NCERT Solutions Class 12 Maths Chapter 13: Probability Ex 13.5

The Foundations of Bernoulli Trials & Binomial Distribution

An experiment is called a Bernoulli trial if it satisfies four key criteria: 1) The number of trials is finite. 2) The trials must be independent of one another. 3) Each trial has exactly two outcomes: success or failure. 4) The probability of success (denoted by p) remains constant in each trial. If any of these conditions fail, we cannot apply the Binomial distribution. Once Bernoulli conditions are met, the random variable X representing the number of successes follows a Binomial distribution. Mathematically, the probability of obtaining exactly r successes in n trials is given by the formula: P(X = r) = nCr p^r q^(n-r), where q = 1 - p represents the probability of failure, and nCr = n! / (r! * (n-r)!) is the binomial coefficient.

Step-by-Step Strategy to Solve Binomial Probability Problems

1. Identify Bernoulli Criteria — Verify if the experiment involves a fixed number of independent trials (n) with constant success probability (p) and only two outcomes.
2. Define Success and Determine p & q — Define clearly what a 'success' is. Find the probability of success, p, from a single trial, and compute q = 1 - p.
3. Formulate the Variable X — Define X as the number of successes. Identify the required conditions, such as X = r, X >= r (at least), or X <= r (at most).
4. Apply the Binomial Formula — Substitute the values of n, p, q, and r into the Binomial Distribution formula P(X = r) = nCr p^r q^(n-r) and simplify carefully.

Step-by-Step Worked Examples

• Example 1: A fair coin is tossed 6 times. Find the probability of getting exactly 4 heads. Step 1: Identify parameters. Here, n = 6 trials. Tosses are independent. Let success be getting a head. Step 2: Calculate p and q. For a fair coin, p = 1/2. Thus, q = 1 - p = 1/2. Step 3: Define X. Let X be the number of heads. We need P(X = 4). Step 4: Use the formula P(X = r) = 6Cr (1/2)^r (1/2)^(6-r). For r = 4: P(X = 4) = 6C4 (1/2)^4 (1/2)^2 = 15 * (1/64) = 15/64. Final Answer: 15/64.
• Example 2: An insurance agent sells policies to 5 individuals. The probability that any individual will live for 30 years is 2/3. Find the probability that at least 4 individuals will live. Step 1: Identify parameters. n = 5 trials. Success is an individual living for 30 years, so p = 2/3. Step 2: Calculate q. q = 1 - 2/3 = 1/3. Step 3: Define 'at least 4' as P(X >= 4) = P(X = 4) + P(X = 5). Step 4: Compute individual probabilities. P(X = 4) = 5C4 (2/3)^4 (1/3)^1 = 5 (16/81) (1/3) = 80/243. P(X = 5) = 5C5 (2/3)^5 (1/3)^0 = 1 (32/243) 1 = 32/243. Step 5: Add the values: P(X >= 4) = 80/243 + 32/243 = 112/243. Final Answer: 112/243.

Board Exam Tips & Pitfalls to Avoid

1. Distinguish 'At Least' vs 'At Most': 'At least r' means we sum from r up to n. 'At most r' means we sum from 0 up to r. Sometimes it is faster to calculate the complement, such as P(X >= 1) = 1 - P(X = 0).
2. Sampling With vs Without Replacement: In Ex 13.5, trials must be independent. If balls are drawn without replacement, the probability changes each time, meaning they are not Bernoulli trials. Ensure the problem states with replacement before blindly applying the binomial formula.
3. Simplifying Binomial Terms: Use properties of combinations like nCr = nC(n-r) to speed up calculation, especially when r is large.

Practice Questions with Solutions

• Q: A die is thrown 6 times. If getting an odd number is a success, find the probability of getting at least 5 successes. A: Step 1: Identify parameters. Number of trials n = 6. Success is getting an odd number on a die. Step 2: Compute single trial probabilities. Odd numbers on a die are {1, 3, 5}, so p = 3/6 = 1/2. Thus, q = 1 - p = 1/2. Step 3: Define target event. 'At least 5 successes' means X >= 5. This includes X = 5 and X = 6. Step 4: Calculate individual terms. P(X = 5) = 6C5 (1/2)^5 (1/2)^1 = 6 (1/64) = 6/64 P(X = 6) = 6C6 (1/2)^6 (1/2)^0 = 1 (1/64) = 1/64 Step 5: Sum the probabilities. P(X >= 5) = 6/64 + 1/64 = 7/64. Final answer: 7/64
• Q: The probability of a shooter hitting a target is 3/4. How many minimum number of times must he fire so that the probability of hitting the target at least once is more than 0.99? A: Step 1: Identify parameters. Let n be the number of trials. Success probability p = 3/4, failure probability q = 1/4. Step 2: Formulate 'hitting at least once'. This is P(X >= 1) = 1 - P(X = 0) > 0.99. Step 3: Substitute formula for P(X = 0). P(X = 0) = nC0 (3/4)^0 (1/4)^n = (1/4)^n. Step 4: Solve inequality 1 - (1/4)^n > 0.99 => (1/4)^n < 0.01. Step 5: Find the smallest integer n. Since 4^1 = 4, 4^2 = 16, 4^3 = 64, 4^4 = 256. 1/256 = 0.0039, which is less than 0.01. For n = 3, (1/4)^3 = 1/64 = 0.0156 (greater than 0.01). Final answer: The shooter must fire at least 4 times.
• Q: Ten eggs are drawn successively with replacement from a lot containing 10% bad eggs. Find the probability that there is at least one bad egg. A: Step 1: Let success be drawing a bad egg. Number of trials n = 10. Step 2: Find p and q. Since 10% are bad, p = 10/100 = 1/10 = 0.1. Thus, q = 1 - 0.1 = 0.9. Step 3: Calculate the probability of 'at least one bad egg', which is P(X >= 1) = 1 - P(X = 0). Step 4: Find P(X = 0) = 10C0 (0.1)^0 (0.9)^10 = (0.9)^10. Step 5: Write the final expression: P(X >= 1) = 1 - (0.9)^10. Final answer: 1 - (0.9)^10
• Q: Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability that all five cards are spades. A: Step 1: Identify parameters. n = 5 trials. Success is drawing a spade. Step 2: Compute single trial probabilities. A deck has 13 spades, so p = 13/52 = 1/4. Thus, q = 3/4. Step 3: Define X as the number of spades. We need P(X = 5). Step 4: Apply formula: P(X = 5) = 5C5 (1/4)^5 (3/4)^0. Step 5: Simplify: P(X = 5) = 1 (1/1024) 1 = 1/1024. Final answer: 1/1024

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