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A consulting firm has $10$ partners in the firm. Mrs Wendytia, the managing partner want to discuss an idea of moving the main office from Central Jakarta to BSD, South Tangerang. Of those $10$ partners, $8$ live in West Jakarta and $2$ live in East Jakarta. She will appoint a committee randomly consist of $3$ partners to have a study about this plan. What is the probability that the committee chosen consist of $2$ partner live in East Jakarta and the others in West Jakarta?

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Find probabilities using the binomial distribution

Full solution

Q. A consulting firm has

10

partners in the firm. Mrs Wendytia, the managing partner want to discuss an idea of moving the main office from Central Jakarta to BSD, South Tangerang. Of those

10

partners,

8

live in West Jakarta and

2

live in East Jakarta. She will appoint a committee randomly consist of

3

partners to have a study about this plan. What is the probability that the committee chosen consist of

2

partner live in East Jakarta and the others in West Jakarta?

Calculate Total Ways: First, calculate the total number of ways to choose $3$ partners from $10$ without regard to where they live. This is a combination problem, so use the combination formula: $C(n, k) = \frac{n!}{k!(n-k)!}$ . $\newline$ $C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10\times9\times8}{3\times2\times1} = 120$ .
Choose Partners from East Jakarta: Next, calculate the number of ways to choose $2$ partners from the $2$ who live in East Jakarta. Again, use the combination formula: $C(n, k) = \frac{n!}{k!(n-k)!}$ . $\newline$ $C(2, 2) = \frac{2!}{(2!(2-2)!)} = \frac{2!}{(2!0!)} = \frac{1}{(1\times1)} = 1$ .
Choose Partner from West Jakarta: Now, calculate the number of ways to choose $1$ partner from the $8$ who live in West Jakarta. Use the combination formula again. $\newline$ $C(8, 1) = \frac{8!}{(1!(8-1)!)} = \frac{8!}{(1!7!)} = \frac{8}{1} = 8$ .
Calculate Probability: To find the probability that the committee consists of $2$ partners from East Jakarta and $1$ from West Jakarta, multiply the number of ways to choose $2$ partners from East Jakarta by the number of ways to choose $1$ partner from West Jakarta, then divide by the total number of ways to choose $3$ partners from $10$ . $\newline$ Probability = $\frac{C(2, 2) \times C(8, 1)}{C(10, 3)} = \frac{1 \times 8}{120} = \frac{8}{120}$ .
Simplify Fraction: Finally, simplify the fraction $8 / 120$ to get the probability. $\newline$ $8 / 120 = 1 / 15$ .

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