# One more question.

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• May 28th 2007, 10:55 AM
3deltat
One more question.
Just looking over some more abstract questions and foundthis..

Four fair coins are tossed. Find the Probability of obtaining 2 heads and two tails

I found the answer already. its 4C2(1/2)^2 x (1/2)^2 = 3/8
But im wondering if there is a different way to get this answer
• May 28th 2007, 11:05 AM
Jhevon
Quote:

Originally Posted by 3deltat
Just looking over some more abstract questions and foundthis..

Four fair coins are tossed. Find the Probability of obtaining 2 heads and two tails

I found the answer already. its 4C2(1/2)^2 x (1/2)^2 = 3/8
But im wondering if there is a different way to get this answer

this is the easiest way to do it i think. the other way i can think of is to do all the combinations manually, that is,

P(getting heads with first coin, tails with second, heads with third, tails with fourth) + P(heads with first coin, heads with second, tails with third, tails with fourth) + .....

and try to make sure you get all combinations of two heads and two tails, which would be a pain, Bernoulli Trials is definatley the way to go here, which is what you did.

Maybe Soroban can come up with some other way, he's somewhat of a genius with probability
• May 28th 2007, 12:18 PM
Plato
Quote:

Originally Posted by 3deltat
Four fair coins are tossed. Find the Probability of obtaining 2 heads and two tails: its 4C2(1/2)^2 x (1/2)^2 = 3/8
But im wondering if there is a different way to get this answer

There are several ways to model this. But all of are equivalent to what you have done.

Basically, there are $2^4$ outcomes. The number of ways to arrange the string “TTHH” is ${4 \choose 2}$.
• May 28th 2007, 12:55 PM
3deltat
Alright thanks.
I asked this question because i found myself trying to answer this question a different way than what i ended up doing, but I agree that this proly would be the easiest way