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Problem from statistics textbook:

Two telephone calls come into a switchboard at random times in a fixed one-hour period. Assume that the calls are made independently of one another. What is the probability that the calls are made

a. in the first half hour?

Answer: $\frac{1}{4}$

b. within five minutes of each other?

Answer: $\frac{23}{144}$

As it is given that they are independent, I was able to calculate part a by simply performing $\frac{1}{2} \times \frac{1}{2}$. However, I am struggling to derive an approach for the latter half of this question. Intuitively, I think of approaching it as an integral with some bounds constrained +/- 5 minutes, but I struggle to come up with values, as well as the joint probability function (which should, I believe, be uniform). Would greatly appreciate some assistance

Red Five
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Diddley4209
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  • Imagine plotting the two call times as $(x,y)$ coordinates in a square, and find the area of the region corresponding to the event in question. – Karl Apr 05 '24 at 19:00
  • I don't entirely follow -- Here's what I've (unsuccessfully) now tried: thinking of the whole area as a square, the denominator becomes 60x60 = 3600. I then thought of this region as the double integral between 0 and 60 (outer) and (x-5) and (x+5) (inner). – Diddley4209 Apr 05 '24 at 19:31
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    When you are imagining the problem as a square, then $x$-axis represents the time of the first call and $y$-axis represents that of the second call. Thus, you are interested in the region $|x -y| \leq 5$, not $x - 5$ and $x + 5$. This graph might be helpful to visualize the area. – sudeep5221 Apr 05 '24 at 20:32
  • Got it, I wasn't accounting for the triangular "tip region" of the inequality. Adding the area of those and dividing by 3600 gives the valid answer – Diddley4209 Apr 05 '24 at 20:42
  • The question does not seem to have enough information to create an answer. Sorry but I can't understand. –  Apr 05 '24 at 23:56
  • Related: https://math.stackexchange.com/questions/103015/chance-of-meeting-in-a-bar and the pages that link to it – Henry Apr 06 '24 at 13:06

1 Answers1

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Lets say first call is made at time $t$, next call is within $[t,t+5)$. Let first call be made at time $T_1$ and next call at time $T_2$.

$$P(|T_1-T_2|\leq 5) = P(T_2 < T_1\leq T_2 + 5) + P(T_1 < T_2\leq T_1 + 5) + P(T_1 = T_2)$$ $$P(T_1 < T_2\leq T_1 + 5) = P(T_2 < T_1\leq T_2 + 5)$$ $$P(|T_1-T_2|\leq 5) = 2 \times P(T_2 < T_1\leq T_2 + 5)$$ $$P(T_1 = T_2) = 0$$ $$P(T_2 < T_1\leq T_2 + 5) = \int_{0}^{55} P(T_2 < T_1 \leq T_2 + 5|T_2 = t ) f_{T_2}(t) dt + \int_{55}^{60} P(T_2 < T_1 \leq T_2 + 5|T_2 = t ) f_{T_2}(t) dt$$

$$f_{T_2}(t) = \frac{1}{60}$$

$$P(T_2 < T_1\leq T_2 + 5) = \int_{0}^{55} \frac{5}{60} \times \frac{1}{60} dt + \int_{55}^{60} \frac{t-55}{60} \times \frac{1}{60} dt$$

$$P(T_2 < T_1\leq T_2 + 5) = \frac{55 \times 5}{3600} + \frac{1}{2}\frac{60^2 - 55^2}{3600} - \frac{55 \times 5}{3600}$$

$$P(T_2 < T_1\leq T_2 + 5) = \frac{1}{2}\frac{60^2 - 55^2}{3600}$$

$$P(|T_1-T_2|\leq 5) = \frac{60^2 - 55^2}{3600} = \frac{575}{3600} = \frac{23}{144}$$