Probability for Ben, Amos and Carl Three men Amos, Ben and Carl share an office at work with a single telephone. Calls call in at random with the proportions of $\dfrac{1}{2}$ for Amos, $\dfrac{1}{3}$ for Ben and $\dfrac{1}{6}$ for Carl. For any incoming, calls, the probabilities that it will be picked up by Amos, Ben, and Carl are $\dfrac{1}{2}$, $\dfrac{3}{10}$ and $\dfrac{1}{5}$ respectively. For calls arriving during working hours, find the probability that (i)  a call is not picked up by the person being called, (ii)
a call is for Ben given that a call is not picked up by the person being called.
For (i), I tried to get P(proportion for Amos & not picked up) or P(proportion for Ben & not picked up) or P(proportion for Carl & not picked up)
$= \dfrac{1}{2} * \dfrac{1}{2} + \dfrac{1}{3} * \dfrac{1}{2} + \dfrac{1}{6} * \dfrac{5}{6} = 0.5546 $
But the answer for (i) is $\dfrac{37}{60}$, what went wrong?
How do I do for (ii)?
Your help is appreciated. Thanks

You got one number wrong. Your idea is correct, we have
\begin{align*}
  \def\P{\mathbf P}\P(\text{call picked up by wrong person}) 
   &=\underbrace{ \frac 12}_{\text{call for A}}\cdot \underbrace{\frac 12}_{\text{call not picked by A}} +
  \underbrace{ \frac 13}_{\text{call for B}}\cdot \underbrace{\frac 7{10}}_{\text{call not picked by B}}+ 
\underbrace{ \frac 16}_{\text{call for C}}\cdot \underbrace{\frac 45}_{\text{call not picked by C}}\\
  &= \frac 14 + \frac 7{30} + \frac 4{30}\\
  &= \frac{30 + 28 + 16}{120}\\
  &= \frac{37}{60}
\end{align*}
For (ii), we have
\begin{align*}
  \P(\text{call for B}\mid\text{wrong person}) &= \frac{\P(\text{call for B}\cap \text{wrong person})}{\P(\text{wrong person})}\\
&= \frac{\frac 12 \cdot \frac 12}{\frac{37}{60}}\\
 &= \frac{15}{37}
\end{align*}
