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A pack of 10 electronic components is known to include 3 defectives. If 4 components are randomly chosen and tested , what is the probability of finding among them not more than one defective ?

I tried using binomial distribution.

$$ P(x=0) + P(x=1) $$

Now , $ P(x=0) = ^4C_0 * p^0 * q^4 $ , where $ p = (\frac{3}{10})$ and $q= (\frac{7}{10}) $

Similarly , $ P(x=1) = ^4C_1 * p^1 * q^3 $ , where $ p = (\frac{3}{10})$ and $q= (\frac{7}{10}) $

Am I doing it wrong ?

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  • $\begingroup$ but I am getting $ (\frac{19}{10}) * (\frac{343}{1000}) $ as answer , while the answer given as $ (\frac{2}{3}) $ $\endgroup$ – DukeLover Oct 22 '15 at 16:18
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The binomial distribution would be applicable if you were sampling with replacement, which is not the case here.

Let $N_0$ be the number of ways of selecting $0$ defective components, $N_1$ be the number of ways of selecting $1$ defective component.

$$N_0=\binom{3}{0}\binom{7}{4}=35$$ and $$N_1=\binom{3}{1}\binom{7}{3}=105$$ while there are $\binom{10}{4}=210$ ways of choosing $4$ components from $10$.

Hence $$\Pr[\le1\text{ defective}]=\frac{35+105}{210}=\frac{2}{3}$$

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You have to use the hypergeometric distribution, because it is without replacement.

$P(X=0)+P(X=1)=\frac{{7\choose 4}\cdot {3\choose 0}}{10\choose 4}+\frac{{7\choose 3}\cdot {3\choose 1}}{10\choose 4}$

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