Question

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

• A. $\frac{1}{7}$
• B. $\frac{2}{7}$
• C. $\frac{3}{7}$
• D. $\frac{5}{7}$

Answer 1

The correct option is A $\frac{1}{7}$

As we know, a non-leap year has $365$ days

$365$ days $=(52\times 7+1)$ days

$\Rightarrow 365$ days $=52$ weeks $+1$ day

$(\because 7\text{days}=1\text{week})$

For $52$ weeks, number of Sundays $=52$

$1$ remaining day can be $Sunday,$ $Monday,Tuesday,Wednesday,$ $Thursday,Friday,Saturday,$ $Sunday.$

$\therefore$ Total number of possible outcomes $=7$

Number of favorable outcomes $=1$

Thus, the probability of getting $53$ Sundays

$=\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$

$=1/7$

Therefore, option (a.) is the correct choice.