbers states that as the number of trials increases, the experimental probability will get closer and closer to the theoretical probability. This is why casino owners are confident they will make money in the long run, even if a few people win big in the short term.

Solved Examples

Let's apply these concepts to solve some problems, moving from simple calculations to more complex reasoning.

Example 1: The Music Survey (Easy)

Given: A radio station asks 200 listeners what type of music they prefer. The results are: 90 prefer Pop, 60 prefer Rock, 30 prefer Hip-Hop, and 20 prefer Classical.

To Find: The experimental probability that the next caller will prefer Rock music.

Solution:

1. Identify the total number of trials. This is the total number of listeners surveyed.

   Total listeners (trials) = 200
   

2. Identify the number of favorable outcomes for the event (preferring Rock).

   Number of listeners who prefer Rock = 60
   

3. Calculate the experimental probability using the formula.

   P(prefers Rock) = (Number who prefer Rock) / (Total listeners)
   

   P(prefers Rock) = 60 / 200
   

4. Simplify the fraction and express it as a decimal or percentage.

   P(prefers Rock) = 6/20 = 3/10 = 0.3
   

Final Answer: The experimental probability that the next caller will prefer Rock is 0.3 or 30%.

Example 2: Factory Quality Control (Medium)

Given: A factory produces 10,000 light bulbs a day. A quality control officer randomly samples 400 bulbs and finds that 8 are defective.

To Find: Approximately how many defective bulbs can the factory expect to produce in a full day?

Solution:

1. First, calculate the experimental probability of a bulb being defective based on the sample.

   P(defective) = (Number of defective bulbs in sample) / (Total bulbs in sample)
   

   P(defective) = 8 / 400
   

2. Simplify this probability.

   P(defective) = 1 / 50 = 0.02
   

   This means there is a 2% chance that any given bulb is defective.

3. Now, apply this probability to the entire population (the total daily production) to estimate the total number of defective bulbs.

   Expected defective bulbs = P(defective) × Total production
   

   Expected defective bulbs = 0.02 × 10000
   

4. Calculate the final number.

   Expected defective bulbs = 200
   

Final Answer: The factory can expect to produce approximately 200 defective bulbs in a day.

Example 3: Coin Toss Reality Check (Hard)

Given: Priya flips a coin. The theoretical probability of getting a 'Head' is ½. She flips the coin 50 times and gets 21 heads.

To Find: a) The experimental probability of getting a head. b) Why the experimental probability is different from the theoretical probability. c) What would likely happen if she flipped the coin 5000 times?

Solution:

1. Part (a): Calculate the experimental probability. Use the data from Priya's experiment.

   Experimental P(Head) = (Number of heads) / (Total flips)
   

   Experimental P(Head) = 21 / 50 = 0.42
   

2. Part (b): Explain the difference. The theoretical probability P(Head) = 0.5 assumes a perfectly fair coin and ideal conditions. The experimental probability 0.42 is based on a small number of actual, random trials. In the short run, random chance can cause the results to deviate from the theoretical ideal. Getting 21 heads instead of the "expected" 25 is normal statistical variation.

3. Part (c): Predict the outcome for more trials. According to the Law of Large Numbers, as the number of trials increases, the experimental probability tends to get closer to the theoretical probability. If Priya flipped the coin 5000 times, the number of heads would likely be very close to 2500.

   The experimental probability would be something like 2510/5000 = 0.502, which is much closer to the theoretical value of 0.5.

Final Answer: a) The experimental probability is 0.42 or 42%. b) The difference is due to random variation in a small number of trials. c) If she flipped it 5000 times, the experimental probability would get much closer to the theoretical probability of 0.5.

Example 4: The Gambler's Mistake (Tricky)

Given: You are playing Snakes and Ladders with a single, fair 6-sided die. To win the game, you need to roll a '6'. In your last 5 turns, you have rolled: 1, 3, 2, 4, 1. You have not rolled a 6 yet. Your friend says, "You are due for a 6! It's much more likely to come up now!"

To Find: Is your friend's reasoning correct? What is the actual probability of rolling a '6' on your next turn?

Solution:

1. Identify the core misconception. Your friend's statement is an example of the Gambler's Fallacy. This is the mistaken belief that past random events influence future independent events.

2. State the principle of independent events. Each roll of a fair die is an independent event. The die has no memory. It does not know that it hasn't landed on '6' for the past five rolls.

3. Determine the theoretical probability for the next roll. The die is fair and has 6 sides. The probability of rolling any specific number, including a '6', remains the same for every single roll.

   Number of favorable outcomes (rolling a '6') = 1
   

   Total possible outcomes = 6
   

4. Calculate the probability for the next turn.

   P(rolling a '6') = 1/6
   

   P(rolling a '6') ≈ 0.167 or 16.7%
   

Final Answer: Your friend's reasoning is incorrect. This is the Gambler's Fallacy. The probability of rolling a '6' on the next turn is still exactly 1/6, regardless of the previous rolls.

{{KEY: type=concept | title=The Gambler's Fallacy | text=The mistaken belief that if a particular random event has occurred more or less frequently than expected in the past, it is more or less likely to happen in the future. Remember: a fair coin or die has no memory!}}

Tips & Tricks

Use these shortcuts to think more clearly about probability in real-world scenarios.

Common Mistakes

Many students stumble on the difference between theory and reality. Here’s how to avoid common pitfalls.

Brain-Teaser Questions

1. A company wants to test a new video game. They conduct a survey at a local gaming cafe from 8 PM to 10 PM and find that 95% of people surveyed love the game. Why might their prediction that "95% of all players will love the game" be flawed?

   💡 Answer: The sample is not representative of the entire population of gamers. They only surveyed people who were already at a gaming cafe late at night, who are likely to be very serious gamers. This sample excludes casual players, younger players who have bedtimes, or people who play on consoles at home. This is called sampling bias.

2. Two friends, Rohan and Alisha, are testing a spinner that is supposed to be split evenly into Red and Blue (50% chance for each). Rohan spins it 10 times and gets Red 8 times (80%). Alisha spins it 500 times and gets Red 260 times (52%). Which result gives you more confidence that the spinner is actually fair?

   💡 Answer: Alisha's result. According to the Law of Large Numbers, a larger number of trials gives an experimental probability that is closer to the true theoretical probability. Rohan's 80% is far from 50%, but this is likely due to randomness in a very small sample. Alisha's result of 52% is very close to the expected 50%, which is strong evidence that the spinner is fair.

3. In a national lottery, 6 numbers are drawn from a pool of 49 balls. The ball #42 has not been drawn in the last 50 games. A television expert claims, "The number 42 is 'cold' and is less likely to be drawn next time." A ticket buyer claims, "The number 42 is 'due' and is more likely to be drawn." Who is correct?

   💡 Answer: Neither is correct. Both are falling for a version of the Gambler's Fallacy. Each lottery draw is an independent event. The machine and the balls have no memory of what numbers were drawn before. The probability of 42 being drawn in the next game is exactly the same as any other number, regardless of its past history.

Mini Cheatsheet

Elements of Probability: Sample Spaces and Events

Elements of Probability: Sample Spaces and Events

Concept Introduction

Imagine you are at a street food stall in Mumbai, waiting to play a simple dice game. The vendor says, "Roll this die once. If you get a 5 or 6, you win a free vada pav!" Before you roll, you naturally think about all the faces that could land up — 1, 2, 3, 4, 5, or 6. This complete list of possibilities is what mathematicians call a sample space.

Now, the winning condition — "getting a 5 or 6" — is just a specific subset of that complete list. This subset is called an event. Understanding sample spaces and events is the foundation of probability theory. Every probability question begins by identifying what can happen (sample space) and what we care about (event). Whether you're predicting cricket match outcomes, weather patterns, or the result of a coin toss, these two concepts anchor your analysis.

In this page, we will systematically explore how to construct sample spaces, identify events within them, and use these structures to calculate probabilities.

{{FORMULA: expr=P(E) = n(E) / n(S) | symbols=P(E):Probability of event E, n(E):Number of outcomes in event E, n(S):Total outcomes in sample space S}}

Definitions & Core Terminology

Building the Logic: From Experiment to Probability

Understanding probability requires a systematic four-step approach:

Step 1: Identify the Random Experiment

A random experiment is any process whose outcome cannot be predicted with certainty. Examples: tossing a coin, rolling a die, drawing a card from a deck.

Step 2: Construct the Sample Space (S)

List every possible outcome. The sample space must be:

• Exhaustive — includes every possible result
• Mutually Exclusive — no outcome appears twice
• Well-defined — clear boundaries for what counts as an outcome

Step 3: Define the Event (E)

Identify which specific outcomes satisfy the condition you care about. An event is always a subset of S. It can be:

• A simple event (single outcome, like "rolling a 3")
• A compound event (multiple outcomes, like "rolling an even number")

Step 4: Calculate Probability

Once S and E are defined, apply the fundamental formula:

P(E) = n(E) / n(S)

Step 5: Verify Logical Consistency

Check that 0 ≤ P(E) ≤ 1. If P(E) = 0, the event is impossible. If P(E) = 1, the event is certain.

Step 6: Interpret the Result

Translate the numerical probability back into real-world language. A probability of 1/6 ≈ 0.167 means "about 16.7% chance" or "roughly 1 in 6 times."

{{KEY: type=concept | title=Sample Space Must Be Complete | text=Every possible outcome must appear in S exactly once. Missing an outcome or listing duplicates invalidates all probability calculations based on that sample space.}}

Solved Examples

Example 1: Tossing a Fair Coin

Given: A fair coin is tossed once.

To Find: Sample space and probability of getting Heads.

Solution:

1. Identify all possible outcomes when tossing a coin.

S = {H, T}

2. Count the total number of outcomes.

n(S) = 2

3. Define the event E = "getting Heads".

E = {H}

4. Count favourable outcomes.

n(E) = 1

5. Apply the probability formula.

P(Heads) = n(E) / n(S) = 1/2 = 0.5

Final Answer: P(Heads) = 1/2 or 50%

Example 2: Rolling a Standard Die

Given: A 6-sided fair die is rolled once.

To Find: (i) Sample space, (ii) Probability of rolling a number greater than 4.

Solution:

1. List all faces of the die.

S = {1, 2, 3, 4, 5, 6}

2. Calculate sample size.

n(S) = 6

3. Define event E = "number greater than 4".

E = {5, 6}

4. Count outcomes in E.

n(E) = 2

5. Calculate probability.

P(E) = n(E) / n(S) = 2/6 = 1/3 ≈ 0.333

Final Answer: (i) S = {1, 2, 3, 4, 5, 6}, (ii) P(E) = 1/3

Example 3: Selecting Fruit from a Basket (Medium)

Given: A basket contains 3 apples, 4 bananas, and 2 oranges. One fruit is drawn at random.

To Find: (i) Sample space, (ii) Probability of selecting a banana, (iii) Probability of selecting a fruit that is NOT an apple.

Solution:

1. Since we care about fruit type (not individual fruits), define sample space by type.

S = {Apple, Banana, Orange}

But wait — we must account for quantity. Total fruits = 3 + 4 + 2 = 9. For probability calculations, we use the count-based approach.

2. Total outcomes (individual fruits).

n(S) = 9

3. For event E₁ = "selecting a banana", count banana pieces.

n(E₁) = 4

P(Banana) = 4/9 ≈ 0.444

4. For event E₂ = "NOT an apple", count non-apple fruits.

n(E₂) = 4 + 2 = 6

P(Not Apple) = 6/9 = 2/3 ≈ 0.667

Final Answer: (i) 9 individual fruits, (ii) P(Banana) = 4/9, (iii) P(Not Apple) = 2/3

Example 4: Tossing Two Coins Simultaneously (Hard)

Given: Two fair coins are tossed together.

To Find: (i) Sample space, (ii) Probability of getting at least one Head, (iii) Probability of getting exactly one Tail.

Solution:

1. Label coins as Coin 1 and Coin 2. List all combinations.

S = {HH, HT, TH, TT}

Here HT means Coin 1 shows H, Coin 2 shows T.

2. Calculate sample size.

n(S) = 4

3. Define event E₁ = "at least one Head" (means one or more Heads).

E₁ = {HH, HT, TH}

n(E₁) = 3

P(At least one H) = 3/4 = 0.75

4. Define event E₂ = "exactly one Tail" (means one H and one T).

E₂ = {HT, TH}

n(E₂) = 2

P(Exactly one T) = 2/4 = 1/2 = 0.5

Final Answer: (i) S = {HH, HT, TH, TT}, (ii) P(At least one H) = 3/4, (iii) P(Exactly one T) = 1/2

Tips & Tricks

{{KEY: type=formula | title=Complement Rule Shortcut | text=P(Event happens) = 1 − P(Event does NOT happen). Use this when counting failures is easier than counting successes.}}

Common Mistakes

Brain-Teaser Questions

Question 1: A bag contains 5 red balls and 3 blue balls. You draw one ball without looking. Your friend says the sample space is S = {Red, Blue}, so P(Red) = 1/2. Is this correct? Why or why not?

💡 Answer: Wrong. While the types are {Red, Blue}, the balls are not equally likely. There are 8 individual balls. Correct sample space has 8 elements (5 red + 3 blue), so P(Red) = 5/8, not 1/2. Sample space must reflect physical count when items are not symmetric.

Question 2: You roll a fair die. Event A = "rolling an even number", Event B = "rolling a prime number". Find P(A), P(B), and P(A and B both happen).

💡 Answer: S = {1, 2, 3, 4, 5, 6}. A = {2, 4, 6}, so P(A) = 3/6 = 1/2. B = {2, 3, 5} (primes), so P(B) = 3/6 = 1/2. "A and B" means outcome is both even AND prime = {2}. P(A and B) = 1/6.

Question 3: In a village fair, you can choose one snack from {Samosa, Pakora, Bhaji} and one drink from {Chai, Lassi}. How many total combinations are possible? List the sample space.

💡 Answer: Use counting principle: 3 snacks × 2 drinks = 6 combinations.

S = {(Samosa, Chai), (Samosa, Lassi), (Pakora, Chai), (Pakora, Lassi), (Bhaji, Chai), (Bhaji, Lassi)}

n(S) = 6. If choices are random and equally likely, each combination has probability 1/6.

Mini Cheatsheet — Screenshot This for Revision!

End of Page 6

Tree Diagrams & Summary & Quick Revision

Page 7: Tree Diagrams & Summary

Welcome to the final page of our journey into the world of probability! So far, we've dealt with single-step experiments. But what happens when events occur in a sequence? How do we keep track of all the possibilities? This is where a powerful visual tool comes into play: the Tree Diagram.

Concept Introduction

Imagine you're at a food stall at a village fair. You can choose one snack (Samosa or Kachori) and one drink (Chai or Lassi). How many different combinations can you make? You could have a Samosa with Chai, a Samosa with Lassi, a Kachori with Chai, or a Kachori with Lassi. It's easy to list them out.

Now, what if there were 3 snacks, 4 drinks, and 2 desserts? Listing all combinations manually becomes messy and prone to errors. A tree diagram is a simple yet brilliant method to visually map out all possible outcomes of such multi-step experiments. Each "branch" of the tree represents a choice, and by following the paths from the "root" to the "leaves," we can clearly see every single possible outcome in the sample space.

{{FORMULA: expr=P(E) = (Number of favourable outcomes) / (Number of possible outcomes) | symbols=P(E):Theoretical Probability of an Event E}}

Definitions & Key Terms

Let's formally define the concepts related to tree diagrams.

How to Construct a Tree Diagram

Building a tree diagram is a logical process. It's like drawing a map of possibilities.

1. Start Point (Root): Draw a single point on the left side of your page. This is the starting point before any event has occurred.

2. First Level Branches: For the first step of the experiment (e.g., the first coin toss), draw one branch extending from the root for each possible outcome. Label the end of each branch with the outcome (e.g., 'H' for Heads, 'T' for Tails).

3. Second Level Branches: From the end of each of the first-level branches, draw a new set of branches representing all possible outcomes for the second step of the experiment.

4. Continue for All Steps: If there are more steps, repeat the process. From the end of each previous branch, draw new branches for all possible outcomes of the next step.

5. Identify Outcomes: Each complete path from the root to the final branch tip represents one unique outcome of the entire experiment.

6. List the Sample Space: Read each path from start to finish and list it. The collection of all these paths is your sample space.

{{KEY: type=concept | title=Visualizing Outcomes | text=The power of a tree diagram is that it transforms a complex counting problem into a simple visual map. The total number of endpoints on the far right of the diagram is equal to the total number of possible outcomes in the sample space.}}

Solved Examples

Let's build some tree diagrams to solve problems, moving from easy to tricky.

Example 1: Tossing a Fair Coin Twice (Easy)

Given: A fair coin is tossed two times.

To Find: The sample space of all possible outcomes.

Solution:

1. First Toss: The first step is the first toss. The possible outcomes are Heads (H) or Tails (T). We draw two branches from our starting point.

   {{VISUAL: diagram: A starting point with two branches extending from it. One branch is labeled 'H' (Heads) and the other is labeled 'T' (Tails).}}

2. Second Toss: The second step is the second toss. From the end of the 'H' branch, we again have two possibilities: H or T. We do the same for the 'T' branch.

3. List the Paths: Now we trace each path from start to finish to find the outcomes.

   • Path 1: H → H gives the outcome HH
   • Path 2: H → T gives the outcome HT
   • Path 3: T → H gives the outcome TH
   • Path 4: T → T gives the outcome TT

4. Write the Sample Space: The set of all these outcomes is the sample space.

S = {HH, HT, TH, TT}

Final Answer: The sample space is {HH, HT, TH, TT}, and there are 4 possible outcomes.

Example 2: The Fruit Basket Selection (Medium)

Given: Basket A has one apple (A) and two oranges (O1, O2). Basket B has one banana (B) and one mango (M). You randomly pick one fruit from each basket.

To Find: (i) Draw a tree diagram showing all possible pairs of fruits. (ii) List the sample space. (iii) The probability of picking one apple and one banana.

Solution:

1. Step 1: Pick from Basket A: The first choice is from Basket A. The outcomes are Apple (A), Orange 1 (O1), or Orange 2 (O2). We start with three branches.

2. Step 2: Pick from Basket B: The second choice is from Basket B. From the end of each of the first three branches, we draw two new branches for the outcomes Banana (B) and Mango (M).

   {{VISUAL: diagram: A tree diagram starting with three initial branches labeled 'A', 'O1', and 'O2'. From the end of each of these, two new branches sprout, one labeled 'B' and one 'M'.}}

3. (i) & (ii) List Outcomes & Sample Space: We trace all the paths.

   • A → B gives AB
   • A → M gives AM
   • O1 → B gives O1B
   • O1 → M gives O1M
   • O2 → B gives O2B
   • O2 → M gives O2M The sample space S is the collection of these 6 outcomes. Since the oranges are distinct items, we list them separately. If the problem treated oranges as identical, the sample space would be smaller, but here we list all combinations. Let's consider the type of fruit, so we can simplify the first step to just Apple (A) or Orange (O). Let's redraw based on fruit type for probability.

   Let's re-approach based on the question's likely intent for probability.

   In Basket A, P(Apple) = ⅓ and P(Orange) = ⅔. In Basket B, P(Banana) = ½ and P(Mango) = ½. The tree diagram of outcomes would be: A-B, A-M, O-B, O-M. Let's assume the question meant one apple and one orange in Basket A for simplicity, as is common.

   Let's stick to the text: one apple, two oranges. Total 3 fruits in A.

   Sample Space S = {AB, AM, O1B, O1M, O2B, O2M}. Total possible outcomes = 6.

4. (iii) Calculate Probability: We need the probability of picking one apple and one banana.

   • The favourable outcome is the path that gives an Apple from Basket A and a Banana from Basket B. This is the outcome AB.
   • Number of favourable outcomes = 1.
   • Total number of possible outcomes = 6.

5. Apply the probability formula.

P(Apple and Banana) = (Number of favourable outcomes) / (Total possible outcomes)

P(Apple and Banana) = 1/6

Final Answer: (i) The tree diagram shows 3 initial branches (A, O1, O2) each splitting into 2 branches (B, M). (ii) The sample space is S = {AB, AM, O1B, O1M, O2B, O2M}. (iii) The probability of picking an apple and a banana is 1/6.

Example 3: Picking Pens with Replacement (Hard)

Given: A box contains 3 red (R), 4 black (B), and 2 green (G) pens. Total pens = 9. You pick a pen, note its color, and put it back. Then your friend does the same.

To Find: The probability that both you and your friend pick pens of the same color.

Solution:

1. Identify the Steps: This is a two-step experiment.

   • Step 1: You pick a pen.
   • Step 2: Your friend picks a pen. The outcomes for each step are Red (R), Black (B), or Green (G).

2. Construct the Tree Diagram: The first level has three branches (R, B, G). Since the pen is replaced, the second level also has three branches (R, B, G) extending from each of the first-level branches.

3. List all Outcomes (Sample Space):

   • From R: RR, RB, RG
   • From B: BR, BB, BG
   • From G: GR, GB, GG The sample space S has 3 × 3 = 9 possible outcomes: {RR, RB, RG, BR, BB, BG, GR, GB, GG}.

4. Identify Favourable Outcomes: We want the event where both pens are the same color.

   • The favourable outcomes are RR, BB, and GG.

5. Calculate Probabilities for Each Path: This is the key step. Since the number of pens of each color is different, the outcomes are not equally likely.

   • P(R) = 3/9 = ⅓
   • P(B) = 4/9
   • P(G) = 2/9

6. Calculate Probability of Favourable Events:

   • P(both Red) = P(You pick R) × P(Friend picks R) = (⅓) × (⅓) = 1/9
   • P(both Black) = P(You pick B) × P(Friend picks B) = (4/9) × (4/9) = 16/81
   • P(both Green) = P(You pick G) × P(Friend picks G) = (2/9) × (2/9) = 4/81

7. Add the Probabilities: Since these are "OR" events (RR or BB or GG), we add their probabilities.

P(same color) = P(RR) + P(BB) + P(GG)

P(same color) = 1/9 + 16/81 + 4/81

8. Find a common denominator (81). 1/9 is 9/81.

P(same color) = 9/81 + 16/81 + 4/81 = 29/81

Final Answer: The probability that both you and your friend pick pens of the same color is 29/81.

Example 4: Tossing Three Coins (Tricky)

Given: Three fair coins are tossed simultaneously.

To Find: The probability of getting exactly two heads.

Solution:

1. Model as a Multi-step Experiment: Tossing three coins at once is equivalent to tossing one coin three times in a row.

   • Step 1: First coin.
   • Step 2: Second coin.
   • Step 3: Third coin.

2. Construct the Tree Diagram: This will have three levels of branches.

   • Level 1: Two branches (H, T).
   • Level 2: From each of the first two branches, two more branches (H, T).
   • Level 3: From each of the four branches at level 2, two more branches (H, T).

   {{VISUAL: diagram: A three-level tree diagram. The first level splits into H and T. The second level shows H splitting into H and T, and T splitting into H and T. The third level shows each of the four previous endpoints splitting again into H and T, resulting in 8 final paths.}}

3. List the Sample Space: Trace all 8 paths from the root to the end.

   • HHH
   • HHT
   • HTH
   • HTT
   • THH
   • THT
   • TTH
   • TTT Total number of possible outcomes = 8.

4. Identify Favourable Outcomes: We need the outcomes with exactly two heads. Let's scan our list.

   • HHT (has 2 heads)
   • HTH (has 2 heads)
   • THH (has 2 heads) Number of favourable outcomes = 3.

5. Calculate the Probability: Use the theoretical probability formula.

P(exactly two heads) = (Number of favourable outcomes) / (Total possible outcomes)

P(exactly two heads) = 3/8

Final Answer: The probability of getting exactly two heads is 3/8.

Tips & Tricks

Use these shortcuts to solve problems faster and more accurately.

Common Mistakes

Avoid these common pitfalls when working with tree diagrams and probability.

Brain-Teaser Questions

Challenge yourself with these slightly tougher problems!

1. A bag contains 4 red balls and 5 blue balls. You draw one ball, lay it aside, and then draw a second ball. What is the probability of drawing 2 blue balls?

   💡 Answer: This involves dependent events. P(1st is Blue) = 5/9. After drawing one blue ball, there are 8 balls left, and only 4 are blue. P(2nd is Blue | 1st was Blue) = 4/8 = ½. P(Blue and then Blue) = P(1st is Blue) × P(2nd is Blue) = (5/9) × (4/8) = 20/72 = 5/18.

2. You spin a spinner with three equal sections (Red, Green, Blue) and roll a standard 6-sided die. What is the probability you get a primary color (Red or Blue) and a number greater than 4?

   💡 Answer: P(Primary color) = P(Red or Blue) = ⅔. P(Number > 4) = P(5 or 6) = 2/6 = ⅓. Since these events are independent, we multiply their probabilities. P(Primary and Number > 4) = (⅔) × (⅓) = 2/9.

3. The final outcomes of a two-step experiment are listed as: {SA, SB, TA, TB, UA, UB}. What were the possible outcomes for the first step and the second step of the experiment?

   💡 Answer: Look for patterns. The first letter in each pair is either S, T, or U. The second letter is always A or B. Step 1 Outcomes: {S, T, U} Step 2 Outcomes: {A, B}

Mini Cheatsheet

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