# Probability of exactly one defective unit

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?

• You can choose the defective in three different ways. Also the number of non-defective units is $95$ not $94$. – Thomas Shelby Dec 9 '18 at 10:49
• 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? – CruZ Dec 9 '18 at 10:53
• Yes. You will get the same answer mentioned below. – Thomas Shelby Dec 9 '18 at 11:03

## 2 Answers

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.

• I don't understand why it's 500 instead of 100? But I didn't think of the binomial distribution, thank you! – CruZ Dec 9 '18 at 10:59
• sorry, the 500 was a typo... – BelowAverageIntelligence Dec 9 '18 at 11:00
• @CruZ This isn't the binomial distribution, it's actually called the hypergeometric distribution. – Quintec Dec 9 '18 at 14:59
• Ah thank you for clearing that up, my mistake! Cheers! – CruZ Dec 9 '18 at 18:07

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: $$\binom{100}{3}$$ $$P(\mbox{"exactly 1 defective"}) = \frac{\binom{5}{1}\cdot \binom{95}{2}}{\binom{100}{3}}$$