Solve the problem.
-At the first tri-city meeting, there were 8 people from town $A, 7$ people from town $B$, and 5 people from town C. If the council consists of 5 people, find the probability of 2 from town $A, 2$ from town $B$, and 1 from town $C$.
A) .076
B) .190
C) .090
D) .038

Question

Quizplus · Accepted Answer

The probability is calculated by finding the number of favorable combinations divided by the total number of combinations. The number of ways to choose 2 people from town A is $ \binom{8}{2} $, from town B is $ \binom{7}{2} $, and from town C is $ \binom{5}{1} $. The total number of ways to choose 5 people from all towns is $ \binom{20}{5} $. The probability is $\frac{\binom{8}{2} \times \binom{7}{2} \times \binom{5}{1}}{\binom{20}{5}} = \frac{28 \times 21 \times 5}{15504} = 0.190$.

Solve the Problem A,7A, 7A,7 People from Town BBB , and 5 People from Town C

Solve the Problem $A, 7$ People from Town $B$ , and 5 People from Town C