Solutions - 5030-08

(1) Define events and count outcomes:

Let $A$ = both selected are less qualified, and $B$ = at least one less qualified is selected.

Total ways to choose 2 from 5: $\left(\genfrac{}{}{0px}{}{5}{2}\right)=10$

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(2) Compute $P[A]$ and $P[B]$:

Ways both are less qualified: $\left(\genfrac{}{}{0px}{}{3}{2}\right)=3$, so $P[A]=\frac{3}{10}$

Ways neither is less qualified (both more qualified): $\left(\genfrac{}{}{0px}{}{2}{2}\right)=1$, so $P[B^c]=\frac{1}{10}$ and $P[B]=\frac{9}{10}$

Since $A\subset B$, we have $P[A\cap B]=P[A]=\frac{3}{10}$.

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(3) Apply conditional probability:

$$\begin{array}{c}P[A|B]=\frac{P[A\cap B]}{P[B]}\\ \\ \gg \gg \frac{3/10}{9/10}\gg \gg \frac{1}{3}\end{array}$$