What is the probability that a leap year will have 53 Sundays?

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$\frac{1}{7}$

$\frac{2}{7}$

$\frac{12}{52}$

$\frac{1}{52}$

answer is A.

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Detailed Solution

Consider the number of complete weeks in a leap year and the scenarios in which a leap year could have 53 Sundays.

The probability of an event E occurring is the ratio of the number of cases in its favour to the total number of equally likely cases.

$P (E) = \frac{n (E)}{n (S)} = \frac{no. of cases favourable to event E}{Total no. of cases}$

A leap year has 366 days $= 52 \times 7 + 2$ days

That means there will be 52 Sunday/Monday/Saturdays plus 2 additional days. Now these 2 additional days would be combination of any of the successive two days.

i)Sunday + Monday

ii)Monday + Tuesday

iii)Tuesday + Wednesday

iv) Wednesday + Thursday

v)Thursday + Friday

vi)Friday + Saturday

vii)Saturday + Sunday

Now the probability of two Sundays will be (i) & (vii) =2 outcomes out of above 7 outcomes. So the probability of having two additional Sundays =2 / 7 .

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