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?A. 3/7B. 5/12C. 27/70D. 2/7E. 9/35

Answer 1

Answer:

D. 2/7

Step-by-step explanation:

The probability is calculate as a division between the number of ways in which Andrew will be selected and karen will not and the total ways in which they can select 4 people from a group of 8 volunteers.

The number of ways in which they can select 4 people from a group of 8 volunteers can be calculated using:

$nCk=\frac{n!}{k!(n-k)!}$

Where nCk give as the number of ways in which we can form groups of k elements from a group of n elements.

So, if we replace n by 8 and k by 4, we get:

$8C4=\frac{8!}{4!(8-4)!}=70$

Then, there are 70 ways to select 4 people from a group of 8 volunteers.

At the same way, the number of ways in which Andrew will be selected and karen will not can be calculate replacing n by 6 and k by 3 in the same equation as:

$6C3=\frac{6!}{3!(6-3)!}=20$

We replace n by 6 because from the 8 volunteers, Andrew is going to be selected and Karen will not, so, the people that affect the result are just the 6 remaining volunteers. Also, k is equal to 3 because Andrew is already selected and we just need to select 3 more people.

Finally, the probability is:

$P=\frac{20}{70} = \frac{2}{7}$