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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