A person has to travel from A to D changing buses at stops B and C en-route. The maximum waiting time at either stop can be 8 minutes each, but any time of waiting up to 8 minutes is equally likely at both places. He can afford up to 13 minutes of total waiting time, if he is to arrive at D on time. What is the probability that the person will arrive late at D?
The options given are: a) 8/13 b) 13/64 c) 119/128 d) 9/128
My logic:
Since the person can wait for max 8 minutes at each stop so he/she will be late if he/she has to wait for either 14, 15 or 16 minutes.
The number of ways this can happen is:
Person waits for 8 minutes at B: Probability for this to occur is 1/8 (since it says that any time of waiting upto 8 minutes is equally likely at both places).
So in order to get late he/she will have to wait for either 6 or 7 or 8 minutes at C. The probability for which will be 3/8.
Person waits for 7 minutes at B: Probability will be 1/8 So probability at C will be 2/8 (either waits for 7 or 8 min)
Person waits for 6 minutes at B: Probability will be 1/8. Probability at C will be 1/8 (can wait for only 8 min in order to get late).
So total probability is: (1/8)(3/8)+(1/8)(2/8)+(1/8)*(1/8) = 6/64.
Now we can consider this same scenario when he/she waits for 8 min at C.
So complete probability is 2*(6/64) = 12/64 = 3/16
And that is nowhere in the options!
Please tell me where am I going wrong.
NOTE: Here I have assumed that he he/she has to wait for atleast 1 min...if I dont assume that and also take into consideration that he/she might not have to wait at all then instead of 1/8 we will have 1/9. But that also is not giving me the correct solution.
You are expected to assume that the waiting time has continuous uniform distribution in the interval $[0,8]$.
Let $X$ be the waiting time at stop B, and $Y$ the waiting time at stop C. We want $\Pr(X+Y)\gt 13$.
Assuming that $X$ and $Y$ are independent, the pair $(X,Y)$ has uniform density in the square with corners $(0,0)$, $(8,0)$, $(8,8)$, $(0,8)$. Draw that square.
Now draw the line $x+y=13$. We want the probability that we land above that line. We want the probability of landing in the little triangle $T$ at the top right of the square.
It turns out that the little triangle $T$ is right-angled, with legs of length $3$. For the line $x+y=13$ meets the line $x=8$ at $(8,5)$.
So $T$ has area $\frac{9}{2}$. The whole square has area $64$, so the required probability is $(9/2)/64$, that is, $\frac{9}{128}$.
Remark: Alternately, you can integrate the joint density function $\frac{1}{64}$ (in the square) over the triangle $T$. Or else integrate the joint density over the part of the square that is not in $T$. That will tell you the probability you are not late.
If our joint density is not constant, we need to integrate. However, in this example, the geometry is enough to give the answer.