Find the probability that their are 53 saturdays in a year which is not a leap year. A year which is not a leap year has 365 days.
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Step-by-step explanation:
so as we know that in a non-leap year there are 365 days
moreover 1 week=7 days
thus 365 days=365/7 weeks
so thus 365 days has 52 weeks+1 odd day
so this means there are 52 Saturdays+1 odd day
so thus odd day can be any day out of 7 days in a week
this now we have to calculate probability of these 7 days for getting 53 Saturdays
so let the sample space be S
thus S={Monday,Tuesdday,Wednesday,
Thursday,Friday,Saturday}
thus n(S)=7
now let A be the event of getting Saturday
A={Saturday}
so thus n(A)=1
now applying
P(A)=n(A)/n(S)
=1/7
thus P(A)=1/7
hence probability of getting 53 saturdays in a non-leap year is 1/7
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