Question

What is the probability of having 53 wednesday in a non leap year (with steps)

Answer 1

As there are 52 complete weeks which contains 364 days .,(since it is non leap year)
so 52 wednesdays  will be there and  remaining 365 th day may be sunday,monday,tuesday ,wednesday,thursday,friday,saturday.
Total number of ways =7 
favourable ways =1 
therefore probability of 53 wednesdays =1/7 

Answer 2

TO FIND,

Probability of having 53 Wednesdays in a nonleap year.

SOLUTION,

As we know that there are 52 weeks in a year with 365 days - a nonleap year.

so as there are 52 weeks there will be 52 Wednesdays i.e. 364 days

the remaining one day can be - Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, or Sunday.

∴ there are 7 ways to layout the days, but we only need Wednesday

hence favorable event= 1

   entire no of events= 7

∴ Probability of 53 Wednesday =$\frac{favorable outcomes}{total outcomes}$

                                                    = $\frac{1}{7}$

HENCE THE PROBABILITY OF 53 WEDNESDAYS IN A NONLEAP YEAR  IS $\frac{1}{7}$.