+1245 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.
+4622 Solution:
We can count the number of ways to choose a group of 4 green and 2 red peppers and the number of ways to choose 5 green and 1 red peppers. These are \(\binom{5}{4}\binom{10}{2}=5\cdot45=225 \) and \(\binom{5}{5}\binom{10}{1}=10\). The total number of ways the cook can choose peppers is \(\binom{15}{6}=5005\). Therefore, the probability that out of six randomly chosen peppers at least four will be green is \(\frac{235}{5005}=\boxed{\frac{47}{1001}}\).
