The probability that a leap year selected at | vedclass

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Question

(c) A leap year consists of $366$ days comprising of $52$ weeks and $2$ days.

There are $7$ possibilities for these $2$ extra days viz.

$(i)$ Su

Vedclass · Accepted Answer

$\frac{3}{7}$

Vedclass · Answer

(c) A leap year consists of $366$ days comprising of $52$ weeks and $2$ days.

There are $7$ possibilities for these $2$ extra days viz.

$(i)$ Sunday, Monday,                 $(ii)$ Monday, Tuesday,

$(iii)$ Tuesday, Wednesday,        $(iv)$ Wednesday, Thursday,

$(v)$ Thursday, Friday,                $(vi)$ Friday, Saturday and

$(vii)$ Saturday, Sunday.

Let us consider two events :

$A:$ the leap year contains $53$ Sundays

$B:$ the leap year contains $53$ Mondays.

Then we have $P(A) = \frac{2}{7},\,\,P(B) = \frac{2}{7},\,\,P(A \cap B) = \frac{1}{7}$

$	herefore $ Required probability $ = P(A \cup B)$

$ = P(A) + P(B) - P(A \cap B) = \frac{2}{7} + \frac{2}{7} - \frac{1}{7} = \frac{3}{7}.$

The probability that a leap year selected at random contains either $53$ Sundays or $53 $ Mondays, is

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