In a non-leap year, the probability of having 53 Tuesdays or 53 Wednesdays is

Question

In a non-leap year, the probability of having 53 Tuesdays or 53 Wednesdays is (A) (1/7) (B) (2/7) (C) (3/7) (D) None of these

Tardigrade Admin · Accepted Answer

In a non-leap year, there are 365 days in which 52 weeks and one day extra. It may be Mon, Tue, Wed, Thu, Fri, Sat, Sun. ∴ Required probability = P(53 Tuesday) + P(53 Wednesday) - P(53 Tuesday ∩ 53 Wednesday) =(1/7)+(1/7)-0 = (2/7)

Q. In a non-leap year, the probability of having $53$ Tuesdays or $53$ Wednesdays is

$\frac{1}{7}$

$\frac{2}{7}$

$\frac{3}{7}$

None of these

In a non-leap year, there are $365$ days in which $52$ weeks and one day extra.
It may be Mon, Tue, Wed, Thu, Fri, Sat, Sun.
$∴$ Required probability
$= P (53$ Tuesday) $+ P (53$ Wednesday)
$- P (53$ Tuesday $\cap 53$ Wednesday)
$= \frac{1}{7} + \frac{1}{7} - 0$
$= \frac{2}{7}$

Q. In a non-leap year, the probability of having 53 Tuesdays or 53 Wednesdays is

71​

72​

73​