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Alice and Bob each arrive at a party at a random time between 1:00 and 2:00. If Alice arrives after Bob, what is the probability that Bob arrived before 1:30?

My thoughts:

I thought that I could create a graph to find the probability of the event occurring, but I'm not exactly sure what to do. I've seen many different variations of this problem, but that only involves Alice or Bob waiting a certain amount of time and then leaving. This, I believe, is a completely different problem. Help is greatly appreciated.

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3 Answers 3

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Sketch a Cartesian coordinate plane. Label the horizontal axis $A$, which represents the number of hours after 1:00 that Alice arrived, and label the vertical axis $B$, which represents the number of hours after 1:00 that Bob arrived. Then the region in the first quadrant for which Alice arrives after Bob is represented by a triangle with vertices $(0,0)$, $(1,0)$, and $(1,1)$, since in the interior of this triangular region, $A > B$. Now, the desired event corresponds to $0 < B < 0.5$, since if Bob arrives before 1:30, he arrived between $0$ and $0.5$ hours after 1:00.

So, within this triangular region, what proportion of the area in this region satisfies $0 < B < 0.5$?

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  • $\begingroup$ Is the answer 1/8? $\endgroup$
    – A Piercing Arrow
    May 12, 2018 at 18:55
  • $\begingroup$ @APiercingArrow A substantial part of becoming proficient at mathematics is learning how to be complete with your reasoning, and that also includes being complete in your communication of such reasoning. Just asking "is the answer $x$?" and expecting others to confirm your result is neither useful nor meaningful for identifying where your gaps in knowledge or reasoning lie, therefore the need for you to explain what you are doing and how you got that result--if you are sincere about wanting help. $\endgroup$
    – heropup
    May 12, 2018 at 19:18
  • $\begingroup$ Thank you. I thought that the answer was 1/8. I drew a 1/2 by 1/2 right triangle. The area of that is 1/8. $\endgroup$
    – A Piercing Arrow
    May 12, 2018 at 19:21
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Let $X$ be the event {Alice arrives after Bob} and $Y$ be the event {Bob arrives before $1:30$}.

Refer to the diagram below:

$\hspace{5cm}$enter image description here

"Alice arrives after Bob" implies that the arriving time for Alice is greater than the arriving time for Bob and it corresponds to the points in the total shaded area (both red and green). So, $P(X)=\frac{1}{2}$.

"Bob arrives before 1:30" implies that the arriving time for Bob is between $1$ and $1:30$ (or $1.5$ in units) and it corresponds to the point in the total shaded area (both red and blue). So, $P(Y)=\frac{1}{2}$.

"Alice arrives after Bob and Bob arrives before $1:30$" correspond to the points in the red area. So, $P(X\cap Y)=\frac{3}{8}$.

"Bob arrived before 1:30 given that Alice arrives after Bob" is what you need to find. So: $$P(Y|X)=\frac{P(Y\cap X)}{P(X)}=\frac{\frac{3}{8}}{\frac{1}{2}}=\frac{3}{4}.$$

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The probability that Bob arrived before 1:30 is 1/4.

You use geometric probability and create a graph. You draw a square starting from the origin, with the x coordinate representing when Bob gets there and the y coordinate representing when Alice gets there. First, using the first condition that Bob arrived before Alice, you can erase half of the square. Erase the right triangle that includes the origin and the opposite corner, and the x axis, because this represents Alice arriving after Bob. Then you need to satisfy the second condition. Erase everything remaining on the right half of the graph. This leaves you with a quadrilateral that is 3/8 of the original square. Because you already know that Alice must have came after Bob, you subtract half from the total to get 1/4.

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  • $\begingroup$ No. The desired probability is conditional, not joint. $\endgroup$
    – heropup
    May 12, 2018 at 19:06
  • $\begingroup$ What do you mean $\endgroup$
    – PotatoLatte
    May 12, 2018 at 19:07
  • $\begingroup$ The question states: "If Alice arrives after Bob, what is the probability that Bob arrived before 1:30?" That means we must compute $$\Pr[B < 0.5 \mid A > B],$$ and not $$\Pr[(B < 0.5) \cap (A > B)].$$ $\endgroup$
    – heropup
    May 12, 2018 at 19:08
  • $\begingroup$ The graph represents the only way both of those conditions can be true. $\endgroup$
    – PotatoLatte
    May 12, 2018 at 19:11
  • $\begingroup$ Oh I get what you mean $\endgroup$
    – PotatoLatte
    May 12, 2018 at 19:11

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