The probability that a non-leap year has $53$ sundays, is.

1 / 7

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2 / 7

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3 / 7

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4 / 7

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Solution

The correct option is B $1 / 7$
A non-leap year contains $365$ days. ie, $52$ weeks and one day. This one day can be any one of seven days of the week.
$∴$ Required probability $= \frac{1}{7}$

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