"The probability of getting heads on a biased coin is 1/3. Sammy tosses the coin 3 times. Find the probability of getting two heads and one tail".

Question

I thought that all you have to do is:
(1/3)(1/3)(2/3)

It makes sense to me, but it's not right...

Answer 1

Here are the possibilities:

HHT
HTH
THH

Each one has a probability of (1/3)(1/3)(2/3) = 2/27. For either-or probabilities, add the probabilities of the individual events.

Answer 2

I WOULD PUT IT AS A SIMPLE 1/3

Answer 3

use 3C2 (1/3)^2(2/3)1 in the calculator. (3c2 as in 3=number of trials, 2= number of successes, plug into the math>prob>ncr function. )