Mathematics, 21.02.2020 06:10 hlannette7005

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?

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Answer from: jennifer9983

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 link

Answer from: sarrivera579

Step-by-step explanation:

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

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

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 from: Quest

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2:

(2 x^3 + 7 x^2 + (x + 4 x) + 2)/(x (x + 2))

x + 4 x = 5 x:

answer: (2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

Answer from: Quest

step-by-step explanation:

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