I'm trying to understand the following conditional probability problem and solution from this page Example 1.19 and it follows the questions here.
Questions:
- What is the sample space? Is it the same for the entire solution? I can't see how the numbers of the elementary events in the sample space add up to 1 in this solution.
- The problem seeks to answer to $P(A_1 \cap A_2 \cap A_3)$, which is $.8560$. How would these elementary events intersect to provide such an answer?
- I have been told that each event here where a unit is chosen is independent from the previous events, so there is no conditional probability in occurring here and in the solution is $P(A|B) = P(A)$ or $P(A \cap B) = P(A) P(B)$. Is this true? I have considered this might be true since computing $P(A_2|A_1)= P(A_2 \cap A_1)/P(A_1)$ does not yield the results in the solution.
- Should I be considering a different approach vs these questions?
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?
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 $P(A_1 \cap A_2 > \cap A_3)$.
$P(A_1) = 95/100$
Given that the first chosen item was good, the second item will be chosen from 94 good units and 5 defective units, thus:
$P(A_2|A_1) = 94/99$
Given that the first and second chosen items were okay, the third item will be chosen from 93 good units and 5 defective units, tus:
$P(A_3|A_2, A_1) = 93/98$
Thus:
$P(A_1 \cap A_2 \cap A_3) = P(A_1)P(A_2|A_1)P(A_3|A_2, A_1)$
$ = 95/100 * 94/99 * 93/98 = 0.8560$