SOLUTION: Three people are selected at random from five females and nine males. Find the probability of the following. At least one is male= At most two are make= process???

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Question 478721: Three people are selected at random from five females and nine males. Find the probability of the following.
At least one is male=
At most two are make=
process???
Answer by Maths68(1474) About Me  (Show Source):
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Female = 5
Male= 9
Total = 14 persons
Selecting 3 persons out of 14 = nCr=14C3
=14!/(14-3)! 3!
=14*13*12*11!/11! 3!
=14*13*12/3*2*1
= 2186/6
=364 ways

Selecting 1 male out of 9 = nCr=9C1=9 ways
Selecting 2 males out of 9 = nCr=9C2=9!/(9-2)!2!=9!/7!2!=9*8*7!/7!2!=9*4=36 ways
Selecting 3 male out of 9 = nCr=9C3=9 ways=9!/(9-3)!3!=9!/6!3!=9*8*7*6!/6!3!=9*8*7/3*2*1=504/6=84 ways
Selecting 1 female out of 5 = nCr=5C1=5 ways
Selecting 2 females out of 5 = nCr=5C2=5!/(5-2)!2!=5!/3!2!=5*4*3!/3!2!=20/2=10 ways
P(At least one is male) = (Selecting 1 male 2 femal + selecting 2 male 1 female + selectiong 3 male 0 female) / selecting 3 out of 14 persons
P(At least one is male) = [(9C1*5C2) + (9C2*5C1) + (9C3*5C0)] / 14C3
P(At least one is male) = [(9*10)+(36*5)+(84*1)] / 364
P(At least one is male) = [(90+180+84)] / 364
P(At least one is male) =354/364
P(At least one is male)=0.9725
P(At most two are male)= (Selecting 1 male 2 femal + selecting 2 male 1 female) / selecting 3 out of 14 persons
P(At most two are male) = [(9C1*5C2) + (9C2*5C1)] /14C3
P(At most two are male)= [(9*10)+(36*5)] / 364
P(At most two are male) = (90+180) / 364
P(At most two are male) =270/364
P(At most two are male) =0.7417