Problem 87 - A Complete Resource Book in Mathematics for JEE Main

Question

Fifteen coupons are numbered \(1,2,3, \ldots 15 .\) Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on the selected coupon is 9 is (A) \(\left(\frac{9}{16}\right)^{6}\) (B) \(\left(\frac{8}{15}\right)^{7}\) (C) \(\left(\frac{3}{5}\right)^{7}\) (D) none of these

Step-by-Step Solution

Answer

(D) none of these

Step 1: Understanding the Problem

We have 15 coupons numbered from 1 to 15. We are selecting 7 coupons one by one, with replacement, and we need to find the probability that the largest number selected is exactly 9. This means all selected coupons should be between 1 and 9, and at least one should be 9.

Step 2: Probability of Selecting a Coupon Numbered ≤ 9

The probability of selecting a coupon numbered between 1 and 9 is \[ P(\text{number} \leq 9) = \frac{9}{15} = \frac{3}{5}. \] Since we're selecting with replacement, each coupon has this same probability.

Step 3: Probability All Selected Coupons are ≤ 9

The probability that all 7 coupons are numbered 9 or less can be calculated by \[ P(\text{all } \leq 9) = \left( \frac{3}{5} \right)^7. \] This covers the total probability for all draws being 9 or less.

Step 4: Probability Largest is 9

To find the probability that the largest coupon is exactly 9, use inclusion-exclusion. Subtract the probability where every selected coupon is less than 9.\[ P(\text{all } < 9) = \left( \frac{8}{15} \right)^7. \] This happens when all selected are from 1 to 8. So, the probability that exactly one or more are 9 is \[ P(\text{largest is 9}) = \left( \frac{3}{5} \right)^7 - \left( \frac{8}{15} \right)^7. \]

Step 5: Matching with Options

Comparing our derived answer to the given options, we observe that none of the expressions in (A), (B), or (C) exactly match the derived probability formula. The correct probability is calculated, which does not fit the provided options.

Key Concepts

Counting Principles

Counting principles are fundamental tools in probability for determining how many ways an event can occur. In the problem of selecting coupons, we need to first calculate the probability of picking a coupon numbered between 1 and 9. This essentially requires an understanding of how many choices we have at each draw.

Every time we draw a coupon, we have 15 possibilities since coupons are replaced after each draw. Out of these 15, 9 coupons are numbered between 1 and 9. This provides the chance to calculate the probability of selecting a number 9 or less at each draw as 9 out of 15. The ability to count accurately in these situations is crucial as it sets the foundational probabilities for solving the problem.

Inclusion-Exclusion Principle

The inclusion-exclusion principle is a useful strategy when finding probabilities or counting events that overlap. In our problem, we need to compute the probability that the largest number among selected coupons is 9, meaning at least one of the chosen coupons must be number 9.

We begin by calculating the probability that all selected coupons are from the group 1 to 9, using the principle of inclusion-exclusion. This involves taking the probability that all selected numbers are less than or equal to 9, and then excluding the probability that all numbers are less than 9. The formula for inclusion-exclusion helps clear the confusion over overlapping scenarios:

• Probability of all numbers ≤ 9: \( rac{3}{5} \)^7
• Probability of all numbers < 9: \( rac{8}{15} \)^7

Subtract the second from the first, and you find the probability for exactly 9 being the largest number.

Probability with Replacement

When dealing with probability scenarios involving replacement, the probabilities remain constant across each trial. In this exercise, since each draw is independent and the coupon is replaced, the given probabilities reset with each draw.

This means that every time you draw a coupon, you consistently face 15 total possibilities, including 9 in our restricted set. This constancy allows the use of simple power calculations to determine probabilities over multiple draws.

For example, the probability of drawing a coupon numbered 9 or less across all seven draws remains at \( (\frac{3}{5})^7 \), illustrating the steadiness provided by replacement.

Binomial Probability

While this exercise is not strictly about binomial probability, the structure of multiple trials with two outcomes draws a parallel. Each draw results in either selecting a number 9 or below or a number 10 or above.

This repetitive pattern makes it fit a binomial-like experiment, where for every drawn coupon the outcome is either "success" (number ≤ 9) or "failure" (number > 9). The ultimate probability actually relies on the combination of these successful draws, reflecting on the overlapping measurements much like binomial experiments.

• "Success" in our context is drawing a number ≤ 9.
• "Failure" would be choosing any number greater than 9.

Understanding this concept helps simplify and facilitate calculations for larger and more complex probability exercises.