Question

Suppose a geologist excavates at Chernobyl and goes with the knowledge that 5% of the rocks are radioactive. She takes home three rocks. Where x is the number of radioactive rocks she took home, what are the probabilities that P[X≤2], P[X = 0], and P[X≥2]? What is E[X], and the standard deviation?

Answer 1

Solution :

Given that,

p = 0.05

q = 1 - p = 1 - 0.05 = 0.95

n = 3

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * p^x * (1 - p)^{n - x}

(a)

P(X 2) = P(X = 0) + P(X = 1) + P(X = 2)

= ((3! / 0! (3)!) * 0.05⁰ * (0.95)³ + ((3! / 1! (2)!) * 0.05¹ * (0.95)² + ((3! / 3! (1)!) * 0.05² * (0.95)¹

= 0.9999

(b)

P(X = 0) = ((3! / 0! (3)!) * 0.05⁰ * (0.95)³ = 0.8574

(c)

P(X 2) = 1 - P(X < 2)

= 1 -  ((3! / 0! (3)!) * 0.05⁰ * (0.95)³ - ((3! / 1! (2)!) * 0.05¹ * (0.95)²

= 1 - 0.9927

= 0.0073

(d) and (e)

Using binomial distribution,

E(X) = Mean = = n * p = 3 * 0.05 = 0.15

Standard deviation = = n * p * q = 3 * 0.05 * 0.95 = 0.3775