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

Options

A. `(3)/(28)`
B. `(2)/(28)`
C. `(7)/(28)`
D. `(5)/(28)`

Answer

D

Explanation

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 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))=(25)/(100), P(E_(2))=(75)/(100)`
`P(A//E_(1))=(2)/(7) " and " P(A//E_(2))=(1)/(7)`
Required probability =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))`
`=(25)/(100)xx(2)/(7)+(75)/(100)xx(1)/(7)=(5)/(28)`