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In a factory there are 100 units of a certain product, 5 of which are defective. We pick three units from the 100 units at random. What is the probability that none of them are defective

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Answer:

Probability of picking all three non-defective units

= 7372/8085  (or 0.911812 to six decimals)

Step-by-step explanation:

Let

D = event that the picked unit is defective

N = event that the picked unit is not defective

Pick are without replacement.

We need to calculate P(NNN) using the multiplication rule,

P(NNN)

= 97/100 * 96/99 * 95/98

=7372/8085

= 0.97*0.969697*0.9693878

= 0.911812

The probability that none of the picked products are defective is;

P(None picked is defective) = 0.856

  • We are told that 5 are defective out of 100.

This means the number of good products that are not defective are 95.

  • Probability of the first picked product not being defective is written as; P(First picked not defective) = 95/100

  • Since the good ones have been picked, there will be 99 left of which the good ones are now 94. Thus, probability of second one not being defective = 94/99

  • Since two good ones have been picked, there will be 98 left and 93 good ones left. Thus, probability of third one not being defective = 93/98

  • Finally, Probability of none of the three being defective is;

95/100 × 94/99 × 93/98 = 0.856

Read more at; https://brainly.com/question/14661097

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