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Assume 5 out of 100 units are defective. We pick 3 out of the 100 units at random.

What is the probability that exactly one unit is defective?


My answer would be

$P(\text{Defect}=1) = P(\text{Defect})\times P(\text{Not defect})\times P(\text{Not defect}) = 5/100 \times 95/99 \times 94/98$

However, I am not sure whether or not this is correct or not. Can someone verify?

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    $\begingroup$ You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$. $\endgroup$ – Thomas Shelby Dec 9 '18 at 10:49
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    $\begingroup$ Sorry, my mistake, I meant 95/99 and 94/98. But then are you saying that I have to calculate like this: P(D=1) = {DDN}U{DND}U{NDD}, where D is af defective unit and N is a non-defective unit? $\endgroup$ – CruZ Dec 9 '18 at 10:53
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    $\begingroup$ Yes. You will get the same answer mentioned below. $\endgroup$ – Thomas Shelby Dec 9 '18 at 11:03
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Your answer should be $$\frac{\binom{5}{1}\binom{95}{2}}{\binom{100}{3}}$$ Since we want the total number of ways to choose 3 meeting the criteria over the total number of ways to choose 3 out of the 100.

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  • $\begingroup$ I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you! $\endgroup$ – CruZ Dec 9 '18 at 10:59
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    $\begingroup$ sorry, the 500 was a typo... $\endgroup$ – BelowAverageIntelligence Dec 9 '18 at 11:00
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    $\begingroup$ @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution. $\endgroup$ – Quintec Dec 9 '18 at 14:59
  • $\begingroup$ Ah thank you for clearing that up, my mistake! Cheers! $\endgroup$ – CruZ Dec 9 '18 at 18:07
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Here is a suggestion how to proceed as ordering does not play a role

  • Choose one defective item: $\binom{5}{1}$
  • Choose two non-defective ones: $\binom{95}{2}$
  • Chose any three: $\binom{100}{3}$ $$P(\mbox{"exactly 1 defective"}) = \frac{\binom{5}{1}\cdot \binom{95}{2}}{\binom{100}{3}}$$
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