The screens used for a certain type of cell phone are manufactured by 3 companies, a, b, and c. the proportions of screens supplied by a, b, and c are 0.5, 0.3, and 0.2, respectively, and their screens are defective with probabilities 0.01, 0.02, and 0.03, respectively. given that the screen on such a phone is defective, what is the probability that company a manufactured it?

The probability is 0.2941

Step-by-step explanation:

Let's call A the event that company A manufactured the screen, B the event that company B manufactured the screen, C the event that the company C manufactured the screen and D the event that the screen is defective.

The probability P(A/D) that the company A manufactures the screen given that is defective is:

P(A/D) = P(A∩D)/P(D)

Where P(D) = P(A∩D) + P(B∩D) + P(C∩D)

So, the probability P(A∩D) that the screen is manufactured by company A and the screen is defective is calculated as:

P(A∩D) = 0.5 * 0.01 = 0.005

Where 0.5 is the proportion of screens manufactured by company A and 0.01 is the probability of defective screens for screens that are manufactured by company A.

At the same way, the probability P(B∩D) that the screen is manufactured by company B and the screen is defective and the probability P(C∩D) that the screen is manufactured by company B and the screen is defective are calculated as:

P(B∩D) = 0.3 * 0.02 = 0.006

P(C∩D) = 0.2 * 0.03 = 0.006

Then, The probability P(D) that the screen is defective is equal to:

P(D) = 0.005 + 0.006 + 0.006 = 0.017

Finally, P(A/D)that the company A manufactures the screen given that is defective is:

P(A/D) = 0.005/0.017 = 0.2941

the range is 19

step-by-step explanation:

the range is biggest number minus the smallest, so in this case it would be

41 - 22 = 19

b

step-by-step explanation: