The following question comes from MITx 6.431x.
Out of five men and five women, we form a committee consisting of four different people. Assume that 1) each committee of size four is equally likely, 2) Alice and Bob are among the ten people being considered.
Calculate the probability that both Alice and Bob are members of the committee.
I know the correct solution to this problem; what I do not understand is why isn't $(1/5)^2*\binom{8}{2}/\binom{10}{4}$ the correct way to calculate.
- $(1/5)^2$ because both Alice and Bob have a 1 out of 5 chances of being picked
- $\binom{8}{2}$ because after Alice and Bob have been picked, there are two spots left to fill out of 4 men and 4 women
- $\binom{10}{4}$ because that's the total amount of combinations possible
Could anyone help please? In particular, $(1/5)^2$ is not necessary - why?
$\left(\frac15\right)^2$ is (incorrectly) the probability of both Alice and Bob being picked. That's it. It is, by itself, a complete (but still incorrect) answer to the problem. It would have been correct if we had wanted the committee to consist specifically of exactly one man and exactly one woman, rather than four people without gender restrictions.
The correct answer to this problem is $$ \frac{\text{Number of committees with Alice and Bob}}{\text{Number of possible committees in total}} $$ There are no probabilities in neither numerator nor denominator here. There is no room for $\frac15$ anywhere.