A factory produces nails and packs them in boxes of 200.If the probability that a nail is substandard is 0.006,find the probability that a box selected at random contains at most two nails which are substandard.

Number of substandard nails in a box.

n = 200

Probability that nails is substandard.

p = 0.006

Calculation λ.

λ = np = 200(0.006) = 1.2

Poisson distribution is written as.

X~P(λ) ⇒ X~P(1.2)

Probability that a box selected at random contains at most two nails is given as.

P(X≤2).

Poisson distribution P(X=x) = \$(e^(-λ) λ^x)/x!\$

P(X≤2).

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

P(X≤2).

\$(e^(-1.2) 1.2^0)/(0!) + (e^(-1.2) 1.2^1)/(1!) + (e^(-1.2) 1.2^2)/(2!)\$

P(X≤2).

\$2.718^(-1.2) + 2.718^(-1.2)×1.2 + (2.718^(-1.2) 1.2^2)/2\$

Probability that a box selected at random contains at most two nails P(X≤2).

0.8795

Answer

C