The probability that in a year of 22nd century chosen at random, there will be $53$ sundays is

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$\frac{3}{28}$

$\frac{2}{28}$

$\frac{7}{28}$

$\frac{5}{28}$

answer is D.

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Detailed Solution

In the 22nd century there are 25 leap years viz.

2100, 2104,…,2196 and 75 non- leap years.

Consider the following events :

$E_{1}$ = Selecting a leap year from 22nd century

$E_{2}$ = Selecting a non-leap year from 22nd century

A = There are 53 Sundays in a year of 22nd century

We have,

$P (E_{1}) = \frac{25}{100}, P (E_{2}) = \frac{75}{100}$

$P (A / E_{1}) = \frac{2}{7}$ and $P (A / E_{2}) = \frac{1}{7}$

Required probability $\begin{array}{l} = P (A) \\ = P ((A \cap E_{1}) \cup (A \cap E_{2})) \\ = P (A \cap E_{1}) + P (A \cap E_{2}) \\ = P (E_{1}) P (A / E_{1}) + P (E_{2}) P (A / E_{2}) \\ = \frac{25}{100} \times \frac{2}{7} + \frac{75}{100} \times \frac{1}{7} = \frac{5}{28} \end{array}$