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ScottTargetTestPrep

Why did we assume the span of 4 years when its no where mentioned that we have to consider a total of 4 years to calculate the probability?


ScottTargetTestPrep
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What is the probability of being born on a leap day? (of a leap year)

A) 1/364
B) 1/365
C) 1/366
D) 1/1460
E) 1/1461

Since a leap day only occurs once every 4 years, which comprise 3 common years and 1 leap year. Since 3 common years = 3 x 365 = 1095 days and 1 leap year = 366 days, there is only 1 leap day in a total of 1095 + 366 = 1461 days. Therefore, the probability of being born on a leap day is 1/1461.

(Note: The solution above does not take into account the negligible difference in probability caused by the fact that not every 4th year is a leap year. Any year evenly divisible by 100 (such as the years 1900 or 2100) is not a leap year, unless the year is divisible by 400 (e.g.,the years 2000 or 2400); then it is a leap year.)

Answer: E

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Thanks for your answer. I have the following doubt.

Since it is mentioned specifically "(of a leap year)"
wouldn't the asnwer be 1/366.

And will the answer change if "(of a leap year)" is not mentioned?
ScottTargetTestPrep
costcosized
What is the probability of being born on a leap day? (of a leap year)

A) 1/364
B) 1/365
C) 1/366
D) 1/1460
E) 1/1461

Since a leap day only occurs once every 4 years, which comprise 3 common years and 1 leap year. Since 3 common years = 3 x 365 = 1095 days and 1 leap year = 366 days, there is only 1 leap day in a total of 1095 + 366 = 1461 days. Therefore, the probability of being born on a leap day is 1/1461.

(Note: The solution above does not take into account the negligible difference in probability caused by the fact that not every 4th year is a leap year. Any year evenly divisible by 100 (such as the years 1900 or 2100) is not a leap year, unless the year is divisible by 400 (e.g.,the years 2000 or 2400); then it is a leap year.)

Answer: E

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I solve it in simple logic.
As they asked it should be a leap day, hence we should calculate total days in 4 years including a leap year. Total days 365 * 3 + 366. From these days, there might be 1 day which is leap day. Hence \(P(Leap Day) = 1/(365 * 3 + 366). \)
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