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?

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

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

answer: no, but if it does it's a obvious pattern.

step-by-step explanation: