   Chapter 15.4, Problem 32E

Chapter
Section
Textbook Problem

Xavier and Yolanda both have classes that end at noon and they agree to meet every day after class. They arrive at the coffee shop independently. Xavier’s arrival time is X and Yolanda’s arrival time is Y, where X and Y are measured in minutes after noon. The individual density functions are f 1 ( x ) =   { e − x i f   x   ≥   0 0 i f   x   <   0     f 2   ( y )   =   { 1 50 y if   0   ≤   y   ≤   10 0 otherwise (Xavier arrives sometime after noon and is more likely to arrive promptly than late. Yolanda always arrives by 12:10 pm and is more likely to arrive late than promptly.) After Yolanda arrives, she’ll wait for up to half an hour for Xavier, but he won’t wait for her. Find the probability that they meet.

To determine

To find: The probability of Xavier and Yolanda meets.

Explanation

Given:

Yolanda came by 12:10 hours and she can wait upto half an hour for Xavier and Xavier doesn’t wait for him.

Let the arrival time of Xavier and Yolanda’s be X and Y, respectively.

The individual density function is denoted by,

f1(x)={ex, if x00   , if x<0f2(x)={150y, if 0y10   0  , otherwise

Calculation:

From the given conditions, it is observed that y varies from 0 to 10 and x varies from y to y+30 . The joint density function is,

f(x,y)=f1(x)f2(y)={150exy , if x0,0y10      0     , otherwise

Therefore, the probability that they both meet each other is,

010yy+30150exydxdy=150010y[ex]yy+30dy=150010y[e(y+30)ey]dy=150010y[eye(y+30)]dy

Apply the technique of integration by parts to integrate the function.

Let u=y .

Then, dv=[eye(y+30)]dy .

Thus, the probability that they both meet each other is obtained below

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