Question

Two friends have decided to meet between $3$ and $4$ 'o clock and they have decided to wait for $10$ min to meet (for eg $A$ have reached by $3$ he will wait for $10$ min for $B$ similarly $B$ also). then what is the probability of both are meeting ?

Answer 1

The time period within which they can arrived is 1 hour. They have to be with 10 minutes of each other.

The probability will vary $\frac{10}{60}$ to $\frac{20}{60}$ and it will remain constant $3:50$ and then decrease to $\frac{10}{60}$.

The required probability will be,

= 1 - unfavorable condition

$=1-2\times [\left(\frac{20}{60}\right)\times \left(\frac{30}{60}\right)\times \left(\frac{40}{60}\right)\times \left(\frac{50}{60}\right)\times \left(\frac{50}{60}\right)]$

$=1-2\times \left[\frac{10}{108}\right]$

$=\frac{(108-20)}{108}$

$=\frac{88}{108}=\frac{11}{27}$