Question

What will be the probability that an ordinary year has 53 sundays?

Answer 1

The answer will be $= \frac{1}{7}$

To find the answer for the sum follow the steps :-

An year actually has = 365 days

So let us check = 365 days =

$52 \times 7 = 364$

and for this 1 day is left.

This one day can be Monday,tuesday,wednesday,thursday,friday,saturday

so if it is sunday the probability of getting 53 sundays will be

$= \frac{1}{7}$

Answer 2

In an ordinary year, there are 52 weeks and 1 day.

$(52 \times 7) = 364 \\ 364 + 1 = 365$

( since there are 365 days in an ordinary year. )

Now,

For the one day, there are these outcomes -

( i.e., the one day can be one of these- )

• sunday
• monday
• tuesday
• wednesday
• thursday
• friday
• saturday

but the favourable outcome is sunday.

Hence the probability will be -

$\frac{1}{7}$