What is the probability that there are 53 Wednesdays in a leap year?

# What is the probability that there are 53 Wednesdays in a leap year?

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

Fill Out the Form for Expert Academic Guidance!l

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

A leap year is given.
We know that the probability is given as the ratio of the number of favorable outcomes with the total number of possible outcomes.
$P\left(E\right)=\frac{\mathit{Number of favourable outcomes n}\left(E\right)}{\mathit{Total possible outcomes n}\left(S\right)}$
We know that a leap year has 52 weeks and 2 extra days.
Therefore 52 Wednesdays are present in 52 weeks.
Total outcomes for the 2 extra days are:   The total possible outcome,  Let E be the event that represents there are 53 Wednesdays in the year.
A number of favourable outcomes for 53rd Wednesday are: $E=\left\{\left(\mathit{Wed},\mathit{Thu}\right),\left(\mathit{Tue},\mathit{Wed}\right)\right\}$  The number of elements n(E) that are contained in E, $n\left(E\right)=2$ The probability of the event E is,
$P\left(E\right)=\frac{\mathit{Number of favourable outcomes n}\left(E\right)}{\mathit{Total number of outcomes n}\left(S\right)}$
$⇒P\left(E\right)=\frac{2}{7}$
Hence, the probability that there are 53 Wednesdays in the year are $\frac{2}{7}$.
Therefore, option 1 is correct.

## Related content

 Matrices and Determinants_mathematics Critical Points Solved Examples Type of relations_mathematics

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)