Question

A company receives equipment from two factories: 38% from factory A, and all other equipment from factory B. Each factory has a percentage of equipment that is defective: 1% of factory A's equipment is defective, while 4% of factory B's equipment is defective. If a piece of the company's equipment is selected at random, what is the probability that it is defective and from factory B?a. 0.0248b. 0.0038c. 0.6012d. 0.6600

Answer 1

Probability will be 0.0286 which is nit given in the bellow option

So none of the given option is correct

Step-by-step explanation:

We have given that company receives equipment from company A is 38 %

So
`p(A)=0.38`

And rest of the equipment is received from company B

So
`p(B)=1-0.38=0.62`

It is given that the equipment which is received from company A is 1 % defective

So
`p(D/A)=0.01`

And equipment which is received from company B is 4 % are defective

So
`p(D/B)=0.04`

So the probability that the equipment is defective
`=p(A)* p(D/A)+p(B)* p(D/B)=0.38* 0.01+0.62* 0.04=0.0286`

So none of the option is correct