Question

From a group of 8 volunteers, including Andrew and Karen, 4 people are to be selected at random to organize a charity event. What is the probability that Andrew will be among the 4 volunteers selected and Karen will not?

Answer 1

Answer:

The probability that Andrew will be among the 4 volunteers selected and Karen will not is 2/7.

Step-by-step explanation:

From the given information it is clear that

The total number of volunteers, including Andrew and Karen = 8

The total number of volunteers, excluding Andrew and Karen = 8-2 = 6

We need to find the probability that Andrew will be among the 4 volunteers selected and Karen will not.

Total number of ways of selecting r volunteers from n volunteers is

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

Total number of ways of selecting 4 volunteers from 8 volunteers is

$\text{Total outcomes}=^8C_4=70$

Total number of ways of selecting 4 volunteers from 8 volunteers, so that Andrew will be among the 4 volunteers selected and Karen will not is

$\text{Favorable outcomes}=^1C_1\times ^6C_3=1\times 20=20$

The probability that Andrew will be among the 4 volunteers selected and Karen will not is

$P=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}$

$P=\frac{20}{70}$

$P=\frac{2}{7}$

Therefore the probability that Andrew will be among the 4 volunteers selected and Karen will not is 2/7.