Question #234414
A University wishes to hire 4 faculty members for its newly established department. They have to select from a group of 8 Male and 7 Female applicants, all of whom are equally qualified to fill in the positions. If University selects 4 at random, what is the probability that:
a) All four will be Men
b) All four will be Women
c) Atleast one will be a Women
d) Atleast one will be a Men
1
Expert's answer
2021-09-08T08:24:04-0400

a)


P(4 Men)=(84)(70)(7+84)P(4\ Men)=\dfrac{\dbinom{8}{4}\dbinom{7}{0}}{\dbinom{7+8}{4}}(84)=8!4!(84)!=70\dbinom{8}{4}=\dfrac{8!}{4!(8-4)!}=70

(70)=1\dbinom{7}{0}=1

(8+74)=15!4!(154)!=1365\dbinom{8+7}{4}=\dfrac{15!}{4!(15-4)!}=1365


P(4 Men)=(84)(70)(7+84)=701365=239P(4\ Men)=\dfrac{\dbinom{8}{4}\dbinom{7}{0}}{\dbinom{7+8}{4}}=\dfrac{70}{1365}=\dfrac{2}{39}

b)


P(4 Women)=(80)(74)(7+84)P(4\ Women)=\dfrac{\dbinom{8}{0}\dbinom{7}{4}}{\dbinom{7+8}{4}}(80)=1\dbinom{8}{0}=1

(74)=7!4!(74)!=35\dbinom{7}{4}=\dfrac{7!}{4!(7-4)!}=35

(8+74)=15!4!(154)!=1365\dbinom{8+7}{4}=\dfrac{15!}{4!(15-4)!}=1365


P(4 Women)=(80)(74)(7+84)=351365=139P(4\ Women)=\dfrac{\dbinom{8}{0}\dbinom{7}{4}}{\dbinom{7+8}{4}}=\dfrac{35}{1365}=\dfrac{1}{39}

(c)


P(at least 1 Women)=1P(0 Women)P(at\ least\ 1\ Women)=1-P(0\ Women)

=1P(4 Men)=1239=3739=1-P(4\ Men)=1-\dfrac{2}{39}=\dfrac{37}{39}

d)


P(at least 1 Men)=1P(0 Men)P(at\ least\ 1\ Men)=1-P(0\ Men)

=1P(4 Women)=1139=3839=1-P(4\ Women)=1-\dfrac{1}{39}=\dfrac{38}{39}



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS