Question

At a random moment within a period of 20 minutes, student 'A' telephones student ' $B$ ', waits for 2 minutes and then puts down the receiver. During the same 20 minutes student ' $B$ ' arrives home at a random moment, stays for 5 minutes and then leaves. Let $P$ denotes the probability that the two will have a talk. If $P=\frac{p}{q}$ as a reduced fraction, then find $(p+q)$.

Answer 1

Answer: 1051

where $0\le x\le 20$ \& $0\le y\le 20$.
Hence $y\le x+2\&y\ge x-5$
or
$y-x\le 2\&x-y\le 5$
$\therefore n(S)=20\times 20=400$
$n(A)=400-\left[\frac{324+225}{2}\right]$
$=400-\frac{549}{2}=\frac{251}{2}$
$\therefore p=\frac{251}{2}\frac{1}{400}=\frac{251}{800}\Rightarrow p+q=1051$