Answer: C. [tex]\dfrac{15}{56}[/tex]
Explanation:
Given : Total people = 8
Number of people are to be selected = 3
The number of combinations of r things taken out of n things is given by :-
[tex]^nC_r=\dfrac{n!}{r!(n-r)!}[/tex]
The total number of ways to select 3 people out of 8 is given by :-
[tex]^8C_3=\dfrac{8!}{3!(8-3)!}\\\\=\dfrac{8\times7\times6\times5!}{3!\times5!}=\dfrac{8\times7\times6}{3\times2\times1}=56[/tex]
If George is included , then one person is confirmed, so we need to selec only 2 people out of 7.
Also, Nina is not selected , so the total number of people left= 6
The total number of ways to select 2 people out of 6 that will include George but not Nina is given by :-
[tex]^6C_2=\dfrac{6!}{2!(6-2)!}\\\\=\dfrac{6\times5\times4!}{2!\times4!}=\dfrac{30}{2}=15[/tex]
i.e. No. of favorable outcomes= 15
Now, the probability that 3 people selected will include George but not Nina :-
[tex]\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}\\\\=\dfrac{15}{56}[/tex]
Hence, the required probability = C. [tex]\dfrac{15}{56}[/tex]
