A contractor purchases a shipment of 99 transistors. The way the business works is that the contractor will randomly select and test 10 of these transistors, and keep the shipment only if at least 9 of the 10 are in working condition. If we know that overall, 20% of the transistors have defects (and thus not in working condition), then what is the probability the contractor will keep all the transistors?

the probability that the contractor keeps the transistors is 37.58%

Step-by-step explanation:

Since each transistor is equally likely to the chosen , then the random variable X= transistors that have defects from a sample of 10 , has a binomial distribution . Thus its probability is calculated through:

P(X=x) = n!/((n-x)!*x!)*p^x*(1-p)^(n-x)

where

n= sample size = 10

p= probability for any transistor to be defective = 0.2 (20%)

x= number of transistors that are defective

P(X=x) = probability that there are x transistors that are defective

thus replacing values and knowing that only 0 or 1 can be defective in order to have at least 9 working transistors:

P(X≤1)= P(X=1)+P(X=0)

this is equivalent to

P(X≤1)= F(X=1)

where F(X) is the cummulative binomial probability distribution. From tables can be found that

P(X≤1)= F(X=1)= 0.3758 (37.58%)

thus the probability that the contractor keeps the transistors is 37.58%

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b

lo98765432wqasdfghjkloi87654567890

step-by-step explanation: