From a group of eight women and seven men, a committee of six is chosen randomly. Find the probability that three women and three men will be chosen.
number of ways to choose 3 women from 8 is C(8,3) or 56
number of ways to choose 3 men from 7 is C(7,3) = 35
number of ways to choose 6 people from 15 is C(15,6) = 5005
so Prob(of stated situation) = 56(35)/5005 = 56/143
To find the probability, we need to determine the total number of possible outcomes and the number of favorable outcomes.
The total number of possible outcomes is obtained by selecting 6 people from a total of 15 (8 women + 7 men). This can be calculated using the combination formula:
C(15, 6) = 15! / (6!(15-6)!) = 5005
The number of favorable outcomes is obtained by selecting 3 women from a group of 8 and 3 men from a group of 7:
C(8, 3) * C(7, 3) = (8! / (3!(8-3)!)) * (7! / (3!(7-3)!)) = 56 * 35 = 1960
Therefore, the probability of selecting three women and three men from the committee is:
P(3 women and 3 men) = favorable outcomes / total outcomes = 1960 / 5005 ≈ 0.3912
Hence, the probability is approximately 0.3912 or 39.12%.