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a city council consists of seven democrats and five republicans. if a c…

Question

a city council consists of seven democrats and five republicans. if a committee of six people is selected, find the probability of selecting four democrats and two republicans. (type a fraction. simplify your answer.)

Explanation:

Step1: Calculate the total number of ways to select 6 people from \(7 + 5=12\) people

The formula for combinations is \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 12\) and \(r = 6\).

$$C(12,6)=\frac{12!}{6!(12 - 6)!}=\frac{12\times11\times10\times9\times8\times7}{6\times5\times4\times3\times2\times1}=924$$

Step2: Calculate the number of ways to select 4 Democrats from 7 and 2 Republicans from 5

For selecting 4 Democrats from 7: \(C(7,4)=\frac{7!}{4!(7 - 4)!}=\frac{7\times6\times5}{3\times2\times1}=35\)
For selecting 2 Republicans from 5: \(C(5,2)=\frac{5!}{2!(5 - 2)!}=\frac{5\times4}{2\times1}=10\)
By the multiplication principle, the number of ways to select 4 Democrats and 2 Republicans is \(C(7,4)\times C(5,2)=35\times10 = 350\)

Step3: Calculate the probability

The probability \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\)

$$P=\frac{350}{924}=\frac{25}{66}$$

Answer:

\(\frac{25}{66}\)