Bernoulli Trials Question involving at least 3 successes

Question

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?

Answer 1

Why did you include the probability of observing $2$ successes? You want only $$\Pr[X \ge 3] = \binom{4}{3} (0.1)^3 (0.9)^1 + \binom{4}{4} (0.1)^4 (0.9)^0.$$

Answer 2

Alternatively, you can calculate the complementary probability:

$$\Pr(X \ge 3) =1-\Pr(X < 3)= 1-\binom{4}{0} (0.1)^0 (0.9)^4 - \binom{4}{1} (0.1)^1 (0.9)^3-\binom{4}{2} (0.1)^2 (0.9)^2$$