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.

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;

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

answer:   n ≤ -3 or n < -5

step-by-step explanation:

-5 + 2n ≥ -11

+5             +5

2n ≥ -6

2/2 ≤ -6/2

n ≤ -3

-6 - 10n > 44

+6           +6

-10n > 50

-10/-10 50/-10

n < -5

total of apples - take away four

6-4 = 2

answer is 2 apples