Fifteen coupons are numbered 1, 2, ..., 15, respectively. Seven coupons are selected at random one a

Q: Wait, why do we subtract $P(X \le 8)$? Can't we just calculate the probability of getting at least one 9 directly?

You absolutely could, but it is much harder! To calculate it directly, you would need to sum the probabilities of getting exactly one 9, exactly two 9s, and so on, up to seven 9s. Using the complement method—subtracting the 'bad' cases (where the maximum is 8 or less) from the 'total' cases (where the maximum is 9 or less)—is a much more elegant and less error-prone way to solve this.

Q: Why does 'with replacement' matter so much here?

It is the foundation of the entire problem! 'With replacement' ensures that every single draw is an independent event. This means the probability of picking any specific number remains $\frac{1}{15}$ for every one of the seven draws. Without this independence, the probability would change after each draw, and we would have to use combinations instead of simple powers.

Q: I got $\left(\frac{3}{5}\right)^7$ as my answer. Why is that wrong?

That is the most common trap! $\left(\frac{3}{5}\right)^7$ represents the probability that all seven coupons are 9 or less. This includes scenarios where you might pick only 1s, 2s, or 8s, and never pick a 9 at all. In those cases, the maximum is not 9; it is something smaller. You must subtract the cases where 9 never appears to ensure that 9 is indeed the maximum.

Question

Fifteen coupons are numbered $1, 2, \dots, 15$, respectively. Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on a selected coupon is $9$, is

Options:

(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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in front of a box containing fifteen coupons, numbered $1$ through $15$. You are tasked with selecting seven coupons, one by one, and after each selection, you place the coupon back into the box.

Because you replace the coupon, the probability of drawing any specific number remains constant at $\frac{1}{15}$ for every single draw. This independence allows us to treat each of the seven draws as a separate, identical event.

### Defining the Maximum

The core of this problem lies in the definition of the 'maximum'. We want the largest number among our seven selections to be exactly $9$.

Let $X$ be the random variable representing the maximum number drawn. For $X$ to be exactly $9$, two conditions must be satisfied simultaneously:

1. Every single coupon drawn must be less than or equal to $9$. If even one coupon were $10$ or higher, the maximum would be greater than $9$.

2. At least one of the coupons drawn must be exactly $9$. If we draw seven coupons and none of them are $9$, the maximum would be $8$ or less.

### The Power of the Cumulative Strategy

Calculating the probability of 'at least one 9' directly can be tedious. Instead, we use the power of the Cumulative Distribution Function (CDF).

We know that the probability that the maximum is exactly $9$ is the probability that the maximum is $9$ or less, minus the probability that the maximum is $8$ or less. Mathematically, this is expressed as:

$$P(X = 9) = P(X \le 9) - P(X \le 8)$$

This subtraction is a surgical strike; it removes all the cases where the maximum is $8, 7, \dots, 1$ from the set of cases where the maximum is $9, 8, \dots, 1$, leaving us with exactly the cases where the maximum is $9$.

### The Calculation

For $P(X \le 9)$, we need all seven draws to result in a number between $1$ and $9$. The probability of drawing a number $\le 9$ in a single trial is $\frac{9}{15}$, which simplifies to $\frac{3}{5}$.

Since we have seven independent trials, the probability that all seven are $\le 9$ is:

$$P(X \le 9) = \left( \frac{3}{5} \right)^7$$

Similarly, for $P(X \le 8)$, we need all seven draws to result in a number between $1$ and $8$. The probability of drawing a number $\le 8$ in a single trial is $\frac{8}{15}$. Thus, the probability that all seven are $\le 8$ is:

$$P(X \le 8) = \left( \frac{8}{15} \right)^7$$

### Avoiding the Trap

Many students stop at $\left( \frac{3}{5} \right)^7$. They see the $9$ and the $15$, simplify to $\frac{3}{5}$, and raise it to the power of $7$.

But as we discussed, this includes the 'bad' outcomes where the maximum is less than $9$. By subtracting $\left( \frac{8}{15} \right)^7$, we effectively filter out those unwanted outcomes.

Our final result is:

$$\left( \frac{3}{5} \right)^7 - \left( \frac{8}{15} \right)^7$$

Looking at our options, we see that none of them match this expression. Therefore, the correct choice is **'none of these'**.

Animated Solution for Mathematics - Probability: Fifteen coupons are numbered $1, 2, \dots, 15$ , respectively. Seven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on a selected coupon is $9$ , is

Select Answer:

Visualized Solution

Understanding the Experiment

Defining the Target Event

The Cumulative Probability Strategy

Calculating $P (X \leq 9)$

Calculating $P (X \leq 8)$

Subtracting the Two Probabilities

Beware of the Common Trap!

Final Conclusion

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

Defining the Maximum

The Power of the Cumulative Strategy

The Calculation

Avoiding the Trap

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