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Q.

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

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a

$\frac{1}{7}$

b

$\frac{2}{7}$

c

$\frac{3}{7}$

d

$\frac{4}{7}$

answer is C.

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Detailed Solution

$t o t a l 365 d a y s$

$52 w e e k s + 1 d a y$

$S u n, T u e, T h u$

$P (E) = \frac{3}{7}$

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