1.

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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Answer:

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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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