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BdMO National 2012: Higher Secondary 08

Posted: Sat Feb 11, 2012 11:22 pm
by Moon
Problem 8:
A decision making problem will be resolved by tossing $2n + 1$ coins. If Head comes in majority one option will be taken, for majority of tails it’ll be the other one. Initially all the coins were fair. A witty mathematician replaced $n$ pairs of fair coins with $n$ pairs of biased coins, but in each pair the probability of obtaining head in one is the same the probability of obtaining tail in the other. Will this cause any favor for any of the options available? Justify with logic.

Re: BdMO National 2012: Higher Secondary 08

Posted: Mon Feb 13, 2012 1:07 am
by abdullah al munim
Yes by symmetry.
I also explained the problem this way during the olympiad.i am not sure whether i was given full mark or not(as the solution appears like literature than math).but i am pretty confident that this completes the solution

Re: BdMO National 2012: Higher Secondary 08

Posted: Mon Feb 13, 2012 1:10 am
by abdullah al munim
Oh, sorry! I wanted to say NO

Re: BdMO National 2012: Higher Secondary 08

Posted: Sat Feb 25, 2012 12:13 am
by turash
amar ans ascilo no change. sahitto likhecilam math er bodole

Re: BdMO National 2012: Higher Secondary 08

Posted: Sun Mar 04, 2012 3:19 pm
by nafistiham
If there were only fair coins, the probability would be $\frac {1}{2}$
as $n$ pairs are changed, the last coin makes the decision. So, as it is fair, the probability will not change.