Question

What is the probability that there are 53 Wednesdays in a leap year?


Answer with complete steps & explanations.

Answer 1

Hello @govind 

Ohk Let's see how many days are there in a Leap Year : 365 + 1
 = 366 Days

And A Leap year has total 52 Weeks and 2 Days..

So that remaining two days can be : 

(i) Sunday & Monday  

(ii) Monday & Tuesday 

(iii) Tuesday & Wednesday 

(iv) Wednesday & Thursday 

(v) Thursday & Friday 

(vi) Friday & Saturday

(vii) Saturday & Sunday

So here we can clearly see that there is 7 possibilities but no. of Favourable outcome is only 2.. { See (iii) and (iv) }

So by Using Formula,

P(E) = $\frac{No. Of .Favourable. Outcomes}{Total. No. Of. Outcomes}$

= $\frac{2}{7}$

Hence the probability is $\frac{2}{7}$

or 0.285

 

Answer 2

Hello! dear,

Your answer goes like this ...

Let Us consider that One year =365 days.
But from the above Question,The leap year Can have "366" days.
Normally a year will have 52 weeks.So,We can conclude that there will be 52 Wednesdays.

So,By the problem,
⇒52 weeks=52x7
⇒366-364=2 days

Actually,In a leap year there will be automatically 52 Wednesdays and 2 days will be left.

So,The remained "2" days may be from the above:-

"Sunday {or} Monday"
"Monday {or} Tuesday"
"Tuesday{or} Wednesday"
"Wednesday{or}Thursday"
"Thursday {or} Friday"
"Friday {or} Saturday"
"Saturday {or} Sunday".

From the above 7 outcomes,Our favourable outcomes are "2".

Hence,The Probability of the 53 wednesdays In a leap year  =2/7.

So,The answer for your Question is "2/7".


Thank you dear,

Hope It helps.