Question

If x is a binomial random variable, compute P(x) for each of the following cases:

(a) P(x≤5),n=7,p=0.3

P(x)=

(b) P(x>6),n=9,p=0.2

P(x)=

(c) P(x<6),n=8,p=0.1

P(x)=

(d) P(x≥5),n=9,p=0.3

 

P(x)=

 

Answer 1

a)

Here, n = 7, p = 0.3, (1 - p) = 0.7 and x = 5
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 5).
P(X <= 5) = (7C0 * 0.3^0 * 0.7^7) + (7C1 * 0.3^1 * 0.7^6) + (7C2 * 0.3^2 * 0.7^5) + (7C3 * 0.3^3 * 0.7^4) + (7C4 * 0.3^4 * 0.7^3) + (7C5 * 0.3^5 * 0.7^2)
P(X <= 5) = 0.0824 + 0.2471 + 0.3177 + 0.2269 + 0.0972 + 0.025
P(X <= 5) = 0.9963

b)

Here, n = 9, p = 0.2, (1 - p) = 0.8 and x = 6
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X > 6).
P(X > 6) = (9C7 * 0.2^7 * 0.8^2) + (9C8 * 0.2^8 * 0.8^1) + (9C9 * 0.2^9 * 0.8^0)
P(X > 6) = 0.0003 + 0 + 0
P(X > 6) = 0.0003

c)

Here, n = 8, p = 0.1, (1 - p) = 0.9 and x = 5
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 5).
P(X <= 5) = (8C0 * 0.1^0 * 0.9^8) + (8C1 * 0.1^1 * 0.9^7) + (8C2 * 0.1^2 * 0.9^6) + (8C3 * 0.1^3 * 0.9^5) + (8C4 * 0.1^4 * 0.9^4) + (8C5 * 0.1^5 * 0.9^3)
P(X <= 5) = 0.4305 + 0.3826 + 0.1488 + 0.0331 + 0.0046 + 0.0004
P(X <= 5) = 0.99998 = 1

d)

Here, n = 9, p = 0.3, (1 - p) = 0.7 and x = 5
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 5).
P(X >= 5) = (9C5 * 0.3^5 * 0.7^4) + (9C6 * 0.3^6 * 0.7^3) + (9C7 * 0.3^7 * 0.7^2) + (9C8 * 0.3^8 * 0.7^1) + (9C9 * 0.3^9 * 0.7^0)
P(X >= 5) = 0.0735 + 0.021 + 0.0039 + 0.0004 + 0
P(X >= 5) = 0.0988