A manufacturer of calculators would like to separate defective calculators from calculators that are

acceptable. Through an error, a shipment of twelve calculators is sent out containing three defective
calculators and nine working calculators. A customer buys five of these calculators, without testing them.
a) What is the probability that all five will be acceptable?
b) What is the probability that two will be defective and three acceptable?
Thanks for your help

prob(defect) = 3/12 = 1/4

prob(working) = 3/4

prob (5 of 5 are working)
= C(5,5) (3/4)^5 (1/4)^0
= 243/1024

prob(3 work, 2 don't work)
= C(5,3)(3/4)^3 (1/4)^2
= 10(27/64)(1/16)
= 270/1024

To solve these probability questions, we need to use the concept of combinations and probability.

First, let's calculate the total number of possible outcomes. Since there are 12 calculators in the shipment, and the customer buys 5 of them, the total number of possible outcomes is the number of ways to choose 5 calculators out of 12, which is denoted as "12 choose 5" or "C(12, 5)".

The formula for combinations is:

C(n, r) = n! / [(n-r)! * r!]

Where "n" represents the total number of objects and "r" represents the number of objects to be chosen.

Now let's calculate the probabilities:

a) Probability that all five calculators are acceptable:

We know that the shipment contains 3 defective calculators and 9 working calculators. Since the customer does not test them, we can assume that the probability of picking a defective calculator is the same for each pick.

The probability of selecting an acceptable calculator is equal to the number of ways to choose 5 working calculators out of 9, divided by the total number of possible outcomes.

P(all acceptable) = C(9, 5) / C(12, 5)
= (9! / (5! * (9-5)!)) / (12! / (5! * (12-5)!))

Simplifying the expression:

P(all acceptable) = (9! / (5! * 4!)) / (12! / (5! * 7!))
= (9! * 7!) / (5! * 4! * 12!)
= (9 * 8 * 7 * 6 * 5! * 7!) / (5! * 4! * 12 * 11 * 10 * 9!)
= (8 * 7 * 6) / (12 * 11 * 10)

Canceling out common factors:

P(all acceptable) = (8 * 7 * 6) / (12 * 11 * 10)
= (8 * 7) / (12 * 11)
= 56 / 132
= 0.4242 (rounded to 4 decimal places)

Therefore, the probability that all five calculators will be acceptable is approximately 0.4242.

b) Probability that two calculators will be defective and three acceptable:

Similarly, we can calculate this probability using combinations.

The probability of selecting two defective calculators and three acceptable calculators is equal to the number of ways to choose 2 defective calculators out of 3, multiplied by the number of ways to choose 3 acceptable calculators out of 9, divided by the total number of possible outcomes.

P(2 defective, 3 acceptable) = (C(3, 2) * C(9, 3)) / C(12, 5)
= ((3! / (2! * (3-2)!)) * (9! / (3! * (9-3)!))) / (12! / (5! * (12-5)!))

Simplifying the expression:

P(2 defective, 3 acceptable) = (3 * 9! * 5!) / (2! * (3-2)! * (9-3)! * 12!)
= (3 * 9! * 5!) / (2! * 12!)

Canceling out common factors:

P(2 defective, 3 acceptable) = (3 * 9! * 5!) / (2! * 12!)
= (3 * 9 * 8 * 7 * 6 * 5!) / (2 * 1 * 12 * 11 * 10 * 9!)

Simplifying the expression:

P(2 defective, 3 acceptable) = (3 * 9 * 8 * 7 * 6) / (2 * 1 * 12 * 11 * 10)
= (3024) / (2640)
= 0.8849 (rounded to 4 decimal places)

Therefore, the probability that two calculators will be defective and three calculators will be acceptable is approximately 0.8849.