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Question
Solve the following:
Find the probability that a year selected will have 53 Wednesdays
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Solution
Let A ≡ the event that a year selected has 53 Wednesdays
L ≡ the event that leap year is selected
N ≡ the event that non-leap year is selected
The required event will happen if any one of L ∩ A and N ∩ A occurs.
These events are mutually exclusive
∴ the required probability
= P(L ∩ A) + P(N ∩ A)
= `"P"("L")*"P"("A"/"L") + "P"("N")*"P"("A"/"N")` ...(1)
There is one leap year in 4 consecutive years
∴ P(L) = `1/4`
`"P"("A"/"L")` = Probability that year has 53 Wednesdays given that it is a leap year
Leap year has 366 day
366 = 7 x 52 + 2
∴ there are 52 full weeks and 2 days.
These days can be Sunday, Monday; Monday, Tuesday; Tuesday, Wednesday; Wednesday, Thursday; Thursday, Friday; Friday, Saturday; Saturday, Sunday.
There are 7 possibilities and favourable cases are 2
∴ `"P"("A"/"L") = 2/7`
Since there are 3 non-leap years in 4 consecutive years,
P(N) = `3/4`
`"P"("A"/"N")` = Probability that year has 53 Wednesdays given that it is a non-leap year
Non-leap year has 365 days
365 = 7 x 52 + 1
∴ there are 52 full weeks and 1 day.
This day could be any day of the week days i.e., any one of 7 days.
The number of favourable case is 1.
∴ `"P"("A"/"N") = 1/7`
∴ from (1), the required probability = `1/4*2/7 + 3/4*1/7`
= `5/28`.
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