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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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Let us define $A_{i}$ as the event that the $i$ th chosen unit is not defective, for $i=1,2,3$. We are interested in $P\left(A_{1} \cap A_{2} \cap A_{3}\right)$. Note that
$$P\left(A_{1}\right)=\frac{95}{100}$$
Given that the first chosen item was good, the second item will be chosen from 94 good units and 5 defective units, thus
$$P\left(A_{2} \mid A_{1}\right)=\frac{94}{99}$$
Given that the first and second chosen items were okay, the third item will be chosen from 93 good units and 5 defective units, thus
$$P\left(A_{3} \mid A_{2}, A_{1}\right)=\frac{93}{98}$$
Thus, we have
\begin{aligned} P\left(A_{1} \cap A_{2} \cap A_{3}\right) &=P\left(A_{1}\right) P\left(A_{2} \mid A_{1}\right) P\left(A_{3} \mid A_{2}, A_{1}\right) \\ &=\frac{95}{100} \frac{94}{99} \frac{93}{98} \\ &=0.8560 \end{aligned}

As we will see later on, another way to solve this problem is to use counting arguments
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