16. Stephen's calculator displays only one digit, as shown in the diagram. Unfortunately, the calculator is broken. Each time he switches it on, each of the seven bars will either illuminate (show up) or not, with probability 0.5 . The resultant display correctly shows one of the ten digits 0 - 9 with probability $\frac{a}{b}$.
Given that $\frac{a}{b}$ is written in its lowest terms, what is the value of $9 a+2 b$ ?
Solution
173

As each bar illuminates with probability 0.5 then each of the $2^{7}$ possible displays occur with equal likelihood. There are ten displays which correctly show a digit (namely, those which show the digits 0-9). The probability of obtaining one of these displays is $\frac{10}{2^{7}}=\frac{10}{128}$ which is $\frac{5}{64}$ in lowest terms. So $a=5$ and $b=64$. The value of $9 a+2 b$ is $9 \times 5+2 \times 64=45+128=173$.