The probability that a leap year will have 53 Fridays or 53 Satur… - Vidyadip

MCQ

The probability that a leap year will have $53$ Fridays or $53$ Saturdays is:

A
$\frac{2}{7}$
✓
$\frac{3}{7}$
C
$\frac{4}{7}$
D
$\frac{1}{7}$

✓

Answer

Correct option: B.

$\frac{3}{7}$

We know that a leap year has $366$ days $($i.e. $7 \times 52 + 2) = 52$ weeks and $2$ extra days .
The sample space for these $2$ extra days is given below:
$S = \{($Sunday, Monday$), ($Monday, Tuesday$), ($Tuesday, Wednesday$), ($Wednesday, Thursday$), ($Thursday, Friday$), ($Friday, Saturday$), ($Saturday, Sunday$)\}$ There are $7$ cases.
$\therefore\text{n(S)}=7$
Let $E$ be the event that the leap year has $53$ Fridays or $53$ Saturdays.
$E = \{($Thursday, Friday$), ($Friday, Saturday$), ($Saturday, Sunday$)\}$
i.e. $n(E) = 3$
$\therefore \text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}=\frac{3}{7}$
Hence, the probability that a leap year has $53$ Fridays or $53$ Saturdays is $\frac{3}{7}$.

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