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 first need to determine the total number of ways to select 2 balls from the box.
Total number of balls = 3 red + 2 green + 6 blue = 11 balls
Total number of ways to select 2 balls from 11 = 11C2 = 55
Next, we need to consider the ways in which we can win R9. Since winning R9 requires selecting a red ball and then a green ball, there are 3 ways to do this:
1. Select a red ball first, then a green ball
2. Select a green ball first, then a red ball
3. Select a red ball and green ball at the same time
The probability of each of these events is as follows:
1. Probability of selecting a red ball first = 3/11
2. Probability of selecting a green ball second = 2/10 (since one red ball has already been selected)
3. Probability of selecting a red and green ball at the same time = 3/11 * 2/10 = 3/55
Therefore, the total probability of winning R9 is:
(3/11 * 2/10) + (3/11 * 2/10) + (3/55) = 6/110 + 6/110 + 3/55 = 12/110 + 3/55 = 15/55 = 3/11
So, the probability of winning R9 is 3/11.