Question

A and B friends. They decide to meet between 1 PM and 2 PM on a given day. There is a condition that whoever arrives first will not wait for the other for more than 15 minutes. The probability that they will meet on that day is

• A. $\frac{1}{4}$
• B. $\frac{1}{16}$
• C. $\frac{7}{16}$
• D. $\frac{9}{16}$

Answer 1

The correct option is C $\frac{7}{16}$

Method I:
Case I: If A come earlier than B then

P(He will not meet with B) $=\frac{45}{60}$

Case II: If B come earlier than A. then

P (He will not meet with A) $=\frac{45}{60}$

So, P (when they wil not meet)

$=\frac{45}{60}\times \frac{45}{60}=\frac{9}{16}$

So, P (when they meet between 1 and 2 p.m.)

= 1 - P (they will not meet)

$=1-\frac{9}{16}=\frac{7}{16}$

Method II:


The shaded graph shows the chances that A and B will meet

$\therefore$ The probability that they will meet

$=\frac{AreaAOBCDEA}{AreaOXDYO}=\frac{1-\frac{1}{2}\times \frac{3}{4}\times \frac{3}{4}\times 2}{1}=1-\frac{9}{16}=\frac{7}{16}$