Find the probability that a leap year selected at random will have 53 sundays
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Hey mate..
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A leap year has 366 days or 52 weeks and 2 odd days. The two odd days can be {Sunday,Monday},{Monday,Tuesday},{Tuesday,Wednesday},{Wednesday,Thursday},{Thursday,Friday},{Friday,Saturday},{Saturday,Sunday}.
So there are 7 possibilities out of which 2 have a Sunday. So the probability of 53 Sundays is 2/7.
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A leap year has 366 days or 52 weeks and 2 odd days. The two odd days can be {Sunday,Monday},{Monday,Tuesday},{Tuesday,Wednesday},{Wednesday,Thursday},{Thursday,Friday},{Friday,Saturday},{Saturday,Sunday}.
So there are 7 possibilities out of which 2 have a Sunday. So the probability of 53 Sundays is 2/7.
rupesh51:
Thank you
Answered by
4
Answer:
In a leap year there are 366 days. In 366 days, we have 522 weeks and 2 days, Thus we can say that leap year ah always 52 Sundays.
The remaining two days can be
(i) Sunday and Mondays
(ii) Mondays and Tuesdays
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(VI) Friday and Saturday
(vii) Saturady 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.
∴ P (E) = n (E)/n (S) = 2/7
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