Suppose that 15 lightbulbs are in a cabinet and that 4 are defective. If three bulbs are chosen at random,a. What is the probability that all three will be defective?b. What is the probability that all three will be good?c.Why do the probabilities from parts (a) and (b) not add up to 1?

Question
Asked Dec 19, 2019
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Suppose that 15 lightbulbs are in a cabinet and that 4 are defective. If three bulbs are chosen at random,

a. What is the probability that all three will be defective?

b. What is the probability that all three will be good?

c.Why do the probabilities from parts (a) and (b) not add up to 1?

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Expert Answer

Step 1

It is given that, 15 lightbulbs are in a cabinet and that 4 are defective.

Then, 15 - 4 = 11 lightbulbs are good.

 If three bulbs are chosen at random,

a) The probability that all three will be defective is,

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11C,4C, 15C3 P(all 3 will be defective) = 4 455

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Step 2

b) The probability that all three will be good is,

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11C,4C, 15C, P(all 3 will be good) E. 165 455

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Step 3

c) If we add the probabilities...

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165 455 455 169 #1 455 |I

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Math

Algebra

Permutations and Combinations