In a manufacturing plant producing LED light bulbs, 80% of

vestirme4

vestirme4

Answered question

2021-09-20

In a manufacturing plant producing LED light bulbs, 80% of the light bulbs produced are non-defective. If 3 light bulbs were examined by a quality inspector, what is the probability that one light bulb is defective? Use the binomial probability function to answer this question.

Answer & Explanation

sweererlirumeX

sweererlirumeX

Skilled2021-09-21Added 91 answers

Step 1
Given that, the 80% of the produced light bulbs are non, defective. Thus, 20 % of the light bulbs are defective.
Thus, the probability that a randomly selected bulb is defective is, p=20100=0.2.
Thus, according to the binomial probability function, the probability that 3 light bulb will be defective among the 3 light bulb is,
P(1)=3C1×p1×(1p)31
=3p(1p)2
=3×0.2×(10.2)2
=0.6×0.82
=0.384
Step 2
Thus, the probability is 0.384

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