Question

Help Please!
Three people are selected at random from six females and eight males. Find the probability of the following
At most two are male.

Answer 1

How many people are there total?

Answer 2

Tell the total number of people.

Answer 3

In total there are 14 people. At most you want 2 men out of the 14 people. Therefore the probability is \[\frac{ 2 }{ 14 }\] and simplified is \[\frac{ 1 }{ 7 }\]. The probability of 2 men is \[\frac{ 1 }{ 7 }\]

Answer 4

You'll need the binomial distribution for this one. The probability of a male being selected is \(p=\cfrac{8}{14}=\cfrac{4}{7}\) and probability of female is \(\cfrac{3}{7}\). If \(X\) represents the number of males selected, then
$$
\large{
P(X\le x)=\sum_0^x\binom{14}{x}\left(\cfrac{4}{7}\right)^x\left(\cfrac{3}{7}\right)^{(14-x)}\\
}
$$
So then,
$$
P(X\le2)=\sum_0^2\binom{14}{x}\left(\cfrac{4}{7}\right)^x\left(\cfrac{3}{7}\right)^{(14-x)}\\
=\binom{14}{0}\left(\cfrac{4}{7}\right)^0\left(\cfrac{3}{7}\right)^{(14-0)}+\\
\binom{14}{1}\left(\cfrac{4}{7}\right)^1\left(\cfrac{3}{7}\right)^{(14-1)}+\\
\binom{14}{2}\left(\cfrac{4}{7}\right)^2\left(\cfrac{3}{7}\right)^{(14-2)}
$$

Does this make sense?

Answer 5

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

Ohh my God.... Typing made me tired a lot..

Well, Hey Sweet @nm61101

Wondering If you are satisfied with this answer, Please close this Question.

Thank You ! Keep in touch with Open Study.. Bye !