1.

The probability that a leap year will have 53 Fridays or 53 Saturdays is A.  \(\frac{2}{7}\)B. \(\frac{3}{7}\)C. \(\frac{4}{7}\) D. \(\frac{1}{7}\)

Answer»

In a leap year we have 366 days. So, every day of a week comes 52 times in 364 days. 

Now we have 2 days remaining. These 2 days can be

S = {MT, TW, WTh, ThF, FSa, SaSu, SuM} 

Where S is the sample space. 

As there are total 7 possibilities. 

∴ n(S) = 7 

Let A denote the event of getting 53 Fridays and B denote event of getting 53 Saturdays. 

We have to find P(A ∪ B) Clearly, 

P(A) = \(\frac{2}{7}\) 

P(B) =  \(\frac{2}{7}\) 

And P(A∩B) =  \(\frac{1}{7}\) 

∴ P(A∪B) =  \(\frac{2}{7}\)  +  \(\frac{2}{7}\)  –  \(\frac{1}{7}\) =  \(\frac{3}{7}\) 

As our answer matches only with option (b) 

∴ Option (b) is the only correct choice.


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