Rework problem 26 from section 1.4 of your text, involving drawing balls from a box. Assume that the box contains 6 balls: 2 white, 3 blue, and 1 red. Balls are drawn in succession without replacement, and their colors are noted until a white ball is drawn. (1) Which of the following is not a valid outcome for this experiment? A. RR B. BBRBW C. BRBW D. BBW (2) How many outcomes are there in the sample space?

To analyze the problem, we need to understand the setup: we have a box containing 6 balls (2 white, 3 blue, and 1 red), and we draw balls in succession without replacement until we draw a white ball. ### (1) Valid Outcomes Let's evaluate each option to determine if it is a valid outcome: - **A. RR**: This outcome indicates that two red balls were drawn before a white ball. However, there is only 1 red ball in the box, so this outcome is **not valid**. - **B. BBRBW**: This outcome indicates that two blue balls were drawn, followed by one red ball, and then a white ball. This is a valid sequence since it does not exceed the number of available balls. - **C. BRBW**: This outcome indicates that a blue ball was drawn first, followed by a red ball, and then a white ball. This is also valid. - **D. BBW**: This outcome indicates that two blue balls were drawn, followed by a white ball. This is valid as well. Thus, the answer to (1) is: **A. RR** is not a valid outcome. ### (2) Counting Outcomes in the Sample Space To find the total number of outcomes in the sample space, we need to consider the different sequences of drawing balls until a white ball is drawn. The possible sequences can be categorized based on how many balls are drawn before the first white ball appears. 1. **Draw 1 white ball**: - W (1 outcome) 2. **Draw 1 non-white ball, then 1 white ball**: - B + W (3 choices for B) → BW (3 outcomes) - R + W (1 choice for R) → RW (1 outcome) - Total: 3 + 1 = 4 outcomes 3. **Draw 2 non-white balls, then 1 white ball**: - BB + W (3 choices for the first B and 2 choices for the second B) → BBW (3 outcomes) - BR + W (3 choices for B and 1 choice for R) → BRW (3 outcomes) - RB + W (1 choice for R and 3 choices for B) → RBW (3 outcomes) - Total: 3 + 3 + 3 = 9 outcomes 4. **Draw 3 non-white balls, then 1 white ball**: - BBB + W (only 1 way to draw 3 blue balls) → BBBW (1 outcome) - BBR + W (3 choices for the first B, 2 choices for the second B, and 1 choice for R) → BBRW (3 outcomes) - BRB + W (3 choices for B and 1 choice for R) → BRBW (3 outcomes) - RBB + W (1 choice for R and 3 choices for B) → RBBW (3 outcomes) - Total: 1 + 3 + 3 + 3 = 10 outcomes Now, we can sum all the valid outcomes: - 1 (W) + 4 (BW, RW) + 9 (BBW, BRW, RBW) + 10 (BBBW, BBRW, BRBW, RBBW) = 1 + 4 + 9 + 10 = 24 outcomes. Thus, the answer to (2) is: **There are 24 outcomes in the sample space.**