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

Answer:

The probability of getting 2 defective items

P(X=2) = 0.0317416

Step-by-step 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

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

$10_{c_{2} } = \frac{10!}{(10-2)!2!} = \frac{10 X 9 X8!}{8!2!} = 45$

P(X =2) = 45 × 0.000705

P(X=2) = 0.0317416