Question
Find the probability that leap year selected at random, will contain 53 Sundays.
✓
Answer
In a leap there are 366 days. In 366 days, we have 52 weeks and 2 days, Thus we can say that leap year has always 52 sundays.
The remaining two days can be
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
From above it is clear that there are 7 elementary events associated with this random experiment.
Clearly, the event A will happen if the last two days of the leap year are either Sunday and Monday or Saturday and Sunday.
n(E) = 2
P(E) = 2
$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{2}{7}$.
The remaining two days can be
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
From above it is clear that there are 7 elementary events associated with this random experiment.
Clearly, the event A will happen if the last two days of the leap year are either Sunday and Monday or Saturday and Sunday.
n(E) = 2
P(E) = 2
$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{2}{7}$.
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