- Home
- Standard 11
- Mathematics
14.Probability
medium
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
$\frac{1}{7}$
Solution
(b) There are $366$ days in a leap year, in which $52$ weeks and two days, The combination of $2$ days –
Sunday -Monday, Monday -Tuesday, Tuesday -Wednesday, Wednesday -Thursday, Thursday -Friday, Friday -Saturday, Saturday -Sunday
$P(53$ Fridays) = $\frac{2}{7}$; $P(53$ Saturdays) $ = \frac{2}{7}$
$P(53$ Fridays and $53$ Saturdays) $ = \frac{1}{7}$
$\therefore$ $P(53$ Fridays or Saturdays) = $P(53$ Fridays$) + P(53$ Saturdays) $-P(53$ Fridays and Saturdays)
$ = \frac{2}{7} + \frac{2}{7} – \frac{1}{7}$ $ = \frac{3}{7}$.
Standard 11
Mathematics
