Problem 87 - A Complete Resource Book in Mathematics for JEE Main 2017

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

(C) \( \left(\frac{3}{5}\right)^7 \).

Step 1: Understanding the Problem

We need to find the probability that the largest coupon number drawn in 7 tries is 9, with replacement, from numbers 1 to 15. The largest number should not exceed 9 and at least one draw should be 9.

Step 2: Calculate Probability of Picking a Number Less Than or Equal to 9

For each draw, the probability of selecting a number less than or equal to 9 is \( \frac{9}{15} = \frac{3}{5} \).

Step 3: Calculate Probability All Draws Are Less Than or Equal to 9

If all 7 draws result in a number \(\leq 9\), then each draw has a probability of \( \frac{3}{5}\), leading to \( \left(\frac{3}{5}\right)^7 \).

Step 4: Calculate Probability No Draws Are Greater Than 9

This probability, \( \left(\frac{3}{5}\right)^7 \), includes all scenarios where 9 is the largest, any smaller number is the largest as long as it is \(\leq 9\). This matches the condition: the largest drawn number is 9.

Key Concepts

Drawing with replacement

When you draw with replacement, every time you draw a coupon, its number is remembered and the coupon is placed back with the others before drawing again. This means each draw is independent because the number of coupons stays the same, and each coupon has the same chance of being picked each time. Think of this like picking a marble from a jar and putting it back before picking another one. Every draw is just like the first. This is very important for probability as it ensures that the probability of picking any specific number remains constant. In this problem, you are drawing 7 coupons one by one, and each time a coupon is drawn, it is replaced, keeping the risk of picking the largest number static throughout each draw.

Sample space of drawn numbers

The sample space refers to all possible outcomes that can occur when drawing coupons. Since there are 15 coupons numbered from 1 to 15, each draw can result in one of these 15 numbers. With replacement, each draw keeps the chance equal for each number. Therefore, each draw can result in any one of these 15 outcomes, which simplifies the sample space to being the same for each of the 7 draws. This consistent sample space is crucial because it allows us to apply the same probability calculations for each draw. Uniform sample space means the probabilities are straightforward to calculate. For instance, the probability of picking a number less than or equal to 9 in any single draw is calculated using the total possible numbers in the sample space.

Conditional probability

Conditional probability helps in calculating the likelihood of an event based on the occurrence of another event. In this scenario, you are concentrating on finding the probability that the largest drawn number is exactly 9. This determines the kind of conditional control we discuss. Here, we condition the probability calculation under two constraints: the probability where each number drawn is \(\leq 9\), and at least one draw results in the number 9. This means that you not only need to choose numbers below 10 but that the number 9 must also appear among the selected coupons, making this a conditional scenario. Conditional probability helps tighten the criteria to what we are searching for exactly, such as ensuring one outcome is included as in this exercise.

Event probability calculation

Calculating the probability of a specific event, such as having the largest number as 9, involves a few steps. First, you determine the probability of each individual draw result matching the desired outcome, which is a number \(\leq 9\) in this exercise. This is done using the fraction of total desired outcomes — numbers 1 through 9 — over total possible outcomes (1 through 15), resulting in \(\frac{3}{5}\).Next, you use the power of the same probability to reflect the repeated actions of 7 draws, resulting in expressions like \(\left(\frac{3}{5}\right)^7\) for scenarios where all draws meet your condition. This approach effectively narrows the number of ways the event (largest number being 9) can occur out of all potential combinations, allowing you to calculate the exact probability of this particular event.