what is the probability that a non leap year selected at random will contain 53 sundays
Answers
Answered by
23
Heya,
According to the question, we should take non leap year.
No. of days in non leap year = 365 days
In a year the no. of weeks = 365/7
= 52×1/7
So, now there will be 52 week in a year. So, there will be surely 52 Sundays in a year.
1 day will be left.
1 day can be Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Total outcome = 7
Favorable outcome = 1
Therefore,
Probability of getting 53 Sundays = 1/7. ..... Answer
Hope this helps....:)
According to the question, we should take non leap year.
No. of days in non leap year = 365 days
In a year the no. of weeks = 365/7
= 52×1/7
So, now there will be 52 week in a year. So, there will be surely 52 Sundays in a year.
1 day will be left.
1 day can be Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday.
Total outcome = 7
Favorable outcome = 1
Therefore,
Probability of getting 53 Sundays = 1/7. ..... Answer
Hope this helps....:)
Answered by
2
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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