You are performing 4 independent Bernoulli trials with p = 0.1 and q = 0.9. Calculate the probability of the stated outcome. Check your answer using technology. (Round your answer to five decimal places.)
At least three successes
P(X ≥ 3) =
So for this one I did:
(4C2)((0.1)^2)((0.9)^2)+(4C3)((0.1)^3)((0.9)^1)+(4C4)((0.1)^4)((0.9)^0)
And the answer I got is .0708 which is not correct, can someone tell me where I went wrong?
Best Answer
Why did you include the probability of observing $2$ successes? You want only $$\Pr[X \ge 3] = \binom{4}{3} (0.1)^3 (0.9)^1 + \binom{4}{4} (0.1)^4 (0.9)^0.$$