A cook has 10 red peppers and 5 green peppers. If the cook selects 8 peppers at random, what is the probability that he selects at least 4 green peppers? Express your answer as a common fraction.
+128460 Note....probability of picking any red = 10/15 = 2/3
And probability of picking any green = (5/15) = (1/3)
P(4 green peppers) = C(8,4) ( 2/3)^4 (1/3)^4 = 1120/6561
P(5 green peppers) = C(8, 5) = (2/3)^3 (1/3)^5 = 560/6561
Adding these ( 1120 + 560) / 6561 = 1680 / 6561 = 560 / 2187
