Thread: 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

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

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 outcomes. The number of ways to arrange the string “TTHH” is .

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