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 Away from home (Posted on 2012-10-23)
Last year John left his home and returned seven whole months to the day.

What is the probability he was absent for exactly 215 days?

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 solution Comment 2 of 2 |

215 = 7*30 + 5

So five of the seven months of duration must include the 31st of a month (Eight calendar months are included, with two--or in one case, the 1st to the 1st, only one--of them being only parts, so we're concerned only with the ends of these months--the first, most likely partial, month, and the next six.)

So of the first seven calendar months involved, only two have only 30 days, and none have less than 30. So it must include a July and August and a December and January.  The only way to do this is to make those first seven calendar months July through January, so as to avoid spanning the end of February to go into March.

He must have left his home in July and returned on the corresponding date the following February. As this is being published in 2012, there are 29 valid dates in February when he could have returned, and thus 29 dates in January he could have left. That was in 2011, which had only 365 days, so the probability is 29/365, the highest probability for any year in which this could have been posted.

If he had left in a leap year and the question posed the following year (presumably after February is over), the probability would have been 28/366, the least that such could be.

Had the interval spanned two ordinary years, the probability would have been 28/365.

Caveats:

I mentioned above that in some arbitrary year the question would be posed "presumably after February is over". Actually, in order to make the situation clean, it would have to be posed after July, the 7th month, is over, so as not to limit the possible starting dates (the denominator) that would keep his return in the past.

 Posted by Charlie on 2012-10-23 13:39:55

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