Probability

Probability — A Theoretical Approach — Part 1

Page 1 of 5: Probability — A Theoretical Approach — Part 1

{{FORMULA: expr=P(E) = (Number of outcomes favourable to E) / (Number of all possible outcomes) | symbols=P(E):Probability of event E}}

Concept Introduction

Have you ever shuffled a playlist and wondered what song will play next? Or checked the weather forecast and seen a "30% chance of rain"? We're surrounded by situations where the outcome is uncertain. Probability is the branch of mathematics that helps us measure this uncertainty.

Instead of just guessing, probability gives us a precise number to describe how likely an event is to happen. It's the logic of chance. In this chapter, we move from just observing what happens (experimental probability) to predicting what should happen based on logic. This is theoretical probability, which is built on a simple, powerful idea: if all outcomes are equally likely, we can calculate the exact chances of a specific event occurring.

Definitions & Formulas

To understand probability, we need a few key terms. Think of these as the building blocks for all our calculations.

The Logic Behind the Formula

The formula for theoretical probability isn't magic; it's based on a single, crucial assumption discussed in your textbook: equally likely outcomes. This means that in a fair experiment, every possible outcome has the exact same chance of occurring. Here’s how this assumption leads directly to the formula.

1. Start with a Fair Experiment: Imagine a fair, six-sided die. The word "fair" is key. It means the die is not weighted, and each side (1, 2, 3, 4, 5, 6) has an equal chance of landing face up.

2. Count All Possibilities: The total number of things that can happen (the sample space) is 6. Each of these outcomes is equally likely.

3. Assign a Chance to Each Outcome: Since there are 6 equally likely outcomes, the probability of any single outcome (like rolling a 5) must be 1 out of 6.

   P(getting a 5) = 1/6
   

4. Define an Event: Let's define an event E as "getting a number greater than 4".

5. Count Favourable Outcomes: Which outcomes satisfy this event? The numbers 5 and 6 are greater than 4. So, there are 2 favourable outcomes.

6. Sum the Individual Probabilities: The total probability of the event E is the sum of the probabilities of all its favourable outcomes.

   P(E) = P(getting a 5) + P(getting a 6) = 1/6 + 1/6 = 2/6
   

   Simplifying this gives 1/3. Notice that this result, 2/6, is simply the number of favourable outcomes (2) divided by the total number of possible outcomes (6). This logic holds for any experiment with equally likely outcomes, giving us our master formula.

{{KEY: type=concept | title=The Assumption of "Equally Likely" | text=Theoretical probability ONLY works when we can assume every individual outcome of an experiment has the same chance of happening. A biased coin or a loaded die would require a different approach (experimental probability).}}

Solved Examples

Let's apply the formula to a few problems, starting from easy and moving to more challenging ones.

Example 1: The Simple Die Roll (Easy)

Given: A single fair six-sided die is rolled once.

To Find: The probability of getting a prime number.

Solution:

1. First, identify the sample space, which is the set of all possible outcomes.

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

   The total number of possible outcomes is 6.

2. Next, identify the event E we are interested in: "getting a prime number".

3. Now, list the favourable outcomes. The prime numbers in our sample space are 2, 3, and 5. (Remember, 1 is not a prime number).

   Favourable Outcomes = {2, 3, 5}
   

   The number of favourable outcomes is 3.

4. Apply the probability formula P(E) = (Favourable Outcomes) / (Total Outcomes).

   P(prime number) = 3 / 6
   

5. Simplify the fraction.

   P(prime number) = 1/2
   

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

Example 2: The Colorful Marble Bag (Medium)

Given: A bag contains 5 red, 8 blue, and 3 green balls. All balls are identical in size and shape. One ball is drawn at random.

To Find: The probability that the ball drawn is not blue.

Solution:

1. First, calculate the total number of balls in the bag. This is the total number of possible outcomes.

   Total Balls = 5 (Red) + 8 (Blue) + 3 (Green) = 16
   

2. Define the event E as "the ball drawn is not blue".

3. Identify the number of favourable outcomes. These are the balls that are not blue, which means they must be either red or green.

   Number of Favourable Outcomes = 5 (Red) + 3 (Green) = 8
   

4. Use the probability formula.

   P(not blue) = (Number of non-blue balls) / (Total number of balls)
   

   P(not blue) = 8 / 16
   

5. Simplify the result.

   P(not blue) = 1/2
   

Final Answer: The probability of drawing a ball that is not blue is 1/2.

Example 3: Tossing Two Coins (Hard)

Given: Two fair coins are tossed simultaneously.

To Find: The probability of getting at least one head.

Solution:

1. When two coins are tossed, we need to list all possible outcomes. Let 'H' be Head and 'T' be Tail. The sample space S is:

   S = {HH, HT, TH, TT}
   

   (HH means Head on the first coin and Head on the second. HT means Head on the first and Tail on the second, and so on.) The total number of possible outcomes is 4.

   {{VISUAL: diagram: A tree diagram showing the four possible outcomes (HH, HT, TH, TT) when two coins are tossed.}}

2. Define the event E as "getting at least one head". This means we can get one head OR two heads.

3. List the outcomes that are favourable to this event.

   Favourable Outcomes = {HH, HT, TH}
   

   The outcome TT is the only one that does not have at least one head. So, the number of favourable outcomes is 3.

4. Apply the probability formula.

   P(at least one head) = (Number of favourable outcomes) / (Total number of outcomes)
   

   P(at least one head) = 3 / 4
   

Final Answer: The probability of getting at least one head is 3/4.

Example 4: The Deck of Cards (Tricky)

Given: One card is drawn from a well-shuffled standard deck of 52 playing cards.

To Find: The probability that the card is a red face card.

Solution:

1. First, understand the composition of a standard deck. There are 52 cards in total. This is our total number of outcomes.

2. The deck is divided into 4 suits: Hearts (♥), Diamonds (♦), Clubs (♣), and Spades (♠). Hearts and Diamonds are red. Clubs and Spades are black. So, there are 26 red cards and 26 black cards.

3. Each suit has 3 face cards: King, Queen, and Jack.

4. The event E is "drawing a red face card". We need to find the cards that are both red and a face card.

   • Red Suits: Hearts and Diamonds.
   • Face cards in Hearts: King, Queen, Jack (3 cards).
   • Face cards in Diamonds: King, Queen, Jack (3 cards).

5. Calculate the total number of favourable outcomes.

   Total Red Face Cards = 3 (from Hearts) + 3 (from Diamonds) = 6
   

6. Apply the probability formula.

   P(red face card) = (Number of red face cards) / (Total number of cards)
   

   P(red face card) = 6 / 52
   

7. Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 2.

   P(red face card) = 3 / 26
   

Final Answer: The probability of drawing a red face card is 3/26.

Tips & Tricks

Use these shortcuts to solve problems faster and more accurately.

Common Mistakes to Avoid

Many students make these simple errors. Be careful to check your work for them!

Brain-Teaser Questions

Test your understanding with these slightly more challenging problems.

1. A standard deck of 52 cards is shuffled. All the face cards (Jacks, Queens, Kings) are removed. What is the probability of drawing an Ace from the remaining cards?

   💡 Answer: There are 3 face cards per suit × 4 suits = 12 face cards. Remaining cards = 52 - 12 = 40. The number of Aces is still 4. So, P(Ace) = 4/40 = 1/10.

2. The letters of the word "ASSASSINATION" are placed in a bag. One letter is chosen at random. What is the probability that the letter is a vowel?

   💡 Answer: Total letters = 13. Vowels are A, I, A, I, A, O. Total vowels = 6. Consonants are S, S, S, S, N, T, N. Total consonants = 7. P(vowel) = (Number of vowels) / (Total letters) = 6/13.

3. You roll two fair six-sided dice. Is it more likely to get a sum of 9 or a sum of 10?

   💡 Answer: Total outcomes = 6 × 6 = 36. Ways to get a sum of 9: (3,6), (4,5), (5,4), (6,3). There are 4 ways. P(sum=9) = 4/36. Ways to get a sum of 10: (4,6), (5,5), (6,4). There are 3 ways. P(sum=10) = 3/36. It is more likely to get a sum of 9.

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Probability — A Theoretical Approach — Part 2

Chapter 14: Probability

Page 2/5: Probability — A Theoretical Approach — Part 2

Concept Introduction

Imagine you're at a funfair, standing in front of a giant spinning wheel divided into 8 equal sections. One section is marked "Grand Prize", while the others are "Try Again". You get one spin. What are your chances of winning? Intuitively, you know it's not very likely. You have only 1 chance out of 8 total possibilities. This simple, intuitive calculation is the heart of theoretical probability.

Instead of spinning the wheel a hundred times and counting the results (which would be experimental probability), we use logic. We assume the wheel is fair and each section is equally likely to be the winner. By analyzing the structure of the experiment before it happens, we can predict the likelihood of an outcome. This is the "classical" or "theoretical" approach, a powerful tool for making predictions in situations governed by chance, from games to genetics.

{{FORMULA: expr=P(E) = (Number of outcomes favourable to E) / (Number of all possible outcomes of the experiment) | symbols=P(E):Theoretical probability of event E}}

Definitions & Formulas

Before we start calculating, let's be crystal clear about the terms we'll be using. These definitions are the building blocks for everything that follows.

Logic: Why the Sum of Probabilities is Always 1

Have you ever wondered why probabilities are always fractions between 0 and 1? The answer lies in understanding elementary events. Let's break down why the probabilities of all possible single outcomes for an experiment must add up to exactly 1.

1. Consider a simple experiment, like rolling a standard six-sided die. The set of all possible outcomes is {1, 2, 3, 4, 5, 6}.

2. The total number of possible outcomes is 6. Since the die is fair, all these outcomes are equally likely.

3. Let's define the elementary events. An elementary event is an event with just one outcome.

   • E₁ = Getting a 1
   • E₂ = Getting a 2
   • E₃ = Getting a 3
   • ...and so on, up to E₆ = Getting a 6.

4. Using the probability formula, we can find the probability of each elementary event. For any single event Eᵢ, the number of favourable outcomes is just 1.

   P(E₁) = 1/6,   P(E₂) = 1/6,   P(E₃) = 1/6,   ... P(E₆) = 1/6
   

5. Now, let's add the probabilities of all these elementary events. This sum represents the probability of getting any of the possible outcomes (a 1, OR a 2, OR a 3, etc.).

   Σ P(Eᵢ) = P(E₁) + P(E₂) + P(E₃) + P(E₄) + P(E₅) + P(E₆)
   

6. Substituting the values gives us the final result.

   Σ P(Eᵢ) = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6 = 1
   

This proves a fundamental rule: The sum of the probabilities of all the elementary events of an experiment is 1. This makes sense, as it's 100% certain that one of the possible outcomes will occur.

Solved Examples

Let's apply the formula to a few problems, starting simple and moving to more complex scenarios.

Example 1: The Classic Coin Toss (Easy)

Given: A fair coin is tossed once.

To Find: The probability of getting a head.

Solution:

1. First, identify all possible outcomes of the experiment. When a coin is tossed, it can land as either a Head (H) or a Tail (T).

   Total possible outcomes = 2 (namely {H, T})

2. Next, identify the event we are interested in. Let E be the event 'getting a head'.

3. Count the number of outcomes that are favourable to event E. The only outcome that is a 'head' is H itself.

   Number of favourable outcomes = 1

4. Apply the theoretical probability formula.

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

   P(head) = 1 / 2
   

Final Answer:

The probability of getting a head is ½.

Example 2: Rolling a Die for an Even Number (Medium)

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

To Find: The probability of getting an even number.

Solution:

1. List all the possible outcomes when a die is rolled. The faces are numbered 1, 2, 3, 4, 5, 6.

   Total possible outcomes = 6 (namely {1, 2, 3, 4, 5, 6})

2. Define the event E as 'getting an even number'.

3. Identify the outcomes from the total set that are favourable to E. The even numbers are 2, 4, and 6.

   Number of favourable outcomes = 3 (namely {2, 4, 6})

4. Use the probability formula to calculate P(E).

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

   P(even number) = 3 / 6
   

5. Always simplify the fraction to its lowest terms.

   P(even number) = 1 / 2
   

Final Answer:

The probability of getting an even number is ½.

Example 3: Balls in a Bag (Hard)

Given: A bag contains 4 red balls, 5 blue balls, and 1 green ball. All balls are of the same size. A ball is drawn at random.

To Find: The probability that the ball drawn is (i) red, (ii) not green.

Solution:

(i) Probability of drawing a red ball

1. First, determine the total number of possible outcomes. This is the total number of balls in the bag.

   Total balls = 4 (red) + 5 (blue) + 1 (green) = 10 Total possible outcomes = 10

2. Let R be the event 'drawing a red ball'. The number of outcomes favourable to R is the number of red balls.

   Number of favourable outcomes for R = 4

3. Calculate the probability P(R).

   P(R) = 4 / 10
   

   P(R) = 2 / 5
   

(ii) Probability of drawing a ball that is not green

1. The total number of possible outcomes remains the same: 10.

2. Let E be the event 'the ball drawn is not green'. The favourable outcomes are the balls that are red or blue.

   Number of favourable outcomes for E = 4 (red) + 5 (blue) = 9

3. Calculate the probability P(E).

   P(E) = 9 / 10
   

Final Answer:

The probability of drawing a red ball is 2/5. The probability of drawing a ball that is not green is 9/10.

Example 4: Prime Numbers on a Die (Tricky)

Given: A single fair die is rolled.

To Find: The probability of getting a prime number.

Solution:

1. The set of all possible outcomes for a die roll is {1, 2, 3, 4, 5, 6}.

   Total possible outcomes = 6

2. This question requires knowledge of prime numbers. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Let's check the outcomes:

   • 1 is not a prime number.
   • 2 is a prime number (only divisible by 1 and 2).
   • 3 is a prime number (only divisible by 1 and 3).
   • 4 is not prime (divisible by 2).
   • 5 is a prime number (only divisible by 1 and 5).
   • 6 is not prime (divisible by 2 and 3).

3. Let E be the event 'getting a prime number'. The favourable outcomes are {2, 3, 5}.

   Number of favourable outcomes = 3

4. Calculate the probability P(E).

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

   P(prime number) = 3 / 6 = 1 / 2
   

Final Answer:

The probability of getting a prime number is ½.

{{KEY: type=concept | title=The Range of Probability | text=The probability of any event E, denoted P(E), will always be a number between 0 and 1, inclusive. So, 0 ≤ P(E) ≤ 1. A probability of 0 means the event is impossible, and a probability of 1 means the event is certain.}}

Tips & Tricks

Here are a few shortcuts to speed up your calculations and improve your intuition.

Common Mistakes

Many students make small, avoidable errors in probability. Here’s a guide to help you spot and fix them.

Brain-Teaser Questions

Ready to test your understanding? These questions require careful thinking.

1. A letter is chosen at random from the word "MATHEMATICS". What is the probability that the letter is a vowel?

💡 Answer: Total letters = 11. The vowels are A, E, A, I. There are 4 vowels. So, P(vowel) = 4/11. (Note: We count repeated letters like 'A' each time they appear).

2. A piggy bank contains one hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin will not be a 50p coin?

💡 Answer: Total coins = 100 + 50 + 20 + 10 = 180. Number of 50p coins = 100. Using the "Not" rule: P(not 50p) = 1 - P(50p) = 1 - (100/180) = 1 - (10/18) = 1 - (5/9) = 4/9.

3. Two friends were born in the year 2004. What is the probability that they have the same birthday? (Ignore the date Feb 29).

💡 Answer: The first friend can have their birthday on any of the 365 days. For them to have the same birthday, the second friend must be born on that exact one day. So, the number of favourable outcomes for the second friend's birthday is 1. The total possible days for their birthday is 365. P(same birthday) = 1/365.

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Probability — A Theoretical Approach — Part 3

Probability — A Theoretical Approach — Part 3

Welcome back! In the previous sections, we learned how to calculate the theoretical probability of a single event. Now, let's explore the fascinating relationship between an event happening and that same event not happening. What if something is guaranteed to happen, or guaranteed not to happen? How do we represent these certainties and uncertainties with numbers?

Imagine a weather forecast. If the meteorologist says there's a 70% chance of rain, your brain instantly processes that there's a 30% chance it won't rain. This intuitive calculation is the heart of today's lesson. The event "rain" and the event "no rain" are complementary events; they are two sides of the same coin, and together they represent all possibilities. We'll also define impossible events (like rolling a 7 on a standard six-sided die) and sure events (like the sun rising in the east).

{{FORMULA: expr=P(E) + P(not E) = 1 | symbols=P(E):Probability of event E occurring, P(not E):Probability of event E NOT occurring}}

Definitions & Formulas

Let's formally define the terms we'll be using throughout this page.

Derivation: The Complement Rule

The relationship P(E) + P(not E) = 1 is one of the most fundamental rules in probability. It's not just a formula to memorize; it comes from a very logical place. Let's break it down.

1. First, let's define the total number of equally likely outcomes in an experiment. We'll call this n.

2. Now, consider an event E. Let the number of outcomes that are favourable to this event E be m.

3. From our basic definition of probability, we know that the probability of event E occurring is the ratio of favourable outcomes to total outcomes.

   P(E) = m/n
   

4. Now, let's think about the complementary event, not E. This event represents all outcomes that are not in E. If there are n total outcomes and m of them are for event E, then the remaining outcomes must be for the event not E. The number of outcomes favourable to not E is therefore n - m.

5. Using the same probability formula, the probability of not E is:

   P(not E) = (n - m) / n
   

6. Finally, let's add the probabilities of E and not E together.

   P(E) + P(not E) = (m/n) + ((n - m) / n)
   

   Since they have a common denominator, we can combine the numerators:

   P(E) + P(not E) = (m + n - m) / n
   

   The m and -m cancel each other out, leaving us with:

   P(E) + P(not E) = n/n = 1
   

This simple proof shows that the probability of an event happening plus the probability of it not happening will always equal 1, or 100%.

{{VISUAL: diagram: A number line from 0 to 1. The point 0 is labeled "Impossible Event (e.g., rolling a 7 on a die)". The point 1 is labeled "Sure Event (e.g., getting a number < 7 on a die)". The space in between shows values like ¼, ½, ¾, representing possible events.}}

{{KEY: type=concept | title=The Range of Probability | text=The probability of any event E must be a value 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.}}

Solved Examples

Let's apply these concepts to some problems, starting from easy and moving to more complex ones.

Example 1: Winning and Losing a Game (Easy)

Given: The probability of Sangeeta winning a tennis match is 0.62.

To Find: The probability of her not winning the match.

Solution:

1. Let E be the event that Sangeeta wins the match. We are given P(E).

   P(E) = 0.62
   

2. Let not E be the event that Sangeeta does not win the match. This is the complementary event. We need to find P(not E).

3. We know that for any complementary events, P(E) + P(not E) = 1. We can rearrange this to find P(not E).

   P(not E) = 1 - P(E)
   

4. Substitute the given value of P(E) into the formula.

   P(not E) = 1 - 0.62
   

   P(not E) = 0.38
   

Final Answer: The probability of Sangeeta not winning the match is 0.38.

Example 2: Rolling a Die (Medium)

Given: A single, fair six-sided die is rolled once.

To Find: The probability of not getting a prime number.

Solution:

1. First, list all possible outcomes when rolling a die. The total number of outcomes is 6.

   Outcomes: {1, 2, 3, 4, 5, 6}

2. Let E be the event of getting a prime number. Identify the prime numbers in our set of outcomes. The prime numbers are 2, 3, and 5. (Note: 1 is not a prime number).

   Favourable outcomes for E: {2, 3, 5} Number of favourable outcomes for E = 3

3. Calculate the probability of event E (getting a prime number).

   P(E) = (Number of favourable outcomes) / (Total number of outcomes) = 3/6 = 1/2
   

4. We need to find the probability of not getting a prime number, which is P(not E). We use the complement rule.

   P(not E) = 1 - P(E)
   

5. Substitute the value of P(E) we found.

   P(not E) = 1 - 1/2
   

   P(not E) = 1/2
   

Final Answer: The probability of not getting a prime number is ½.

Example 3: Drawing Marbles (Hard)

Given: A bag contains 3 red marbles, 5 black marbles, and 4 white marbles. A marble is drawn at random.

To Find: The probability that the marble drawn is (i) white (ii) not black.

Solution:

(i) Probability of drawing a white marble

1. First, calculate the total number of marbles in the bag. This is our total number of possible outcomes.

   Total marbles = 3 (red) + 5 (black) + 4 (white) = 12
   

2. Let W be the event of drawing a white marble. The number of outcomes favourable to W is the number of white marbles.

   Number of white marbles = 4

3. Calculate the probability P(W).

   P(W) = (Number of white marbles) / (Total number of marbles) = 4/12
   

   Simplify the fraction.

   P(W) = 1/3
   

(ii) Probability of drawing a marble that is not black

1. Let B be the event of drawing a black marble. The number of black marbles is 5.

2. First, calculate the probability of drawing a black marble, P(B).

   P(B) = (Number of black marbles) / (Total number of marbles) = 5/12
   

3. We need to find the probability of not drawing a black marble, which is P(not B). We use the complement rule.

   P(not B) = 1 - P(B)
   

4. Substitute the value of P(B).

   P(not B) = 1 - 5/12 = 12/12 - 5/12
   

   P(not B) = 7/12
   

Final Answer: (i) The probability of drawing a white marble is ⅓. (ii) The probability of not drawing a black marble is 7/12.

Example 4: Deck of Cards (Tricky)

Given: A card is drawn from a well-shuffled standard deck of 52 playing cards.

To Find: The probability that the card drawn is neither a heart nor a king.

Solution:

1. This question is tricky. Finding the probability of "neither A nor B" is easier by finding the probability of "either A or B" and then subtracting from 1. Let E be the event that the card is either a heart or a king.

2. Count the total number of outcomes. For a standard deck, this is 52.

3. Count the number of outcomes favourable to E.

   • There are 13 hearts.
   • There are 4 kings.
   • However, the King of Hearts is counted in both groups. To avoid double-counting, we must subtract it once.
   • Favourable outcomes = (Number of hearts) + (Number of kings) - (King of Hearts)

   Number of outcomes for E = 13 + 4 - 1 = 16
   

4. Calculate the probability of event E (getting a heart or a king).

   P(E) = 16/52
   

   Simplify the fraction by dividing both by 4.

   P(E) = 4/13
   

5. The question asks for the probability of neither a heart nor a king, which is the complement of event E, i.e., P(not E).

   P(not E) = 1 - P(E)
   

6. Substitute the value of P(E).

   P(not E) = 1 - 4/13 = 13/13 - 4/13
   

   P(not E) = 9/13
   

Final Answer: The probability that the card is neither a heart nor a king is 9/13.

Tips & Tricks

Use these shortcuts to solve problems faster and more accurately.

Common Mistakes

Many students make these simple errors. Be careful to avoid them!

Brain-Teaser Questions

Test your understanding with these slightly more challenging problems.

1. It is given that in a group of 3 students, the probability of 2 students NOT having the same birthday is 0.992. What is the probability that the 2 students HAVE the same birthday?

   💡 Answer: This is a direct application of complementary events. Let E be the event that 2 students have the same birthday. The given probability is for the event "not E". P(not E) = 0.992 P(E) = 1 - P(not E) = 1 - 0.992 = 0.008.

2. A bag contains lemon-flavored candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out an orange-flavored candy?

   💡 Answer: This is an impossible event. Since the bag only contains lemon-flavored candies, it is impossible to draw an orange-flavored one. The probability is 0.

3. Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on them is not a multiple of 4?

   💡 Answer: Total outcomes = 6 × 6 = 36. Let E be the event that the sum is a multiple of 4. The possible sums are 4, 8, 12. Favourable pairs: (1,3), (2,2), (3,1) for sum 4. (2,6), (3,5), (4,4), (5,3), (6,2) for sum 8. (6,6) for sum 12. Total favourable outcomes for E = 3 + 5 + 1 = 9. P(E) = 9/36 = 1/4. The probability that the sum is not a multiple of 4 is P(not E) = 1 - P(E) = 1 - 1/4 = 3/4.

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Probability — A Theoretical Approach — Part 4

Probability — A Theoretical Approach — Part 4

Welcome back! So far, we've explored probability with simple tools like coins and dice. But the real power of probability shines when we apply it to more complex situations with many possible outcomes. How do casinos calculate odds for card games? How does a weather forecast predict a 70% chance of rain?

These questions involve analyzing larger sets of possibilities. In this section, we'll master this by diving into one of the most classic tools in probability: a standard deck of playing cards. We'll also learn a powerful technique involving complementary events—sometimes, the easiest way to find the chance of something happening is to first find the chance of it not happening.

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

Definitions & Formulas

Before we start solving problems, let's formally define the key terms and formulas for this section. Understanding these is crucial for building a strong foundation.

Derivation: The Rule of Complementary Events

One of the most useful ideas in probability is the relationship between an event happening and the same event not happening. The logic is simple and elegant. Let's derive the formula P(E) + P(E') = 1.

1. An experiment has a total of n(S) equally likely outcomes. This is our entire universe of possibilities.

2. Let E be our event of interest. The number of outcomes that are "favourable" to E is n(E).

3. If an outcome is not favourable to E, it must be favourable to the complement, E'. Therefore, the number of outcomes for E' is the total minus the outcomes for E.

   n(E') = n(S) - n(E)
   

4. Now, let's find the probability of E' using the fundamental probability formula.

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

5. Substitute the expression for n(E') from Step 3 into this formula.

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

6. We can split this fraction into two parts.

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

7. We know that n(S)/n(S) = 1 and n(E)/n(S) = P(E). Substituting these back gives us the final relationship.

   P(E') = 1 - P(E)
   

   Rearranging this gives the famous identity: P(E) + P(E') = 1. This tells us that the probability of an event happening plus the probability of it not happening must always equal 1 (or 100%).

{{KEY: type=concept | title=The Power of Complements | text=For any event E, the probability that E will not occur is 1 minus the probability that it will occur. This is extremely useful for 'at least one' or 'none' type problems, which are often harder to calculate directly.}}

Solved Examples

Let's apply these concepts. We'll start with the basics of playing cards and move to more complex scenarios. For all card problems, assume a standard, well-shuffled 52-card deck with no jokers.

{{VISUAL: diagram: A chart showing the breakdown of a 52-card deck into 4 suits (Spades, Hearts, Clubs, Diamonds), 2 colours (Red/Black), and 13 ranks per suit (A, 2-10, J, Q, K), highlighting the 12 face cards (J, Q, K of each suit).}}

Example 1: Drawing a Specific Suit (Easy)

Given: A single card is drawn from a standard deck of 52 playing cards.

To Find: The probability of drawing a Spade.

Solution:

1. First, identify the total number of possible outcomes. A standard deck has 52 cards.

   n(S) = 52
   

2. Next, identify the number of favourable outcomes. The event E is 'drawing a Spade'. There are 13 spade cards in a deck.

   n(E) = 13
   

3. Apply the theoretical probability formula, P(E) = n(E) / n(S).

   P(Spade) = 13 / 52
   

4. Simplify the fraction to its lowest terms.

   P(Spade) = 1 / 4
   

Final Answer: The probability of drawing a Spade is 1/4.

Example 2: Drawing a Card with Multiple Properties (Medium)

Given: A single card is drawn from a standard deck of 52 playing cards.

To Find: The probability that the card is a red face card.

Solution:

1. The total number of possible outcomes remains the same.

   n(S) = 52
   

2. Define the event E as 'the card is a red face card'. We need to count how many cards fit this description.

   • The red suits are Hearts and Diamonds.
   • The face cards are Jack (J), Queen (Q), and King (K).
   • Combining these, the favourable cards are: J♥, Q♥, K♥, J♦, Q♦, K♦.

3. Count the number of these favourable outcomes.

   n(E) = 6
   

4. Use the probability formula.

   P(Red Face Card) = 6 / 52
   

5. Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 2.

   P(Red Face Card) = 3 / 26
   

Final Answer: The probability of drawing a red face card is 3/26.

Example 3: Using Complementary Events (Hard)

Given: A bag contains 5 red marbles, 8 white marbles, and 4 green marbles. One marble is drawn at random.

To Find: The probability that the marble drawn is not green.

Solution:

1. This can be solved in two ways. Let's first use the complement rule, which is more efficient. First, calculate the total number of marbles in the bag.

   n(S) = 5 (Red) + 8 (White) + 4 (Green) = 17
   

2. Let the event G be 'the marble is green'. The event we want is G' ('the marble is not green'). Let's first find the probability of G.

   n(G) = 4
   

   P(G) = n(G) / n(S) = 4 / 17
   

3. Now, use the complement rule: P(G') = 1 - P(G).

   P(not Green) = 1 - (4 / 17)
   

4. Calculate the final probability.

   P(not Green) = 17/17 - 4/17 = 13 / 17
   

   Alternative Method (Direct Calculation): The event 'not green' means the marble is either red or white. n(not Green) = n(Red) + n(White) = 5 + 8 = 13. P(not Green) = n(not Green) / n(S) = 13 / 17. Both methods yield the same result, but understanding the complement rule is essential for more complex problems.

Final Answer: The probability that the marble is not green is 13/17.

Example 4: Multiple Conditions and Complements (Tricky)

Given: A card is drawn from a standard deck of 52 cards.

To Find: The probability that the card is neither a king nor a queen.

Solution:

1. The total number of outcomes is the full deck.

   n(S) = 52
   

2. Let event E be 'the card is a king or a queen'. The event we want to find is E', 'the card is neither a king nor a queen'. It is easier to calculate P(E) first.

3. Count the number of outcomes favourable to E. There are 4 kings and 4 queens in a deck.

   n(E) = 4 (Kings) + 4 (Queens) = 8
   

4. Calculate the probability of event E.

   P(E) = n(E) / n(S) = 8 / 52
   

5. Simplify this fraction.

   P(E) = 2 / 13
   

6. Now, use the complement rule P(E') = 1 - P(E) to find the probability of the card being neither a king nor a queen.

   P(E') = 1 - (2 / 13)
   

7. Perform the subtraction to get the final answer.

   P(E') = 13/13 - 2/13 = 11 / 13
   

Final Answer: The probability that the card is neither a king nor a queen is 11/13.

Tips & Tricks

Mastering probability involves recognizing patterns and choosing the quickest path to the solution. Here are some shortcuts.

Common Mistakes

Many students lose marks not because the concept is hard, but due to small oversights. Here are some common pitfalls to avoid.

Brain-Teaser Questions

Ready to test your skills? These questions require you to apply the concepts in slightly unfamiliar ways.

1. A number is selected at random from the integers 1 to 30. What is the probability that it is a prime number?

   💡 Answer: The prime numbers between 1 and 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. There are 10 prime numbers. Total numbers = 30. So, P(Prime) = 10/30 = 1/3.

2. A piggy bank contains one hundred 50p coins, fifty ₹1 coins, twenty ₹2 coins and ten ₹5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, what is the probability that the coin will not be a 50p coin?

   💡 Answer: Total coins = 100 + 50 + 20 + 10 = 180. Number of 50p coins = 100. P(50p coin) = 100/180 = 10/18 = 5/9. P(not a 50p coin) = 1 - P(50p coin) = 1 - 5/9 = 4/9.

3. From a standard deck of 52 cards, all face cards are removed. A card is then drawn from the remaining deck. What is the probability of drawing a red card?

   💡 Answer: Total cards = 52. Total face cards = 12. Remaining cards = 52 - 12 = 40. So, n(S) = 40. The number of red cards in a full deck is 26. The face cards removed include 6 red face cards (J, Q, K of Hearts and Diamonds). Remaining red cards = 26 - 6 = 20. So, n(Red) = 20. P(Red card) = 20/40 = 1/2.

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Summary & Quick Revision

Page 5 of 5: Summary & Quick Revision

Concept Introduction

Welcome to the final summary of our journey into probability! Think about a simple, everyday question: "What are the chances of rain tomorrow?" When a weather app says there's a 70% chance of rain, it's using probability to predict the future based on data and patterns. This is an example of empirical probability, based on observations.

In this chapter, we've focused on a different kind: Theoretical Probability. Instead of running experiments thousands of times, we use logic and reasoning to calculate the chances. We build a perfect, "fair" model in our minds—like a perfectly balanced coin or a non-loaded die. By understanding all the possible things that could happen, we can precisely calculate the likelihood of the one specific thing we're interested in, without ever tossing a single coin. This powerful idea, assuming fairness and equal chances, is the foundation of classical probability.

{{FORMULA: expr=P(E) = (Number of outcomes favourable to E) / (Number of all possible outcomes) | symbols=P(E):Probability of event E}}

Definitions & Formulas

Here are the essential terms and the core formula you'll need to master probability.

The Logic Behind the Formula

The famous probability formula isn't magic; it's built on a simple, logical foundation. Here’s a step-by-step breakdown of why it works, using the example of rolling a fair six-sided die.

1. Identify All Possibilities: First, we list every single thing that can happen. For a die roll, the possible outcomes are {1, 2, 3, 4, 5, 6}. There are a total of 6 possible outcomes.

2. Assume Fairness: The crucial assumption is that the die is fair. This means every outcome is equally likely. The chance of getting a 1 is the same as getting a 2, and so on.

3. Calculate Individual Probability: If there are n total equally likely outcomes, then the probability of any single specific outcome happening is 1/n. For our die, the probability of rolling a 3 is 1/6. The probability of rolling a 5 is also 1/6.

4. Define Your Event: Now, let's define an event we care about. Let Event E be "getting a number greater than 4".

5. Count the "Wins": We look at our list of total outcomes and pick out the ones that count as a "win" for our event E. The numbers greater than 4 are {5, 6}. So, there are 2 favourable outcomes.

6. Sum the Individual Probabilities: The total probability of event E is the sum of the probabilities of each favourable outcome. P(E) = P(getting a 5) + P(getting a 6) P(E) = (1/6) + (1/6) = 2/6 = 1/3.

This final result, 2/6, is exactly what the formula gives us: (Number of favourable outcomes) / (Total number of outcomes). The formula is just a shortcut for this summing process!

Solved Examples

Let's apply the formula to solve problems, moving from easy to tricky.

Example 1: Easy (Drawing a specific color ball)

Given: A bag contains 4 red balls, 5 blue balls, and 1 yellow ball. All balls are of the same size. A ball is drawn at random.

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

Solution:

1. First, calculate the total number of possible outcomes. This is the total number of balls in the bag.

   Total balls = 4 (red) + 5 (blue) + 1 (yellow) = 10
   

2. Next, identify the number of outcomes favourable to the event "drawing a blue ball".

   Number of blue balls = 5
   

3. Apply the probability formula.

   P(blue ball) = (Number of favourable outcomes) / (Total number of outcomes)
   

   P(blue ball) = 5 / 10
   

4. Simplify the fraction to its lowest terms.

   P(blue ball) = 1/2
   

Final Answer: The probability of drawing a blue ball is 1/2.

Example 2: Medium (Using number properties with a die)

Given: A single, fair six-sided die is rolled once.

To Find: The probability of getting a prime number.

Solution:

1. List all possible outcomes when rolling a die.

   Total outcomes = {1, 2, 3, 4, 5, 6}
   

   The total number of possible outcomes is 6.

2. Identify the event. The event E is "getting a prime number".

3. List the favourable outcomes. We need to find the prime numbers between 1 and 6. Remember, 1 is not a prime number.

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

   The number of favourable outcomes is 3.

4. Use the probability formula.

   P(prime number) = (Number of favourable outcomes) / (Total number of outcomes)
   

   P(prime number) = 3 / 6
   

5. Simplify the fraction.

   P(prime number) = 1/2
   

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

{{KEY: type=concept | title=Theoretical vs. Empirical Probability | text=Theoretical probability is what we expect to happen in an ideal experiment (e.g., a coin giving heads 1/2 of the time). Empirical probability is what actually happens when you perform the experiment many times. In this chapter, we only focus on the theoretical approach.}}

Example 3: Hard (Divisibility rules with numbered cards)

Given: A box contains 15 cards, numbered from 1 to 15. One card is drawn at random from the box.

To Find: The probability that the number on the card is divisible by 3.

Solution:

1. Determine the total number of possible outcomes. Since the cards are numbered 1 to 15, there are 15 possible outcomes.

   Total outcomes = {1, 2, 3, ..., 14, 15}
   

   Total number of outcomes = 15.

2. Define the event E as "the number is divisible by 3".

3. List all the outcomes that are favourable to this event. These are the multiples of 3 between 1 and 15.

   Favourable outcomes = {3, 6, 9, 12, 15}
   

4. Count the number of favourable outcomes. There are 5 such numbers.

   Number of favourable outcomes = 5
   

5. Calculate the probability using the formula.

   P(divisible by 3) = (Number of favourable outcomes) / (Total number of outcomes)
   

   P(divisible by 3) = 5 / 15
   

6. Simplify the resulting fraction.

   P(divisible by 3) = 1/3
   

Final Answer: The probability of drawing a card with a number divisible by 3 is 1/3.

Example 4: Tricky (Counting letters in a word)

Given: A single letter is chosen at random from the word "MATHEMATICS".

To Find: The probability that the chosen letter is a vowel.

Solution:

1. First, carefully count the total number of letters in the word "MATHEMATICS". This will be the total number of possible outcomes.

   M, A, T, H, E, M, A, T, I, C, S
   

   Total number of letters = 11.

2. Define the event E as "the chosen letter is a vowel". The vowels in English are A, E, I, O, U.

3. Now, count the number of vowels in the word "MATHEMATICS". Be careful with repeated letters.

   Vowels are: A, E, A, I
   

   The number of favourable outcomes (vowels) is 4.

4. Apply the probability formula.

   P(vowel) = (Number of favourable outcomes) / (Total number of outcomes)
   

   P(vowel) = 4 / 11
   

Final Answer: The probability of choosing a vowel is 4/11.

Tips & Tricks

Use these shortcuts to solve problems faster and more accurately.

Common Mistakes to Avoid

Many students lose marks due to small, careless errors. Here’s a guide to what to watch out for.

Brain-Teaser Questions

Ready to test your skills? These questions require careful thinking.

1. A standard deck of 52 playing cards is shuffled. What is the probability of drawing a card that is a black face card? (Hint: Face cards are Jack, Queen, King).

   💡 Answer: Total cards = 52. The black suits are Spades and Clubs. Each suit has 3 face cards (J, Q, K). So, total black face cards = 2 suits × 3 cards/suit = 6. P(black face card) = 6/52 = 3/26.

2. Two fair coins are tossed simultaneously. What is the probability of getting at least one head?

   💡 Answer: First, list all possible outcomes: {HH, HT, TH, TT}. Total outcomes = 4. The event "at least one head" includes the outcomes {HH, HT, TH}. There are 3 favourable outcomes. P(at least one head) = 3/4. (Alternatively, using complements: The only outcome with no heads is TT. P(no heads) = 1/4. So, P(at least one head) = 1 - P(no heads) = 1 - 1/4 = 3/4).

3. A number x is selected at random from the numbers 1, 2, 3. Another number y is selected at random from the numbers 1, 4, 9. What is the probability that the product x × y is less than 9?

   💡 Answer: List all possible pairs of (x, y). There are 3 choices for x and 3 for y, so 3 × 3 = 9 total possible outcomes. The pairs are: (1,1), (1,4), (1,9), (2,1), (2,4), (2,9), (3,1), (3,4), (3,9). Now find their products: 1, 4, 9, 2, 8, 18, 3, 12, 27. Count the products that are less than 9: {1, 4, 2, 8, 3}. There are 5 such products. P(product < 9) = 5/9.

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