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

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none of these

$\frac{1}{7}$

$\frac{3}{7}$

$\frac{2}{7}$

answer is B.

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Detailed Solution

Given a non-leap year

$∴$ A non leap year has 365 days

Which includes 52 whole weeks

$∴$ 52 X 7 = 364

Remaining is 1 day which could be either of the day out of 7 days.

$∴$ Probability of having 53 Tuesdays or 53 2 Wednesday $= \frac{2}{7}$

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