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Question: After your complaint about their service, a representative of an insurance company promised to call you "between $7$ and $9$ this evening". Assume that this means that the time T of the call is uniformly distributed in the specified interval. Assume that you know in advance that the call will last exactly $1$ hour. From $9$ to $9:30$, there is a game show on $TV$ that you wanted to watch. Let $M$ be the amount of time of the show that you miss because of the call. Compute the expected value of $M$.

What I have understood is $P(M | X < 8:00) = 0$, i.e. probability that show will be missed is $0$ when call is received before $8:00$. If I consider time b/w $8:00$ to $8:30$, then expected value is $(8:30-8:00)/2 = 15.$ Is it the right way to proceed. I don't know what is actual answer.

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    $\begingroup$ What you say in English is almost right. You just need to also consider the case that the phone call starts between 8:30-9:00. Note that what you write in maths is incorrect: you should say that given $X$ is the start time of the call, then $E(M|X<8:00)=0$. $\endgroup$ Dec 6, 2020 at 15:48
  • $\begingroup$ @BenjaminWang thanks for correcting. After 8:30, getting show missed will become a sure event. i.e. $P(M | X > 8:30) = 1$. Right? $\endgroup$
    – Dae Hyun
    Dec 6, 2020 at 16:34
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    $\begingroup$ Again, the english is right, and the maths is off by a little. You should say that $P(M=30|X>8:30)=1$. $\endgroup$ Dec 6, 2020 at 16:50

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If the call arrives before 8, M=0

If call arrives in the interval $[8;8:30]$ your M is uniform in $[0;30]$. This happens with probability $p=\frac{1}{4}$

If the call arrives after 8:30 you miss all your TV show. This happens with probability $p=\frac{1}{4}$

Thus

$$E(M)=15\frac{1}{4}+30\frac{1}{4}=11.25$$

You can expect to miss about 11 minutes of your show

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