# Answer on Statistics and Probability Question for Farhan

Question #3254

A company manager wants to hire 4 people from a group of 10 applicants, 4 men and 6 women, all of whom are equally qualified to fill the position. If he selects the four at random, what is the probability that

i. All four will be women

ii. At least two will be men?

i. All four will be women

ii. At least two will be men?

Expert's answer

Let’s calculate the number N of all options to choose 4 applicants from 10:

N=C

Let A be the event that all four will be women, N_A = number of all options to choose 4 women from 6. Then using the classic definition of probability:

P(A) = N

Let B be the event that at least two will be men. Using the formula P(B)+P(B ̅ )=1, let’s find P(B ̅).

B ̅=A∪C, where C is the event that exactly 3 women were chosen. As events A and C are incompatible,

P(B ̅ )=P(A)+P(C).

For event C we should choose 3 women from 6 and 1 man from 4. So:

P(B) = 1-P(B ̅ ) = 1-P(A)-P(C) = 1-1/14-(C

= 13/14-(6!*4!)

P(B) = 23/42

N=C

_{10}^{4}.Let A be the event that all four will be women, N_A = number of all options to choose 4 women from 6. Then using the classic definition of probability:

P(A) = N

_{A}/N = (C_{6}^{4})/(C_{10}^{4}) = (6!/(4!*2!)) / (10!/(4!*6!)) = 1/(2*7) = 1/14.Let B be the event that at least two will be men. Using the formula P(B)+P(B ̅ )=1, let’s find P(B ̅).

B ̅=A∪C, where C is the event that exactly 3 women were chosen. As events A and C are incompatible,

P(B ̅ )=P(A)+P(C).

For event C we should choose 3 women from 6 and 1 man from 4. So:

P(B) = 1-P(B ̅ ) = 1-P(A)-P(C) = 1-1/14-(C

_{6}^{3}C_{4}^{1})/N = 13/14-(6!/(3!*3!)*4!/3!)/(10!/(4!*6!)) == 13/14-(6!*4!)

^{2}/(10!*(3!)^{3}) = 13/14-(4*5*6)^{2}*4!/(5*6*7*8*9*10*6) = 13/14 - 8/21 = 39/42 - 16/42 = 23/42P(B) = 23/42

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