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I'm working on a problem and came up with two different solutions! I'm not sure which is correct.

Problem: Two clerks with service time $exp(1)$ are helping 2 customers while a third waits. What is the probability that the next customer will be the second to finish? (Leaving one of the two initial customers still being helped)

Solution 1: Exponentials are memoryless, so when the customer in line starts getting served -- the process renews. As such, the probability that one finishes before the other is 1/2.

Solution 2: Let $T_i$ be the time it takes for customer $i$ to be served.

By law of total probability: $P($ Person 3 leaves second $) = P(T_1 + T_3 < T_2 \mid T_1 < T_2) P(T_1<T_2) + P(T_2 + T_3 < T_1 \mid T_2 < T_1) P(T_2<T_1)$

This keeps evaluating to 1/4!

Any thoughts? Thanks

Update: $P(T_1<T_2) = \int_{0}^{\infty}\int_{0}^{T_1}e^{-(t_1+t_2)} dt_1 dt_2 = 1/2$ and $P(T_1 + T_3 <T_2 \mid T_1 < T_2) = \int_{0}^{\infty}\int_{0}^{T_2}e^{-t_2}ze^{-z} dz dt_2 = 1/4 $

Update: Ah -- I see what happened. I was computing the wrong probability: $P(T_1 + T_3 <T_2) = 1/4$ but $P(T_1 + T_3 <T_2 \mid T_1 < T_2) = P(T_1 + T_3 <T_2 \cap T_1 < T_2)/P(T_1 < T_2) = P(T_1 + T_3 <T_2)/P(T_1 < T_2) =(1/4)/(1/2) = 1/2$

The 2nd and 3rd equality is established by noting all $T_i$ are positive so $T_1 + T_3 <T_2$ implies $T_1 < T_2$

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The first solution is right. Without knowing how you evaluate the expression in the second solution to $\frac14$, it's hard to say what's going wrong, but it should evaluate as $\frac12\cdot\frac12+\frac12\cdot\frac12=\frac12$.

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  • $\begingroup$ Hi @joriki, see my work above $\endgroup$
    – yoshi
    Feb 22, 2016 at 15:42
  • $\begingroup$ @yoshi: You calculated $P(T_1 + T_3 <T_2)$, but you wanted $P(T_1 + T_3 < T_2 \mid T_1 < T_2)$. $\endgroup$
    – joriki
    Feb 22, 2016 at 15:44
  • $\begingroup$ @yoshi: By the way, the vertical line in conditional probabilities is best rendered as \mid instead of | -- that gives the right spacing. $\endgroup$
    – joriki
    Feb 22, 2016 at 15:45
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    $\begingroup$ thanks, I didnt know that -- fixed $\endgroup$
    – yoshi
    Feb 22, 2016 at 15:47
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    $\begingroup$ maybe im skipping too many steps, ill come back from work and add more details $\endgroup$
    – yoshi
    Feb 22, 2016 at 16:10

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