Problem 28 - A First Course in Probability

Question

An urn contains 5 red, 6 blue, and 8 green balls. If a set of 3 balls is randomly selected, what is the probability that each of the balls will be (a) of the same color? (b) of different colors? Repeat under the assumption that whenever a ball is selected, its color is noted and it is then replaced in the urn before the next selection. This is known as sampling with replacement.

Step-by-Step Solution

Answer

In summary, for the given problem, we have: 1. Without Replacement: (a) Probability of all 3 balls being of the same color: \( \frac{86}{969} \) (b) Probability of all 3 balls being of different colors: \( \frac{240}{969} \) 2. With Replacement: (a) Probability of all 3 balls being of the same color: (5/19)^3 + (6/19)^3 + (8/19)^3 (b) Probability of all 3 balls being of different colors: \( 6 * (\frac{5}{19})(\frac{6}{19})(\frac{8}{19}) \)

Step 1: Calculate the total number of ways to select 3 balls from the urn

There are a total of 5 (red) + 6 (blue) + 8 (green) = 19 balls in the urn. We can choose 3 balls in C(19,3) ways, where C(n,r) denotes the number of possible combinations of choosing 'r' items from a set of 'n' items. Using the combination formula, we have: C(19,3) = \( \frac{19!}{3!(19-3)!} = \frac{19!}{3!16!} = 969\).

Step 2: Calculate the probability of (a) all 3 balls being of the same color

This event can occur in 3 ways: all red, all blue, or all green. - All red: We can select 3 red balls in C(5,3) ways, i.e. \( \frac{5!}{3!2!} = 10\). - All blue: We can select 3 blue balls in C(6,3) ways, i.e. \( \frac{6!}{3!3!} = 20\). - All green: We can select 3 green balls in C(8,3) ways, i.e. \( \frac{8!}{3!5!} = 56\). The probability of this event is the sum of these possibilities divided by the total number of ways to select 3 balls, i.e., \( P(A) = \frac{10+20+56}{969} = \frac{86}{969}\).

Step 3: Calculate the probability of (b) all 3 balls being of different colors

For this event to occur, we must choose one ball of each color. The number of ways to choose one ball from each color is: 5 (for red) * 6 (for blue) * 8 (for green) = 240. The probability of this event is the number of ways to choose one ball of each color divided by the total number of ways to select 3 balls, i.e., \( P(B) = \frac{240}{969}\). #Case 2: With Replacement#

Step 4: Calculate the probability of (a) all 3 balls being of the same color, with replacement

For each ball, we have a probability of choosing: - Red: \( \frac{5}{19} \) - Blue: \( \frac{6}{19} \) - Green: \( \frac{8}{19} \) Since the balls are replaced after noting their color, the probability of choosing all 3 balls of the same color is the sum of probabilities for all red, all blue, or all green: (5/19)^3 + (6/19)^3 + (8/19)^3.

Step 5: Calculate the probability of (b) all 3 balls being of different colors, with replacement

We can have 3! = 6 arrangements for the 3 balls of different colors. For each arrangement, the probability of that arrangement occurring is the product of the probabilities for each color, i.e., \( (\frac{5}{19})(\frac{6}{19})(\frac{8}{19}) \). Since there are 6 such arrangements, the overall probability for this case is 6 times the above expression: \( 6 * (\frac{5}{19})(\frac{6}{19})(\frac{8}{19}) \).