The probability that a non-leap year selected | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) non-leap year has $365$ days,which is $52$ weeks and $1$ extra day. The extra day can be any one of the $7$ days of the week: {Monday,Tuesday,Wedn

Vedclass · Accepted Answer

$\frac{2}{7}$

Vedclass · Answer

(C) non-leap year has $365$ days,which is $52$ weeks and $1$ extra day. The extra day can be any one of the $7$ days of the week: {Monday,Tuesday,Wednesday,Thursday,Friday,Saturday,Sunday}. Let $A$ be the event that the year has $53$ Saturdays and $B$ be the event that the year has $53$ Sundays. Since the year has $52$ weeks,it already contains $52$ Saturdays and $52$ Sundays. For the year to have $53$ Saturdays,the extra day must be a Saturday. $P(A) = \frac{1}{7}$. For the year to have $53$ Sundays,the extra day must be a Sunday. $P(B) = \frac{1}{7}$. Since the extra day cannot be both a Saturday and a Sunday simultaneously,$A$ and $B$ are mutually exclusive events. The probability that the year has $53$ Saturdays or $53$ Sundays is $P(A \cup B) = P(A) + P(B) = \frac{1}{7} + \frac{1}{7} = \frac{2}{7}$. Note: The question asks for $52$ Saturdays $OR$ $53$ Sundays. Since every non-leap year has $52$ Saturdays,the event 'contains $52$ Saturdays' is a certain event with probability $1$. However,interpreting the standard problem format for this specific question type,it asks for the probability of having $53$ Saturdays or $53$ Sundays. Given the options,the intended answer is $\frac{2}{7}$.

The probability that a non-leap year selected at random will contain $52$ Saturdays or $53$ Sundays is

Similar Questions

Suppose $A$ and $B$ are two events such that $P(A \cap B) = \frac{3}{25}$ and $P(B - A) = \frac{8}{25}$. Then, $P(B)$ is equal to

Suppose $3$ bulbs are selected at random from a lot. Each bulb is tested and classified as defective $(D)$ or non-defective $(N)$. Write the sample space of this experiment.

$A$ problem in statistics is given to three students $A, B$ and $C$. Their probabilities of solving the problem are $\frac{1}{2}, \frac{1}{3}$ and $\frac{1}{4}$ respectively. If all of them try independently,then the probability that the problem is solved is

The vertices of a regular tetrahedron are marked $1, 2, 3, 4$. If three such tetrahedra are thrown simultaneously,what is the probability that the sum of the numbers on the faces is $5$?

Three coins are tossed once. Find the probability of getting no head.

Vedclass Test Series

Exam Paper Generator

Online Exam Module