Question

A batch of 400 containers for frozen orange juice contains 5 that are defective. Two are selected, at random, without replacement from the batch. 26. What is the probability that the second one selected is defective given that the first one was defective? HINT: Note that you are only calculating a probability for the second draw conditional on what happened in the first draw. 27. What is the probability that both are acceptable? 28. If three are selected without replacement, what is the probability that all three are defective?

Answer 1

Answer:

0.0100250

0.0001566

0.0000009445555

Step-by-step explanation:

Given the following :

Total number of containers = 400

Number of defective oranges = 5

2 are randomly selected without replacement :

A.) What is the probability that the second one selected is defective given that the first one was defective?

Since selection is without replacement and given that the first pick is defective ; number of defective oranges left = 4 ; total oranges left in container = (400 - 1) = 399

Hence,

Probability = (required outcome / Total possible outcomes)

P(2nd defective given 1st is defective) = 4 / 399

27. What is the probability that both are acceptable?

This means probability that both picks aren't defective.

Ist pick = 1 - P(defective) = (1 - (395/400) = 0.0125

2nd pick = 1 - P(defective) 1 - (394/399) = 0.0125313

.Hence, 0.0125 × 0.0125313 = 0.0001566

28. If three are selected without replacement, what is the probability that all three are defective?

1st pick = 5 / 400 = 0.0125

2nd pick = 4 / 399 = 0.0100250

3rd pick = 3 / 398 = 0.0075376

Hence, (0.0125 × 0.0100250 × 0.0075376) = 0.0000009445555