4 men and 6 women are ranked according to their scores on an exam. Assume that no two scores are alike, and that all 10! 10 ! possible rankings are equally likely. Let ???? X denote the highest ranking achieved by a man (so ????=1 X = 1 indicates that a man achieved the highest score on the exam). Find each of the following:
Find each of the following:
(a) P(X=2).
(b) P(P(X=3).
(c) P(P(X=6).
(d) P(P(X=9).

We know that are 4 men and 6 women are ranked according to their scores on an exam. X = 1 indicates that a man achieved the highest score on the exam.

a) We calculate P(X=2).

We calculate the number of possible combinations

[tex]C^{10}_{2}=\frac{10!}{2! (10-2)!}=\frac{10\cdot 9\cdot 8!}{2\cdot 1 \cdot 8!}=45[/tex]

We calculate the number of favorable combinations

[tex]C_2^4=\frac{4!}{2!(4-2)!}=6[/tex]

We get that is

[tex]\boxed{P(X=2)=\frac{6}{45}=\frac{2}{15}}[/tex]

b) We calculate P(X=3).

We calculate the number of possible combinations

[tex]C^{10}_{3}=\frac{10!}{3! (10-3)!}=\frac{10\cdot 9\cdot 8\cdot 7!}{3\cdot2\cdot 1 \cdot 7!}=120[/tex]

We calculate the number of favorable combinations

[tex]C_3^4=\frac{4!}{3!(4-3)!}=4[/tex]

We get that is

[tex]\boxed{P(X=3)=\frac{4}{120}=\frac{1}{30}}[/tex]

c) We calculate P(X=6). This case is not possible because 6 men cannot be selected because we have been given 4 men.

We conclude P(X=6)=0.

d) We calculate P(X=9). This case is not possible because 9 men cannot be selected because we have been given 4 men.

We conclude P(X=9)=0.