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Q.

The probability that a non leap year selected at random will contain 53 Sundays is

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a

$\frac{5}{7}$

b

$\frac{1}{7}$

c

$\frac{3}{7}$

d

$\frac{2}{7}$

answer is A.

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Detailed Solution

Dividing 365 by 7 we get remainder 1 which can be anyone of 7 day of week

$P (E) = \frac{1}{7}$

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