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I remember now that I had 1/8 as the answer in my mind when i framed the question.........it only mentions about the first three heads......so essentially the last two throws are useless.....

the answer will always be 1/8 irrespective of the number of throws.......i wonder why i did not reply to stolyar's posting........or maybe i did thru a PM.

Say, if instead of 5 coins we had 1,241,352 coins. What is the p that first 3 coins will have heads? It is still 1/8, but i doubt you can calculate 2^1,241,352!

Best is to use 1/2^3. Three independent events.

I guess this is what brstorewala means when saying the answer is

hey....I got 1/8 too!!! [#permalink]
06 Jun 2003, 12:00

Well,Here is how I woked it out.....
P(getting a head)=1/2 so P(getting a head first three times)=1/2*1/2*1/2=1/8!!!!!!!Ofcourse.the p(of getting a tail )= 1/2 too.I am not sure about this logic!!!! _________________

Its not about you getting knocked down,Its about how fast you get back up again!!!!

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