A manufacturer of calculators would like to separate defective calculators from calculators that are acceptable. Through an error, a shipment of 12 calculators is sent out containing 3 defective calculators and 9 working calculators. A customer buys 5 of these calculators without testing them.
a) what is the probability that all 5 will be acceptable?
b)what is the probability that 2 will be defective and three acceptable?
Please help. Is my solution correct 3/12x 5/9=0.14 Thanks in advance
No
To calculate the probabilities correctly, we need to consider the information given in the problem.
a) To find the probability that all 5 calculators are acceptable, we can calculate the probability of picking one acceptable calculator from the shipment and multiply it by the probability of picking another acceptable calculator each time we select one. Since it is mentioned that the customer buys 5 calculators without testing them and there are 9 acceptable calculators out of the total 12, the calculation will be as follows:
P(1st acceptable) = 9/12
P(2nd acceptable) = 8/11
P(3rd acceptable) = 7/10
P(4th acceptable) = 6/9
P(5th acceptable) = 5/8
To find the probability that all 5 will be acceptable, we multiply these probabilities together:
P(all 5 acceptable) = (9/12) * (8/11) * (7/10) * (6/9) * (5/8) ≈ 0.1909
Therefore, the probability that all 5 calculators will be acceptable is approximately 0.1909, or 19.1%.
b) To find the probability that 2 calculators will be defective and 3 will be acceptable, we need to consider the possible combinations of picking 2 defective calculators and 3 acceptable calculators from the shipment. There are 3 defective calculators out of 12 and 9 acceptable calculators in the shipment. We can use the concept of combinations to calculate the probability as follows:
Number of ways to select 2 defective calculators from 3: C(3, 2) = 3
Number of ways to select 3 acceptable calculators from 9: C(9, 3) = 84
Number of total possible combinations of selecting 5 calculators from 12: C(12, 5) = 792
Therefore, the probability that 2 calculators will be defective and 3 will be acceptable is:
P(2 defective, 3 acceptable) = (3 * 84) / 792 ≈ 0.3182
Hence, the probability that 2 calculators will be defective and 3 will be acceptable is approximately 0.3182, or 31.8%.
Your solution for part (a) is incorrect. The correct probability is approximately 0.1909.