Answer
$\frac{4}{9}.$
Work Step by Step
We know that $\text{probability}=\frac{\text{number of favourable outcomes}}{\text{number of all outcomes}}=\frac{n(E)}{n(S)}.$
Hence here because we have $4$ $\$1$ bills, $n(E)=4$, in total we have $n(S)=4+3+2=9$ bills, therefore $\text{probability}=\frac{4}{9}.$
