Mathematics: 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.

sorry im in6th grade i dont know but i dont get what the q is asking so im realy sorry...

An electronic company manufactures a specific component

bellalemieux

30.01.2022

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

Step-by-step 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;

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

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

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

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

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

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

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