Question

The probability that a non leap year selected at random will have $53$ Sundays is

• A. $0$
• B. $1/7$
• C. $2/7$
• D. $3/7$

Answer 1

Answer: (B)

In a pon-leap year, total number of days is $365$. Out of them, there are $52$ weeks and $1$ day extra. Thus, a non-leap year always has $52$ Sunday. The remaining $1$ day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday.
Out of these $7$ cases, we have Sunday in one case.
$\therefore$ Total number of outcomes $=7$
Number of favourable outcomes $=1$
Hence, required probability $=\frac{1}{7}$