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
Best Answer
Part (i) should be $\frac{3\cdot5+2\cdot7+1\cdot8}{60}=\frac{37}{60}$.
Part (ii) is the probability that Amos or Carl picked up the phone, times the probability that the phone call was for Ben, divided by the probability that the phone call was picked up by the wrong person.
$$\frac{\frac{1}{3}(\frac12+\frac15)}{\frac{37}{60}}$$
$$=\frac{\frac{7}{30}}{\frac{37}{60}}$$
$$=\frac{14}{37}$$