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question 12
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a committee of 3 people is selected randomly from a group of 5 men and 4 women.
what is the probability that the committee has at least one woman?
a. none of these
b. (\frac{1}{3})
c. (\frac{5}{6})
d. (\frac{4}{9})
e. (\frac{2}{3})
1:15:53
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Explanation:

Step1: Calculate total number of ways to form committee

The total number of people is \(5 + 4=9\). The number of ways to choose a committee of \(3\) people from \(9\) is given by the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 9\) and \(r=3\).

$$C(9,3)=\frac{9!}{3!(9 - 3)!}=\frac{9\times8\times7}{3\times2\times1}=84$$

Step2: Calculate number of ways with no women (all - men)

The number of ways to choose \(3\) men from \(5\) men. Using the combination formula with \(n = 5\) and \(r = 3\)

$$C(5,3)=\frac{5!}{3!(5 - 3)!}=\frac{5\times4}{2\times1}=10$$

Step3: Calculate probability of no - women

The probability of having no women \(P(\text{no women})=\frac{C(5,3)}{C(9,3)}=\frac{10}{84}=\frac{5}{42}\)

Step4: Calculate probability of at least one woman

The probability of at least one woman \(P(\text{at least one woman})=1 - P(\text{no women})\)

$$P(\text{at least one woman})=1-\frac{5}{42}=\frac{42 - 5}{42}=\frac{37}{42}\approx0.881$$

Let's check using another approach:

$$P(\text{at least one woman})=\frac{C(4,1)\times C(5,2)+C(4,2)\times C(5,1)+C(4,3)\times C(5,0)}{C(9,3)}$$

$$C(4,1)\times C(5,2)=\frac{4!}{1!(4 - 1)!}\times\frac{5!}{2!(5 - 2)!}=4\times10 = 40$$

$$C(4,2)\times C(5,1)=\frac{4!}{2!(4 - 2)!}\times\frac{5!}{1!(5 - 1)!}=6\times5=30$$

$$C(4,3)\times C(5,0)=\frac{4!}{3!(4 - 3)!}\times1 = 4$$

$$C(4,1)\times C(5,2)+C(4,2)\times C(5,1)+C(4,3)\times C(5,0)=40 + 30+4=74$$

$$\frac{74}{84}=\frac{37}{42}\approx0.881$$

Now, \(\frac{5}{6}\approx0.833\), \(\frac{4}{9}\approx0.444\), \(\frac{2}{3}\approx0.667\), \(\frac{1}{3}\approx0.333\)

Answer:

a. None of these