# Probability of balls in the box

In a box there are 12 balls; 4 defective, 8 not defective.

What is the probability that when 3 balls are drawn, at least two of them are defective.

I know the answer is $$\frac{{4 \choose 2}{8 \choose 1} + {4 \choose 3}}{{12 \choose 3}}$$

But why isn't the answer also $$\frac{{4 \choose 2}{10 \choose 1}}{{12 \choose 3}}$$?

Because I choose 2 balls from the 4 defective, and 1 from the remaining 10 (it will either be defective or not).

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Why would it be the latter? – Craig Sep 7 '11 at 20:01
@Craig check out the edits. – Kraken Sep 7 '11 at 20:02
You could check whether 4C2*10C1=4C2*8C1+4C3. – Ross Millikan Sep 7 '11 at 20:17
@Ross Millikan the question says "why is it not equal to the latter"..obviously the two wont match.. i am asking the reason for this mismatch? what wrong logic am i applying here? – Kraken Sep 7 '11 at 20:19
Because the cases with all three balls defective are getting counted three times. You are counting defective, defective, something; defective, something, defective; something, defective, defective. So 4C2*10C1-4C2*8C1+4C3=2*4C3 – Ross Millikan Sep 7 '11 at 20:35

## 2 Answers

The second approach is over-counting. To illustrate why, denote the defective balls by $D_1,D_2,D_3,D_4$ and the non-defective by $N_1,...,N_8$.

If we begin by choosing two of the defective balls, say $D_1$ and $D_2$, then yes, there are ten other balls total. However, if we pick another of the defective balls, say $D_3$, then we end up with the same combination as we'd have if we had started with $D_1$ and $D_3$ and then chosen $D_2$--or the same combination that we'd have if we had started with $D_2$ and $D_3$ and then chosen $D_1$.

Generalizing the subscripts above, there are $4\cdot 2=8$ extra combinations that we're counting twice. So, subtracting 8 from ${4 \choose 2}{10 \choose 1}$ we get 52, which is the same as ${4 \choose 2}{8\choose 1} + {4 \choose 3}$. :)

This is an excellent question that illustrates how even basic probability problems can be counter-intuitive.

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Another way is to look at the possible sequences. Let's say $D$ is a defect ball and $G$ is a nondefect (good ball). If you sample 3 balls without replacement the event 'to get at least 2D' is $\{DDD,GDD,DGD,DDG\}$. The probability of the first event is $P(DDD)= \frac{4 \cdot 3 \cdot 2}{12 \cdot 11 \cdot 10}$ It seems that the probability of the other 3 events has to be different, but it is actually not: $P(GDD)=P(DGD)=P(DDG)=\frac{4 \cdot 3 \cdot 8}{12 \cdot 11 \cdot 10}$ because the denominator does not change and the numerator is the product of the same 3 numbers (and multiplication is commutative), therefore: $P(D \geq 2)=P(DDD)+P(GDD)+P(DGD)+P(DDG)=\frac{4 \cdot 3 \cdot 2}{12 \cdot 11 \cdot 10}+3 \cdot \frac{4 \cdot 3 \cdot 8}{12 \cdot 11 \cdot 10}=.2364$ Control:$1-P(D<2)=1-\Big( \frac{8 \cdot 7 \cdot 6}{12 \cdot 11 \cdot 10}+3 \cdot \frac{8 \cdot 7 \cdot 4}{12 \cdot 11 \cdot 10} \Big)=.2364$

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