What is the probability that a leap year will have 53 Fridays or 53 Sa

1.	What is the probability that a leap year will have 53 Fridays or 53 Saturday ?
Answer» In a leap year we have 366 days. So everyday 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) = \(\frac{2}{7}\) + \(\frac{2}{7}\) – \(\frac{1}{7}\) = \(\frac{3}{7}\)

What is the probability that a leap year will have 53 Fridays or 53 Saturday ?

Answer»

In a leap year we have 366 days. So everyday 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) = \(\frac{2}{7}\) + \(\frac{2}{7}\) – \(\frac{1}{7}\) = \(\frac{3}{7}\)

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