I'm trying to understand the following problem and solution from this page Example 1.19. What does the sample space and $A_1 $event look like? Also, how is $P(A_2|A_1)$ (conditional probability) established? It's my understanding conditional probability uses the form $P(A|B) = P(A \cap B) / P(B)$. I don't see that occurring here. Thanks!
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$
This solution agrees with the text. I believe these are dependent events. The sample space or elementary events of the units chosen is:
$E = \{U_1,U_2 ... U_{100}\} x \{U_1,U_2 ... U_{99}\} x \{U_1,U_2 ... U_{98}\}$
The first event and subsequent events are calculated:
$A_1 = E \cap \{U_1,U_2..U_{95}\} x \{U_1,U_2..U_{99}\} x \{U_1,U_2..U_{98}\}$
The first result:
$P(A_1) = |A_1|/|E| = (95*99*98) / (100*99*98) = 95/100$
$A_2 = E \cap \{U_1,U_2..U_{100}\} x \{U_1,U_2..U_{94}\} x \{U_1,U_2..U_{98}\}$
$P(A_2 \cap A_1) = |A_2 \cap A_1|/|E| = (95*94*98) / (100*99*98) = (95*94)/(100*99)$ we also note $A_2$ and $A_1$ are independent because: $P(A_2 \cap A_1) = P(A_1)P(A_2) = (95*94)/(100*99) = 95/100 * 94/99$
The second result: $P(A_2|A_1) = P(A_2 \cap A_1)/P(A_1) = (95*94)/(100*99) * 100/95 = 94/99$
because $A_2$ and $A_1$ are independent, $P(A_2|A_1) = P(A_1)P(A_2)/P(A_1) = P(A_2) = 94/99$.
$A_3 = E \cap \{U_1,U_2..U_{100}\} x \{U_1,U_2..U_{99}\} x \{U_1,U_2..U_{93}\}$
$P(A_3 \cap A_2 \cap A_1) = |A_3 \cap A_2 \cap A_1|/|E| = (95*94*93) / (100*99*98)$
$P(A_2 \cap A_1) = |A_2 \cap A_1|/|E| = (95*94*98)/(100*99*98) = (95*94)/(100*99)$
we note $P(A_1 \cap A_2 \cap A_3) = (95*94*93)/(100*99*98) = P(A_1)P(A_2|A_1)P(A_3|A1,A2) = P(A_1)P(A_2)P(A_3) = 95/100 * 94/99 * 93/98$, so $A_1, A_2, A_3$ are also independent.
The third result: $P(A_3|A_2 \cap A_1) = P(A_3 \cap A_2 \cap A_1)/P(A_2 \cap A_1) = (95*94*93)/(100*99*98) * (100*99)/(95*94)= 93/98$
alternative as $A_3, A_2, A_1$ are independent, $P(A_3|A_2 \cap A_1) = P(A_3) = 93/98$.