A cook has 10 red peppers and 5 green peppers. If the cook selects 6 peppers at random, what is the probability that he selects at least \(4\) green peppers? Express your answer as a common fraction.
+130071 P (4 green peppers) = C(5,4) * C (10,2) / C(15 ,6) = 45 /1001
P(5 green pepers) = C(5,5) * C(10,1) /C(15,6) = 2 /1001
Total probability = [45 + 2] /1001 = 47 /1001
+9674 Note that he can choose (i) 4 green peppers and 2 red peppers, or (ii) 5 green peppers and 1 red pepper. Then,
\(\begin{array}{rcl} \text{Probability} &=&\dfrac{\displaystyle \binom{5}4\binom{10}2+\binom{5}5\binom{10}1}{\displaystyle \binom{10+5}6}\\ &=& \dfrac{47}{1001} \end{array}\)
