Question

An electronic company manufactures a specific component for a tuner amplifier. It finds that out of every 1,000 components produced, 8 are defectives. If the components are packed in boxes of 250, find a. the probability of 0,1,2 and 3 defectives in a box b. the probability that the box contains at least 3 defectives.

Answer 1

a. P(X = 0) = 0.134

P(X = 1) = 0.271

P(X = 2) = 0.272

P(X = 3) = 0.1812

b. P(X > 3) = 0.142

Explanation:

The given parameters are

Number of defective component per 1000 = 8

Therefore, the probability that the specific component is defective = 8/1000 = 0.008

The binomial probability is given as follows;

`P(X = r) = \dbinom{n}{r}p^(r)\left (1-p \right )^(n-r)`

`P(X = 0) = \dbinom{250}{0} \left ((1)/(125) \right )^(0)\left ((124)/(125) \right )^(250-0) = 0.134`

`P(X = 1) = \dbinom{250}{1} \left ((1)/(125) \right )^(1)\left ((124)/(125) \right )^(249) = 0.271`

`P(X = 2) = \dbinom{250}{2} \left ((1)/(125) \right )^(2)\left ((124)/(125) \right )^(248) = 0.272`

`P(X = 3) = \dbinom{250}{3} \left ((1)/(125) \right )^(3)\left ((124)/(125) \right )^(247) = 0.1812`

The probability that the box contains at least 3 defectives is the probability P(X > 3) = 1 - (P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0))

Which gives;

P(X > 3) = 1 - (0.1812 - 0.271 - 0.272 - 0.134) = 0.142