Find the probability that a non-leap year selected at random will contain either 53 Sundays or 53 Mondays.
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7
a non leap year consist 365 days
means 52 week and one day
so it is compulsory that there will be 52 Sundays and 52 Mondays
for last one day
it can be any one day out of 7
but required event is Sunday and Monday
only 2 days
thus probability = no. of favourable outcome/total outcome
= 2/7
means 52 week and one day
so it is compulsory that there will be 52 Sundays and 52 Mondays
for last one day
it can be any one day out of 7
but required event is Sunday and Monday
only 2 days
thus probability = no. of favourable outcome/total outcome
= 2/7
JinKazama1:
^_^ √√
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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