Question

Circuit boards are assembled by selecting 4 computer chips at random from a large batch of chips. In this batch of chips, 90 percent of the chips are acceptable. Let X denote the number of acceptable chips out of a sample of 4 chips from this batch. What is the least probable value of X?

Answer 1

Answer:

the least probable value of X is X=0 , with probability P(X=0)=0.0001 =0.01%

Step-by-step explanation:

denoting X=number of acceptable chips out of a sample of 4 chips. If the probability of choosing one acceptable chip is 0.90 , then the probability of choosing one chip with failures is 1-0.90 = 0.1

since is less likely to take one failed chip ( 0.1 is smaller than 0.9) , and each chip is independent from the others , the least probable value of X is when there are 4 failed chips or 0 acceptable chips. then the probability for P(X=0) will be

P(X=0) = 0.1*0.1*0.1*0.1 = 0.0001

Note:

Strictly speaking , X has a binomial distribution , therefore

P(X) =  4!/[(4-x)!*x!]*0.9^x *0.1^(4-x)

that has a minimum value for X=0 → P(X=0)= 0.0001

Answer 2

Answer:

0

Step-by-step explanation:

I got this question on AP and got it right