Question

Two persons A and B agree to meet at a place between 11 am to 12 noon. The first one to arrive waits for 20 min and then leave. If the time of their arrival be independent and at random, then the probability that A and B meet is

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{4}{9}$
• D. $\frac{5}{9}$

Answer 1

The correct option is D $\frac{5}{9}$

Let A and B arrive at the place of their meeting x minutes and y minutes after 11 noon.

The given condition $\Rightarrow$ their meeting is possible only if
$|x–y|\le 20$
OABC is a square, where $A\equiv (60,0)$ and $C\equiv (0,60)$
Considering the equality part of (i)
ie, |x – y| = 20
$\therefore$ The area representing the favourable cases
= Area OPQBRSO
= Area of square $OABC-Areaof\Delta PAQ$ - Area of $\Delta SRC$
$=\left(60\right)\left(60\right)-\frac{1}{2}\left(40\right)\left(40\right)-\frac{1}{2}\left(40\right)\left(40\right)$
= 3600 - 1600
= 2, 000 sq unit
Total way = Area of square OABC
= (60) (60) = 3,600 sq unit
$\therefore$ Required probability = $\frac{2,000}{3,600}=\frac{5}{9}$