Algebra - probability basic problem

11

vaibhav sharma ·2010-06-11 10:34:35

fun with maths:
One day two mathematicians, Igor and Pavel, meet in the street.

"How are you? How are your sons?" asks Igor. "You have three sons as I remember, don't you? But I have forgotten their ages."

"Yes, I do have three sons," replies Pavel. "The product of their ages is equal to 36." Looking around and then pointing to a nearby house, Pavel says, "The sum of their ages is equal to the number of windows in the building over there."

Igor thinks for a minute and then responds, "Listen, Pavel, I cannot find the ages of your sons."

"Oh, I am very sorry", says Pavel; "I forgot to tell you that my oldest son has red hair."

Now Igor is able to find the ages of the brothers. Can you do it?

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11

vaibhav sharma ·2010-06-11 10:55:29

no answers till now .......nobody interested in maths?????

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1

arkadipta sarkar ·2010-06-12 23:48:50

is the answer 2/3??
here is how I did it:
the probable cases are:
1)W₁ i,e the added marble is drawn leaving the previously present black marble B
2)W₁ i,e the added marble is drawn leaving the previously present white marble W₀
3)W₀ i,e the previously present marble is drawn leaving the added white marble W₁

So the answer comes out to be 2/3.

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11

vaibhav sharma ·2010-06-12 23:55:52

@arkadipta
yes totally right

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arkadipta sarkar ·2010-06-13 00:34:01

how to do the second problem??
i have no idea.

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106

Asish Mahapatra ·2010-06-13 02:38:19

18,2,1 ?

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11

vaibhav sharma ·2010-06-29 20:26:31

no
2,2,9

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13

Avik ·2010-06-29 22:27:25

2) :O Took me quite a while to figure out,...but still here goes my solution-

The Clues-
1-Product of ages 36.
2-Sum of ages equal to the no. of windows in a building.
3-There is only one elder son. (matlab two older brothers cannot be twins.)

Now, using clue 1, there are 8 possible cases of age distribution possible-
(1,2,18) (1,3,12) (1,4,9) (1,1,36) (2,1,18) (2,2,9) (2,3,6) (3,3,4) (1,6,6)

Using 2, if we find the sum of ages fr each of the possible options, we find tht each gives a distinct sum, except fr (2,2,9) & (1,6,6) which sum up to 13 each.
This explains why Igor culdn't answer straightaway after counting the building windows.

The third clue will now make us discard off (1,6,6) 'coz there is only one eldest among the 3.
So, it is (2,2,9).

P.S. A better sol.n may exist, this was all manual work, summinp up each and then cancelling the unappropriate...
Earlier i evn wondered if red hair were a sign of a particular age group smwhere :P

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