In a non-leap year, the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays is

Question

In a non-leap year, the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays is (A) (1/7) (B) (2/7) (C) (3/7) (D) (4/7)

Tardigrade Admin · Accepted Answer

Days in a non leap year = 365 So there are 52 weeks and one day. This one day may be anyone in sample space [Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday] So, P (53 Sundays or 53 Tuesdays or 53 Thursdays) =(1/7)+(1/7)+(1/7) = (3/7).

Q. In a non-leap year, the probability of getting $53$ Sundays or $53$ Tuesdays or $53$ Thursdays is

$\frac{1}{7}$

$\frac{2}{7}$

$\frac{3}{7}$

$\frac{4}{7}$

Days in a non leap year $= 365$
So there are $52$ weeks and one day.
This one day may be anyone in sample space [Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday]
So, $P$ ( $53$ Sundays or $53$ Tuesdays or $53$ Thursdays)
$= \frac{1}{7} + \frac{1}{7} + \frac{1}{7} = \frac{3}{7}$ .

Q. In a non-leap year, the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays is

71​

72​

73​

74​