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A box contains 12 transistors, 5 of which are defective. If 5 are selected at random, find the probability of the statements below.

a. All are defective
b. None are defective

2 Answers

2 votes

Answer:

a. The probability that all 5 selected transistors are defective is 0.0024 or 0.24%.

To calculate this probability, we use the formula for the probability of the intersection of independent events:

P(all 5 defective) = P(defective) x P(defective) x P(defective) x P(defective) x P(defective)

where P(defective) = 5/12 for each transistor.

Plugging in the values, we get:

P(all 5 defective) = (5/12) x (5/12) x (5/12) x (5/12) x (5/12) = 0.0024

b. The probability that none of the selected transistors are defective is 0.163 or 16.3%.

To calculate this probability, we use the complement rule:

P(none defective) = 1 - P(at least one defective)

To calculate P(at least one defective), we can use the complement rule again:

P(at least one defective) = 1 - P(none defective)

So, we need to find P(none defective) first:

P(none defective) = P(not defective) x P(not defective) x P(not defective) x P(not defective) x P(not defective)

where P(not defective) = 7/12 for each transistor.

Plugging in the values, we get:

P(none defective) = (7/12) x (7/12) x (7/12) x (7/12) x (7/12) = 0.163

Then, we can find P(at least one defective):

P(at least one defective) = 1 - P(none defective) = 1 - 0.163 = 0.837

Finally, we can find P(none are defective) using the complement rule:

P(none defective) = 1 - P(at least one defective) = 1 - 0.837 = 0.163.

User Ajoy Sinha
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4 votes
A) all are defective
User JwJosefy
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