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Chapter 3: Problem 29

There are 15 tennis balls in a box, of which 9 have not previously been used. Three of the balls are randomly chosen, played with, and then returned to the box. Later, another 3 balls are randomly chosen from the box. Find the probability that none of these balls has ever been used.

Short Answer

Expert verified
The probability that none of the balls have ever been used when choosing two consecutive groups of three balls from a box of 15 balls is approximately 3.39%.

Step by step solution

01

Find the number of ways to choose the first group

We can use the combination formula to find the number of ways to choose 3 balls from 15. The combination formula is given by: \[\binom{n}{k} = \frac{n!}{k!(n-k)!}\] In this case, \(n = 15\) and \(k = 3\): \[\binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15!}{3!12!} = 455\] So there are 455 ways to choose the first group of three balls.
02

Find the number of ways to choose the first group so that none of the balls have been used

We need to choose three balls from the nine that have not been used. Using the combination formula with \(n = 9\) and \(k = 3\): \[\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = 84\] So 84 ways to choose the first group without any used balls.
03

Find the number of ways to choose the second group

After returning the first group of balls, we are still choosing from 15 balls. So there are still 455 ways to choose the second group of 3 balls.
04

Find the number of ways to choose the second group so that none of the balls have been used

Since the balls in the first group were returned, we still have 9 unused balls to choose from. Using the combination formula with \(n = 9\) and \(k = 3\): \[\binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = 84\] So again, there are 84 ways to choose the second group without any used balls.
05

Calculate the probability that none of the balls have ever been used

The probability of both events happening is the product of the individual probabilities. We find the probability for each event by dividing the number of ways to choose a group without any used balls (84) by the total number of ways to choose a group (455): Probability of the first group with no used balls = \(\frac{84}{455}\) Probability of the second group with no used balls = \(\frac{84}{455}\) So the probability that none of the balls have ever been used is: \[\frac{84}{455} \times \frac{84}{455} \approx 0.0339\] Thus, there is approximately a 3.39% chance that none of the balls have ever been used.

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