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Chapter 2: Problem 35

Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue, and 18 green balls. Find the probability that (a) 3 red, 2 blue, and 2 green balls are withdrawn; (b) at least 2 red balls are withdrawn; (c) all withdrawn balls are the same color; (d) either exactly 3 red balls or exactly 3 blue balls are withdrawn.

Short Answer

Expert verified
In summary: (a) The probability of withdrawing 3 red, 2 blue, and 2 green balls is \(\frac{\binom{12}{3} \times \binom{16}{2} \times \binom{18}{2}}{\binom{46}{7}}\). (b) The probability of withdrawing at least 2 red balls is \(\frac{\sum\limits_{k=2}^7 {\binom{12}{k} \times \binom{34}{7-k}}}{\binom{46}{7}}\). (c) The probability of withdrawing all balls of the same color is \(\frac{\binom{12}{7} + \binom{16}{7} + \binom{18}{7}}{\binom{46}{7}}\). (d) The probability of withdrawing either exactly 3 red balls or exactly 3 blue balls is \(\frac{\left(\binom{12}{3}\times\binom{34}{4} + \binom{16}{3}\times\binom{30}{4}\right) - \left(\binom{12}{3}\times\binom{16}{3}\times\binom{18}{1}\right)}{\binom{46}{7}}\).

Step by step solution

01

Calculate Total Possible Outcomes

First, we need to find the total number of ways to draw 7 balls from the urn containing 46 (12 red + 16 blue + 18 green). We will use combinations, denoted as C(n,k) or \(\binom{n}{k}\), where n is the number of items and k is the number of items to choose. So, the total possible outcomes are \(\binom{46}{7}\).
02

(a) Find Probability of Withdrawing 3 Red, 2 Blue, 2 Green Balls

We have to find the number of ways to draw 3 red balls, 2 blue balls, and 2 green balls. For each color, we will use combinations: - Red: \(\binom{12}{3}\) - Blue: \(\binom{16}{2}\) - Green: \(\binom{18}{2}\) We will then multiply these three numbers together to get the total number of ways to draw this specific combination. Next, we will divide the result by the total possible outcomes, \(\binom{46}{7}\), to find the probability.
03

(b) Find Probability of Withdrawing At Least 2 Red Balls

We have to find the probability of withdrawing at least 2 red balls. We will consider the cases of 2, 3, 4, 5, 6, or 7 red balls, and sum the probabilities of each case. For each case, we have to count the ways to draw other colored balls as well: - 2 Red, 5 non-Red: \(\binom{12}{2} * \binom{34}{5}\) - 3 Red, 4 non-Red: \(\binom{12}{3} * \binom{34}{4}\) - 4 Red, 3 non-Red: \(\binom{12}{4} * \binom{34}{3}\) - 5 Red, 2 non-Red: \(\binom{12}{5} * \binom{34}{2}\) - 6 Red, 1 non-Red: \(\binom{12}{6} * \binom{34}{1}\) - 7 Red: \(\binom{12}{7}\) We add these probabilities together (considering each case is mutually exclusive), and then divide by the total possible outcomes, \(\binom{46}{7}\), to find the probability.
04

(c) Find Probability of Withdrawing All Balls of the Same Color

To find this probability, we need to consider the case where all balls are either red, blue, or green and sum their probabilities: - 7 Red: \(\binom{12}{7}\) - 7 Blue: \(\binom{16}{7}\) - 7 Green: \(\binom{18}{7}\) Then, we add these three probabilities together, and divide by the total possible outcomes, \(\binom{46}{7}\), to find the probability.
05

(d) Find Probability of Withdrawing Exactly 3 Red or Exactly 3 Blue Balls

We need to consider the probability of either withdrawing exactly 3 red balls or exactly 3 blue balls, being careful not to double-count the cases where both conditions are met. - Exactly 3 Red, 4 non-Red: \(\binom{12}{3} * \binom{34}{4}\) - Exactly 3 Blue, 4 non-Blue: \(\binom{16}{3} * \binom{30}{4}\) Next, we subtract the cases where both 3 red and 3 blue balls are withdrawn (counted twice, once in each case above). This case has 1 green ball: - Exactly 3 Red, 3 Blue, 1 Green: \(\binom{12}{3} * \binom{16}{3} * \binom{18}{1}\) We sum the probabilities of the first two cases, and subtract the probability of the overlapping case. Then, we divide by the total possible outcomes, \(\binom{46}{7}\), to find the probability.

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