Question

reese, greg, and brad meet once a week for coffee. they each have their favorite café and, to be fair, they use randomization to choose where they will meet. each person has a colored marble: red (r) for reese, green (g) for greg, and blue (b) for brad. each week, all three marbles are mixed well in a bag and a marble is selected. the favorite café of the person associated with the selected marble is chosen for that week’s meeting. what is the probability that greg will not get to pick the café for either of the first two weeks?

Explanation:

Step1: Determine Greg's marble color

Greg's favorite café is associated with the green (G) marble. So, we want the probability that in the first two weeks, the selected marble is not green (i.e., it's either red (R) for Reese or blue (B) for Brad).

Step2: Probability of not picking Greg's marble in one week

There are 3 marbles: R, G, B. The number of non - green marbles is 2 (R and B). The probability of not picking Greg's marble in one week is the number of favorable outcomes (non - green) divided by the total number of outcomes. So, $P(\text{not G in one week})=\frac{2}{3}$.

Step3: Probability for two independent weeks

Since the selection in each week is independent, the probability of not picking Greg's marble in two consecutive weeks is the product of the probabilities for each week. So, $P(\text{not G in two weeks}) = P(\text{not G in week 1})\times P(\text{not G in week 2})$. Substituting the value from step 2, we get $\frac{2}{3}\times\frac{2}{3}=\frac{4}{9}$.

Answer:

$\frac{4}{9}$