A batch of 330 samples of rejuvenated mitochondria contains 6 that are mutated (or defective). Two are selected from the batch, at random, without replacement. (a) What is the probability that the second one selected is defective given that the first one was defective? Round your answer to five decimal places (e. g. 98.76543). Enter your answer in accordance to the item a) of the question statement (b) What is the probability that both are defective? Round your answer to five decimal places (e. g. 98.76543). Enter your answer in accordance to the item b) of the question statement (c) What is the probability that both are acceptable? Round your answer to five decimal places (e. g. 98.76543). Enter your answer in accordance to the item c) of the question statement

Question

Mathematics

Adrien Frank

24 October, 15:20

A batch of 330 samples of rejuvenated mitochondria contains 6 that are mutated (or defective). Two are selected from the batch, at random, without replacement. (a) What is the probability that the second one selected is defective given that the first one was defective? Round your answer to five decimal places (e. g. 98.76543). Enter your answer in accordance to the item a) of the question statement (b) What is the probability that both are defective? Round your answer to five decimal places (e. g. 98.76543). Enter your answer in accordance to the item b) of the question statement (c) What is the probability that both are acceptable? Round your answer to five decimal places (e. g. 98.76543). Enter your answer in accordance to the item c) of the question statement

+1

Answers (1)

Know the Answer?

Jocelyn Gibson · Answer 1

(a) Probability = 0.01520

(b) Probability = 0.00028

(c) Probability = 0.96391

Step-by-step explanation:

Total mitochondria = 330

Defective = 6

(a) Without replacement, if the first selection contains a defective mitochondria, we have a total of 5 defective remaining from a total of 329 mitochondria. Thus the probability the second one is defective Given the first one was defective will be:

Probability second is defective given the first one is not = 5 / 329 = 0.01520

(b) The probability the first is defective is : 6 / 330 = 0.01818

The probability the second is defective (without replacement) = 5 / 329 = 0.01520

The probability both are defective = 0.01818 * 0.01520 = 0.00028

(c) Probability the first one is acceptable = 324 / 330 = 0.98182

Probability the second one is acceptable = 323 / 329 = 0.98176

Probability both are acceptable = 0.98182 * 0.98176 = 0.96391