Question

A committee consists of 10 women and 7 men. Three members are chosen as officers. What is the probability that all three officers are women? What is the probability that one of the officers chosen is a woman and the other two are men? What is the probability that at lease one officer is a man?

Answer 1

Answer:

Step-by-step explanation:

1.

$P=\frac{10C3}{17 C3} =\frac{\frac{10*9*8}{3*2*1} }{\frac{17*16*15}{3*2*1} } =\frac{10*9*8}{17*16*15} =\frac{3}{17}$

2.

$P=\frac{10C1 * 7C2}{17 C3} =\frac{\frac{10}{1}*\frac{7*6}{2*1} }{\frac{17*16*15}{3*2*1} } =\frac{210}{680} =\frac{21}{68}$

3.

so possibility can be 1 man and two women,

two men and 1 women,

three men.

$P_{1}(1 ~man~and~two~women)=\frac{7C1*10C2}{17C3} }=\frac{\frac{7}{1}*\frac{10*9}{2*1} }{17C3} =\frac{315}{17C3} \\P_{2}(two~men~and~1~woman)=\frac{7C2*10C1}{17C3} =\frac{\frac{7*6}{2*1}*\frac{10}{1} }{17C3} =\frac{210}{17C3} \\P_{3}(three~men)=\frac{7C3}{17C3} =\frac{\frac{7*6*5}{3*2*1} }{17C3} =\frac{35}{17 C3} \\17 C3=\frac{17*16*15}{3*2*1} =680\\reqd.~probability=P_{1}+P_{2}+P_{3}=\frac{315}{680}+\frac{210}{680}+\frac{35}{680}=\frac{560}{680}=\frac{14}{17}$

Answer 2

Using the hypergeometric distribution, it is found that there is a:

0.1765 = 17.65% probability that all three officers are women.

0.3088 = 30.88% probability that one of the officers chosen is a woman and the other two are men.

0.8235 = 82.35% probability that at lease one officer is a man.

The members are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.
• N is the size of the population.
• n is the size of the sample.
• k is the total number of desired outcomes.  

In this problem:

• Total of 10 + 7 = 17 members, thus $N = 17$
• 10 women, thus $k = 10$
• Sample of 3, thus $n = 3$.

The probability that all are women is P(X = 3), thus:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 3) = h(3,17,3,10) = \frac{C_{10,3}C_{7,0}}{C_{17,3}} = 0.1765$

0.1765 = 17.65% probability that all three officers are women.

From this, also 1 - 0.1765 = 0.8235 = 82.35% probability that at lease one officer is a man.

The probability that one of the officers chosen is a woman and the other two are men is P(X = 1), thus:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}$

$P(X = 1) = h(1,17,3,10) = \frac{C_{10,1}C_{7,2}}{C_{17,3}} = 0.3088$

0.3088 = 30.88% probability that one of the officers chosen is a woman and the other two are men.

A similar problem is given at https://brainly.com/question/24670062