Question

Large Consignments of computer components are inspected for defectives by means of a sampling system. Ten components are examined and the lot is to be rejected if two or more are found to be defective. If a consignments contains exactly 10% defectives.
Find the probability of the consignment by using the technique of Binomial probability distribution that the consignment is: i) Accepted ii) Rejected

Answer 1

n=10

p=0.1

q=0.9

X~Binomial(10,0.1)

$P(X) = C^{10}_x(0.1)^x(0.9)^{10-x}$

i) Accepted if

$P(X<2) = P(X=0)+P(X=1) \\ P(X=0) = \frac{10!}{0!(10-0)!} \times (0.1)^0 \times (0.9)^{10-0} \\ = 1 \times 1 \times 0.3486 \\ = 0.3486 \\ P(X=1) = \frac{10!}{1!(10-1)!} \times (0.1)^1 \times (0.9)^{10-1} \\ = 10 \times 0.1 \times 0.3874 \\ = 0.3874 \\ P(X<2) = 0.3486 + 0.3874 = 0.736$

ii) Rejected if

$P(X≥2) = 1 -P(X<2) \\ = 1 -0.736 \\ = 0.264$