Question

The probability that a leap year selected at random contains either $53$ Sundays or $53$ Mondays, is

• A. $\frac{7}{13}$
• B. $\frac{2}{23}$
• C. $\frac{3}{7}$
• D. None of these

Answer 1

The correct option is C $\frac{3}{7}$

Total number of days in a leap year is $366$.

It will contain $52$ weeks and $2$ days.

These two days can be

$S=\left\{\right(Sun,Mon);(Mon,Tues);(Tues,Wed);(Wed,Thurs);(Thurs,Fri);(Fri,Sat);(Sat,Sun\left)\right\}$

Therefore $n\left(S\right)=7$

Let $A$ be the event of getting $53$ sundays.

Therefore $A=\left\{\right(Sun,Mon);(Sat,Sun\left)\right\}$

For $53$ Sundays , probability is $P\left(A\right)=\frac{2}{7}$

Let $B$ be the event of getting $53$ mondays.

Therefore $B=\left\{\right(Sun,Mon);(Mon,Tues\left)\right\}$

For $53$ Mondays , probability is $P\left(B\right)=\frac{2}{7}$

=27

This includes one ways where sunday and monday simultaneously Occur

(i.e) $A\cap B=\{Sun,Mon\}$

Probability for this is $P(A\cap B)=\frac{1}{7}$.

=17

Hence required probability that a leap year selected at random contain $53$ sundays or $53$ mondays is

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

$=\frac{2}{7}+\frac{2}{7}-\frac{1}{7}=\frac{2+2-1}{7}$

Therefore the required probability is $\frac{3}{7}$