Question

A production process produces 2% defective parts. A sample of five parts from the production process is selected. What is the probability that the sample contains exactly two defective parts?

Answer 1

Answer:

Step-by-step explanation:

Hi there,

 

= 5 choose 2 * 0.02^2 * 0.98^3

= 10 * 0.02^2 * 0.98^3

= 0.0038

Answer 2

Answer:

0.0038 ( approx )

Step-by-step explanation:

∵ The probability of defective parts, p = 2% = 0.02,

So, the probability of non defective parts, q =  1 - 0.02 = 0.98,

The binomial distribution formula,

$P(X)=\sum_{r=0}^{n} ^nC_r p^r q^{n-r}$

Where,

$^nC_r=\frac{n!}{r!(n-r)!}$

s.t. n = number of samples,

Here. n = 5,

Thus the probability that the sample contains exactly two defective parts would be,

$P(X=2)=^5C_2 (0.02)^2 (0.98)^3$

$=10 (0.02)^2(0.98)^3$

$=0.003764768$

$\approx 0.0038$