find the probability that a leap year selected at random will have 53 Sunday for 5 mark
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Answered by
36
No. of days in a leap year = 366
No. of weeks = 366/7 = 52 weeks, 2 days
Since, a week contains all days, 52 weeks mean 52 days of week
Since, the remaining 2 days could be any day,
S ={Su Mo, Mo Tu, Tu We, We Th, Th Fr, Fr Sa, Sa Su}(All possible days written in short-form)
n(S) = 7
n(E) = 2
P(E) = n(E)/n(S) = 2/7
Hope that you understand this (^_^)'
No. of weeks = 366/7 = 52 weeks, 2 days
Since, a week contains all days, 52 weeks mean 52 days of week
Since, the remaining 2 days could be any day,
S ={Su Mo, Mo Tu, Tu We, We Th, Th Fr, Fr Sa, Sa Su}(All possible days written in short-form)
n(S) = 7
n(E) = 2
P(E) = n(E)/n(S) = 2/7
Hope that you understand this (^_^)'
yakumoreyucurry:
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Answered by
1
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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