Question

probability of getting two sunday in a leap year??​

Answer 1

Answer:

There are 7 possibilities out of which 2 have a Sunday. So the probability of 53 Sundays is 2/7.

Answer 2

Answer:

$\frac{2}{7}$

Step-by-step explanation:

In a leap year there are 366 days.

So there are 52 weeks and 2 extra days as follows:

No. of days =366

so no.of weeks =

$\frac{366}{7} = 52 \frac{2}{7}$

so we will have all days 52 times. So we calculate probability for the rest 2 days and these two days will be consecutive days as shown...

Now, Sample Space,

S={(Sun,Mon),(Mon,Tue),(Tue,Wed),(Wed,Thu),(Thu,Fri),(Fri,Sat),(Sat,Sun)}

so n(S)=7

now event favouring Sunday,E={(Sun,Mon),(Sat,Sun)}

so n(E) = 2

and probability of E, P(E)=n(E)/n(S)=2/7

so P(E)

$= \frac{2}{7}$