Question

In a non-leap year, the probability of having $53$ Fridays or Saturdays is:

• A. $\frac{3}{7}$
• B. $\frac{4}{7}$
• C. $\frac{2}{7}$
• D. $\frac{1}{7}$

Answer 1

The correct option is C

$\frac{2}{7}$

Explanation for the correct option:

Finding the probability:

In a non-leap year, there are $52$ weeks and an extra day, which can be any of the seven days.

Let, $A$ be the event that the extra day is Friday, so $P\left(A\right)=\frac{1}{7}$

Let, $B$ be the event that the extra day is Saturday, so $P\left(B\right)=\frac{1}{7}$

The probability that the extra day is both Friday and Saturday is null, so $P(A\cap B)=0$

Therefore, the probability that the extra day is Friday or Saturday is

$\begin{array}{rcl}P\left(A\cup B\right)& =& P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)\\ & =& \frac{1}{7}+\frac{1}{7}-0\\ & =& \frac{2}{7}\end{array}$

Hence, option C is correct.