1. P(X ≤ 1)when n = 9, p = 0.3 2. P(two or fewer successes)
1. P(X ≤ 1)when n =
9, p = 0.3
2. P(two or fewer successes)
Homework Answers
1)
Here, n = 9, p = 0.3, (1 - p) = 0.7 and x = 1
As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)
We need to calculate P(X <= 1).
P(X <= 1) = (9C0 * 0.3^0 * 0.7^9) + (9C1 * 0.3^1 * 0.7^8)
P(X <= 1) = 0.0404 + 0.1556
P(X <= 1) = 0.1960
2)
Here, n = 9, p = 0.3, (1 - p) = 0.7 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)
We need to calculate P(X <= 2).
P(X <= 2) = (9C0 * 0.3^0 * 0.7^9) + (9C1 * 0.3^1 * 0.7^8) + (9C2
* 0.3^2 * 0.7^7)
P(X <= 2) = 0.0404 + 0.1556 + 0.2668
P(X <= 2) = 0.4628
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