What is the probability that a leap year selected at random will contain either 53 Thursday Friday?

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Text Solution

`(3)/(7)``(2)/(7)``(5)/(7)``(1)/(7)`

Answer : A

Solution : A leap year consists of 366 days, i.e., 52 full weeks and two extra days. These two extra days can be any one of the following possible outcomes, (i) Monday and Tuesday (ii) Tuesday and Wednesday (iii) Wednesday and Thursday (iv) Thursday and Friday (v) Friday and Saturday (vi) Saturday and Sunday (vii) Sunday and Monday. Let A and B be the events that a leap year contains 53 Thursday and 53 Friday, respectively. Then, `P(A)=(2)/(7),P(B)=(2)/(7)` and `P(A nn B)=(1)/(7)` `therefore` Required probability is given by `P(A uu B) = P(A)+P(B)-P(A nn B)` `=(2)/(7)+(2)/(7)-(1)/(7)=(3)/(7)`

The probability that a leap year selected at random will contain either 53 thursdays or 53 fridays is:

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