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Got it~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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closed as unclear what you're asking by Davide Giraudo, MathOverview, Shailesh, C. Falcon, carmichael561 Apr 21 '17 at 1:02

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Let's divide the problem up into several possible outcomes:

  • $X_1$: A arrives between 9am and 10am (B at any time).
  • $X_2$: A arrives between 10am and 11am and B arrives between 11am and 12am.
  • $X_3$: Both A and B arrive between 10am and 11am.

These three are disjoint and together they give all possible results. So let now $R$ be the result that $A$ arrives before $B$. Then we have $$P(R) = P(R | X_1) + P(R | X_2) + P(R | X_3).$$

Are you ok with it so far? Now let's look at the $X_i$: The chance of $X_1$ should better be $1/2$, if we are assuming the time to be completely random. The chance of $X_2$ happening is $1/4$, as we have $1/2$ chance that $A$ arrives in this time slot and the same for $B$. As in both $X_1$ and $X_2$, we always have $A$ arriving before $B$, we can say $$P(R) = 1/2 + 1/4 + P(R | X_3).$$

As the first two terms already sum up to $0.75$, for your result to be true you would need $P(R | X_3) = 0$. But it is well possible that $A$ arrives before $B$ and they still both arrive between 10am and 11am...

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  • $\begingroup$ I see, and P(R|X3) is 1/2*1/2*1/2, so P(R) is 1/2+1/4+1/8 = 7/8??? $\endgroup$ – sarah Apr 19 '17 at 15:24
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The pair of arrival times is a random point in the square $Q:=[9,11]\times[10,12]$ with probability measure ${\rm d}P={1\over4}{\rm d}({\rm area})$. The line $x=y$ cuts off a small triangle at the lower right of $Q$. If the random point is above this line truck $A$ is first, and if the random point is below this line truck $B$ is first. It follows that the probability in question is ${7\over8}$.

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  • $\begingroup$ This solution is really straightforward. $\endgroup$ – sarah Apr 19 '17 at 15:37

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