In a certain game, a ball is randomly selected from a box that contains 3 red balls, 2 green balls and 6 blue balls. If a red ball is selected R5 is won, if green you win R4 and if blue is selected you win nothing. Suppose that two balls are selected at random with replacement, what is the probability of wining R8.
To find the probability of winning R8, we need to find the probability of selecting a red ball and a green ball in any order in the two draws.
The probability of selecting a red ball in the first draw is 3/11, and the probability of selecting a green ball in the second draw is 2/11. Since the draws are made with replacement, the total probability of selecting a red ball followed by a green ball is:
(3/11) * (2/11) = 6/121
Similarly, the probability of selecting a green ball followed by a red ball is also 6/121.
Therefore, the total probability of winning R8 is the sum of the probabilities of these two events:
(6/121) + (6/121) = 12/121
So, the probability of winning R8 is 12/121.