Question

A company that manufactures computer components has a 3% defective rate. If 10 components are randomly selected, find the probability of getting 2 defective components. Binomial Probability Formula: P(x) = nCx px qn-x

Answer 1

The probability of getting 2 defective items

P(X=2) = 0.0317416

Explanation:

Step(i):-

Given that size of the components n = 10

Given that the probability of getting the defective item

p = 3% = 0.03

q = 1-p =1- 0.03 = 0.97

Let 'X' be a random variable in a binomial distribution

`P(X=r) =n_{C_(r) } p^(r) q^(n-r)`

Step(ii):-

The probability of getting 2 defective items

`P(X=2) =10_{C_(2) } (0.03)^(2) (0.97)^(10-2)`

`10_{c_(2) } = (10!)/((10-2)!2!) = (10 X 9 X8!)/(8!2!) = 45`

P(X =2) = 45 × 0.000705

P(X=2) = 0.0317416