Answer:
a) [tex]P(X=2)=\frac{2}{15}[/tex]
b) [tex]P(X=3)=\frac{1}{30}[/tex]
c) P(X=6)=0
d) P(X=9)=0
Step-by-step explanation:
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.
