what is the probability that a leap year, selected at random, will have 33 Sunday's ?
Explanation ke sath btao anybody
Answers
Answer:
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.
Answer:
In a leap year there are 366 days .
We have,
366 days = 52 weeks and 2 days
Thus, a leap year has always 52 Sundays.
The remaining 2 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.
Clearly, there are seven elemetary events associated with this random experiment,
Let A be the event that a leap year has 53 Sundays.
Clearly, the event A will happen if the last two days of the leap year are either Sunday and Monday or Saturday and Sunday.
∴ Favourable number of elementary events = 2
Hence, required probability = 2/7