Question

Suppose that a man leaves for work between 8:00 a.m and 8:30 a.m and takes between 40 and 50 minutes to get to the oce. Let X denote the time of departure and let Y denote the time of travel. If we assume that these random variables are independent and uniformly distributed, find the probability that he arrives at the office before 9:00 a.m

Answer 1

This is task on geometric probability. We consider random vector Z =(X,Y) which has uniform distribution in rectangle .and interpret random event Q of arriving to work before 9.00 as subset of universe  with definition $Q=\{(x,y)\in [0;30]\times [40;50]:x+y\le 60\}$  . Then probability of Q will be $p(Q)=\frac{|Q|}{|U|}=\frac{|Q|}{300}$ . For finding measure  of favorable event Q we use geometrical considerations:

we see that favorable event Q trapezoid with base AC=10 and two heights AB=60-50=10 and CD=60-40=20. Thus area of Q is |Q|=$10\cdot \frac{10+20}{2}=150$ . Therefore the probability to find is $p(Q)=S(Q)/S(U)=\frac{150}{300}=0.5$

Thus answer is 0.5.