Probability that a passenger will catch the bus

Question

A bus and a passenger arrive at a bus stop at a uniformly distributed time over the interval $0$ to $1$ hour. Assume the arrival times of the bus and passenger are independent of one another and that the passenger will wait up to $15$ minutes for the bus to arrive. What is the probability that the passenger will catch the bus?

Hint: Let $Y1$ denote the bus arrival time and $Y2$ the passenger arrival time. Determine the joint density of $Y1$ and $Y2$ and determine $$P(Y2 \leq Y1 \leq Y2+\frac{1}{4}).$$

We already know that the answer to this question is $\frac{7}{32}$. Any help would be much appreciated.

Answer 1

Draw an isosceles triangle with base 1, and height 1. Make a congruent triange with the same base pointed down.

The area of this figure equals figure equals 1.

For any two arrival times of man and bus, mark the points on the joined base. Now draw a similar triangle that is point up if the man beats the bus, and point down if the bus beats the man to the stop.

Now every pair of arrival times has been mapped to a point in this figure. And every point in the figure maps to a pair of arrival times. The shaded region mark the times the man arrives before the bus, and within 15 minutes.

Area of a tapeziod $\frac 12 (b_1 + b_2)h$

$(\frac 12)(1+\frac 34)(\frac 14) = \frac {7}{32}$

Answer 2

So basically you will draw a unit square on $[0,1]\times[0,1]$. This is your joint density. The x axis corresponds to $y_1$ and the $y$ axis to $y_2$. Then $Y_2<Y_1$ corresponds to the region where you are below the $y_2=y_1$ line inside $[0,1]\times[0,1]$. The area of this region is of course $1/2$. Then you will draw another line, which corresponds to $y_2=y_1-1/4$. you are interested in another region which is above this line and inside $[0,1]\times[0,1].$

Finally the desired region is the intersection of both regions. In order to find this area, you can first find the area below $y_2=y_1-1/4$ and inside $[0,1]\times[0,1]$. This is $$\frac{3}{4}\frac{3}{4}\frac{1}{2}=\frac{9}{32}.$$ The area over this line and intersected by the area below $y_2=y_1$ is then $$\frac{1}{2}-\frac{9}{32}=\frac{7}{32}.$$

Answer 3

There is a $\frac{3}{4}$ chance the bus will be later than 10:15. Therefore there is a $\frac{1}{4}$ of an hour window of the person $Y2$, arriving on time. If $Y1 < \frac{1}{4}$ and $Y2 > \frac{1}{4}$ the probability of the person missing the bus is certainly $1$.

If $Y1 (\frac{1}{4})$ AND $Y2 (\frac{1}{4})$ arrives before 10:15, then $\frac{1}{4}+\frac{1}{4} = \frac{1}{2}$ chance of catching the bus.

So we have: $$\frac{3}{4} \cdot \frac{1}{4} + (\frac{1}{4}\cdot \frac{1}{4} \cdot \frac{1}{2}) = \dfrac{3}{16} + \dfrac{1}{32} = \dfrac{7}{32} $$