Final answer:
The probability of drawing a white ball twice (event F) from a box with one white and one red ball, sampling with replacement, is 1/4.
Step-by-step explanation:
The question pertains to the subject of probability, which is a branch of mathematics concerned with the likelihoods of different outcomes. Specifically, it addresses the concept of sampling with replacement wherein the outcome of one trial does not affect the subsequent ones.
For event F, the probability of getting the white ball twice (P(F)) is calculated by multiplying the probability of getting a white ball on the first draw by the probability of getting a white ball on the second draw. Since there is one white ball and one red ball, the probability of drawing a white ball in either draw is 1/2. Therefore:
P(F) = (1/2) × (1/2) = 1/4 or 0.25
For event G, the event of getting two balls of different colors, we have two favorable outcomes (white then red, or red then white), so:
P(G) = (1/2) × (1/2) + (1/2) × (1/2) = 1/2 or 0.50
For event H, the probability of getting white on the first pick (P(H)) is simply:
P(H) = 1/2 or 0.50
To answer the questions about mutual exclusivity, events F and G are mutually exclusive because they cannot happen at the same time; you cannot get two white balls and two balls of different colors in the same two draws. However, events G and H are not mutually exclusive as getting a white ball on the first pick could be part of both events.
Regarding independence, event F (getting the white ball twice) and the subordinate events of getting the white ball on the first or second draw (events F and S) are independent because the outcome of the first draw does not affect the probability of the second draw when sampling with replacement.