Asked by Stuck
"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".
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...
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...
Answers
Answered by
PsyDAG
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.
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.
Answered by
Anonymous
I WOULD PUT IT AS A SIMPLE 1/3
Answered by
John
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. )
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