Question

In a non-leap year the probability of getting $53$ Sundays or $53$ Tuesdays or $53$ Thursday is.
A) $\dfrac{1}{7}$
B) $\dfrac{2}{7}$
C) $\dfrac{3}{7}$
D) $\dfrac{4}{7}$

Answer 1

In this problem we have to find the probability for a non-leap year getting the required attributes. By using the given in probability relation we will get the required result.
We know that, in a year $364$ days contain $52$ Sundays, Mondays, and Tuesdays and so on.
We have to find the probability of the remaining $1$ day.
Using those probability we can find the required answer.

Complete step-by-step answer:
We have to find that in a non-leap year the probability of getting $53$ Sundays or $53$ Tuesdays or $53$ Thursday.
We know that the number of days in a non-leap year is $365$ days.
Again, $1$ year $= 52$weeks $= 52 \times 7 = 364$ days
These $364$ days contain $52$ Sundays, Mondays, and Tuesdays and so on.
The remaining $1$ day will be left sample space for this.
1 day = {Mon, Tue, Wed, Thurs, Fri, Sat, Sun}
Probability of Sunday $= P[53$ Sundays$] = \dfrac{1}{7}$
Probability of Tuesday $= P[53$ Tuesdays $] = \dfrac{1}{7}$
Probability of Thursday $= P[53$ Thursdays$] = \dfrac{1}{7}$
Now, $P(53$ Sundays or $53$ Tuesdays or $53$ Thursday$)$
$\Rightarrow \dfrac{1}{7} + \dfrac{1}{7} + \dfrac{1}{7}$
Simplifying we get,
$\Rightarrow \dfrac{3}{7}$

$\therefore$ The probability of getting $53$ Sundays or $53$ Tuesdays or $53$ Thursday is $\dfrac{3}{7}$.

Note: Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e. how likely they are to happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.
The probability of all the events in a sample space adds up to 1.
The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.