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

To find the probabilities, we need to use the concept of probability and the principle of counting. The principle of counting helps us determine the number of favorable outcomes and the total number of possible outcomes.

In this case, we have a shipment of twelve calculators, consisting of three defective and nine working calculators. We are interested in finding the probability of different scenarios when a customer buys five calculators without testing them.

a) Probability that all five calculators are acceptable:
To find this probability, we need to calculate the ratio of favorable outcomes (five acceptable calculators) to the total number of possible outcomes (any five calculators from the twelve).

Step 1: Determine the number of favorable outcomes.
Since there are nine acceptable calculators in the shipment, we need to choose all five of them. This can be calculated using the combination formula: C(n, r) = n! / (r!(n-r)!)
Here, n is the total number of acceptable calculators (9) and r is the number of calculators we want to choose (5).

Combination of 9 choose 5: C(9, 5) = 9! / (5!(9-5)!) = (9 * 8 * 7 * 6 * 5!) / (5! * 4 * 3 * 2 * 1) = 126.

Therefore, there are 126 favorable outcomes, i.e., the customer gets five acceptable calculators.

Step 2: Determine the total number of possible outcomes.
As mentioned earlier, the customer can choose any five calculators from the shipment of twelve.
This can be calculated using the combination formula: C(n, r) = n! / (r!(n-r)!)
Here, n is the total number of calculators in the shipment (12) and r is the number of calculators the customer chooses (5).

Combination of 12 choose 5: C(12, 5) = 12! / (5!(12-5)!) = (12 * 11 * 10 * 9 * 8!) / (5! * 7 * 6 * 5!) = 792.

Therefore, there are 792 total possible outcomes, i.e., all combinations of choosing five calculators from the twelve.

Step 3: Calculate the probability.
Finally, we can calculate the probability of all five calculators being acceptable by dividing the number of favorable outcomes (126) by the total number of possible outcomes (792):

P(all five acceptable) = 126 / 792 = 0.1591 (rounded to four decimal places) or 15.91% (rounded to two decimal places).

b) Probability that two calculators are defective and three are acceptable:
To find this probability, we will follow a similar approach as in part a. Again, we need to calculate the ratio of favorable outcomes to total possible outcomes.

Step 1: Determine the number of favorable outcomes.
The customer needs to choose two defective calculators (out of the three in the shipment) and three acceptable calculators (out of the nine in the shipment).
Using the combination formula:

Favorable outcomes = Combination of 3 choose 2 * Combination of 9 choose 3
Favorable outcomes = C(3, 2) * C(9, 3)
Favorable outcomes = (3! / (2!(3-2)!)) * (9! / (3!(9-3)!))
Favorable outcomes = (3 * 2!) * ((9 * 8 * 7) / (3 * 2!))
Favorable outcomes = (3 * 2) * (9 * 8 * 7 / 3)
Favorable outcomes = 6 * (9 * 8 * 7 / 3)
Favorable outcomes = 6 * (6 * 8 * 7)
Favorable outcomes = 6 * 336
Favorable outcomes = 2016

Therefore, there are 2016 favorable outcomes, i.e., the customer gets two defective calculators and three acceptable calculators.

Step 2: Determine the total number of possible outcomes.
As in part a, there are 792 total possible outcomes (the customer can choose any five calculators from the twelve in the shipment).

Step 3: Calculate the probability.
Finally, we can calculate the probability of getting two defective and three acceptable calculators by dividing the number of favorable outcomes (2016) by the total number of possible outcomes (792):

P(2 defective and 3 acceptable) = 2016 / 792 = 2.5455 (rounded to four decimal places) or 254.55% (rounded to two decimal places).

I hope this explanation helps you! If you have any further questions, feel free to ask.