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 without replacement, what is the probability of wining R9.
To calculate the probability of winning R9, we need to consider the different ways in which this can happen.
There are a total of 11 balls in the box, so the probability of selecting a red ball on the first draw is 3/11. This leaves 10 balls in the box, of which 2 are green, so the probability of selecting a green ball on the second draw is 2/10.
Therefore, the probability of winning R9 by selecting a red ball and then a green ball is:
(3/11) * (2/10) = 6/110 = 3/55
Similarly, the probability of winning R9 by selecting a green ball and then a red ball would also be:
(2/11) * (3/10) = 6/110 = 3/55
Adding these two probabilities together, we get:
3/55 + 3/55 = 6/55
Therefore, the probability of winning R9 by selecting a red ball first and then a green ball is 6/55.