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

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

1. A

$\frac{3}{28}$

2. B

$\frac{2}{28}$

3. C

$\frac{7}{28}$

4. D

$\frac{5}{28}$

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### 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,

$P\left(A/{E}_{1}\right)=\frac{2}{7}$ and $P\left(A/{E}_{2}\right)=\frac{1}{7}$

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

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