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Question

Find the probability of having 53 Saturdays in:
i)  a non - leap year
ii) a leap year
[4 MARKS]
 


Solution

Each subpart: 2 Marks each

(i) In a non-leap year, there are 365 days and 364 days make 52 weeks. In a week, there is one Saturday.
It means we have to find the probability of having a Saturday out of the remaining 1 day.
Probability (having 53 Saturdays)
=n(E)n(S)=17

ii) We know that in a leap year, there are 366 days and 364 days make 52 weeks.  In each week, there is one Saturday. It means we have to find the probability of having a Saturday out of the remaining 2 days.
The remaining two days can either be a (Sunday-Monday),(Monday-Tuesday),(Tuesday-Wednesday),(Wednesday-Thursday),(Thursday-Friday),(Friday-Saturday),(Saturday-Sunday)
Probability(having 53 Saturdays)
=n(E)n(S)=27

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