Question

Two people are selected at random from a group of 6 pilots and 4 engineers. What is the probability that both of them are engineers?

Answer 1

Probability is the ratio of: the number of observations of an event n(A) compared to the total number of observation n(S). Generally it could be written as
P(A) = $\dfrac{n(A)}{n(S)}$

Use combination to solve this problem
The event being observed n(A) in this problem is 2 engineers being selected from 4 engineers. The observation n(S) is about 2 people being selected from 10 people (engineers + pilots).

n(A) could be determined using combination of 2 engineers from 4 engineers
n(A) = $C^4_2$
n(A) = $\dfrac{4!}{2!(4-2)!}$
n(A) = $\dfrac{4!}{2!2!}$
n(A) = $\dfrac{4 \times 3 \times 2 \times 1}{2 \times 1 \times 2 \times 1}$
n(A) = $\dfrac{24}{4}$
n(A) = 6

n(S) could be determined using combination of 2 people from 10 people
n(S) = $C^10_2$
n(S) = $\dfrac{10!}{2!(10-2)!}$
n(S) = $\dfrac{10!}{2!8!}$
n(S) = $\dfrac{10 \times 9 \times 8!}{2 \times 1 \times 8!}$
n(S) = $\dfrac{10 \times 9}{2}$
n(S) = $\dfrac{90}{2}$
n(S) = 45

The probability
P(A) = $\dfrac{n(A)}{n(S)}$
P(A) = $\dfrac{6}{45}$
P(A) = $\dfrac{2}{15}$

The probability is 2/15