There's something wrong with your question. The first number should be the number of trials and the second number should be the probability of success. A probability can't be more than 1.
X ∼ Binomial(n,p)
I'm going to assume you really meant to say X ∼ binomial(10, 0.3)
P(X = k) = C(n,k) * p^k * q^(n-k)
In your case, you want the probability that the number of successes is less than or equal to 9. Instead of calculating the cases for k = 0 to 9, it's much easier to just calculate the one case for k=10 and then subtract from certainty (1).
Answers & Comments
You meant (10,.3) not (10,3)
P(x <= 9) = 1-P(x=10)
P(x=10) = 10C10 (0.3)^10 (0.7)^0 = (0.3)^10
P( x <= 9) = 1- (0.3)^10
= 0.9999941
There's something wrong with your question. The first number should be the number of trials and the second number should be the probability of success. A probability can't be more than 1.
X ∼ Binomial(n,p)
I'm going to assume you really meant to say X ∼ binomial(10, 0.3)
P(X = k) = C(n,k) * p^k * q^(n-k)
In your case, you want the probability that the number of successes is less than or equal to 9. Instead of calculating the cases for k = 0 to 9, it's much easier to just calculate the one case for k=10 and then subtract from certainty (1).
P(X = 10) = C(10,10) * 0.3^10 * 0.7^0
= 1 * 0.3^10
≈ 0.0000059049
And therefore:
P(X ≤ 9) = 1 - P(X = 10)
≈ 1 - 0.0000059049
≈ 0.9999940951
Answer:
~99.9994%