Question

In a group of twelve people, four had been selected for a transfer. If two of the twelve are selected, what is the probability that both of those selected are selected for a transfer?

about 5%
about 9%
about 12%
about 15%

Answer 1

`\displaystyle\\|\Omega|=\binom{12}{2}=(12!)/(2!10!)=(11\cdot12)/(2)=66\\|A|=\binom{4}{2}=(4!)/(2!2!)=(3\cdot4)/(2)=6\\\\P(A)=(6)/(66)=(1)/(11)\approx9\%`

Answer 2

9%

Explanation:

Probability of the first person being selected is 4/12 and the probability of the second person being selected is 3/11

The probability of both is:

4/12 x 3/11

= 1/11 which is 9%