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The probability that a leap year selected at random contains either $53$ Sundays or $53 $ Mondays, is
$\frac{2}{7}$
$\frac{4}{7}$
$\frac{3}{7}$
$\frac{1}{7}$
Solution
(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 $P(A) = \frac{2}{7},\,\,P(B) = \frac{2}{7},\,\,P(A \cap B) = \frac{1}{7}$
$\therefore $ 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}.$
