Question

From a Group of 8 People, Including George and Nina, 3 people are to be selected at random to work on a certain project. What is the probability that 3 people selected will include George but not NinaA 5/56B 9/56C 15/56D 21/56E 25/56

Answer 1

Answer:  C. $\dfrac{15}{56}$

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 :-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

The total number of ways to select 3 people out of 8 is given by :-

$^8C_3=\dfrac{8!}{3!(8-3)!}\\\\=\dfrac{8\times7\times6\times5!}{3!\times5!}=\dfrac{8\times7\times6}{3\times2\times1}=56$

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 :-

$^6C_2=\dfrac{6!}{2!(6-2)!}\\\\=\dfrac{6\times5\times4!}{2!\times4!}=\dfrac{30}{2}=15$

i.e. No. of favorable outcomes= 15

Now, the probability that 3 people selected will include George but not Nina :-

$\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}\\\\=\dfrac{15}{56}$

Hence, the required probability = C. $\dfrac{15}{56}$