Conditional probability - sampling light bulbs There is a box with 12 light bulbs inside and only 4 of them work. 3 light bulbs are sampled one after another. I'm asked to find the probability that every light bulb works.
Let $A_n$ be the event in which the nth light bulb works, then we need to find $P(A_1 \cap A_2 \cap A_3)$. I calculated the probability for each event and then multiplied them: $P(A_1 \cap A_2 \cap A_3) = \frac{4}{12}\frac{3}{11}\frac{2}{10} = \frac{1}{55}$. The correct answer uses conditional probability and the result is different, why? I don't understand it intuitively.
EDIT: I'll copy paste the solution here:
$P(A_1 \cap A_2 \cap A_3) = P(A_3 | A_1 \cap A_2) P(A_2 | A_1) P(A_1) = \frac{6}{10}\frac{7}{11}\frac{8}{12} = \frac{14}{55}$
This solution seems to match the solution for the opposite problem (probability of the three light bulbs not working), so I think they made a mistake.
 A: Let $N =$ Total # light bulbs and $W =$ # working light bulbs at any point of time.


*

*Before $A_1$ occurred: $N=12,\; W=8\; \Rightarrow P(A_1)=\frac{W}{N}=\frac{8}{12}$. 

*After $A_1$ occurred and before $A_2$ occurred: $N=11,\; W=7\; \Rightarrow P(A_2|A_1)=\frac{W}{N}=\frac{7}{11}$ (since one working blub was drawn, both $N$ and $W$ will decrease by $1$).

*After $A_2$ occurred and before $A_3$ occurred: $N=10,\; W=6\; \Rightarrow P(A_3|A_1 \cap A_2)=\frac{W}{N}=\frac{6}{10}$ (since another working blub was drawn).


Combining, with chain rule, $P(A_1 \cap A_2 \cap A_3) = P(A_1)P(A_2|A_1)P(A3|A1 \cap A_2)=\frac{8}{12}.\frac{7}{11}.\frac{6}{10}$.
A: There are 12 light bulbs, eight of which work.  The probability the first bulb drawn works is 8/12= 2/3.  If you do draw a working light bulb the first time, that leaves 11 bulbs 7 of which work so the probability of drawing a working light bulb now is 7/11.  If you do draw a working light bulb on the second draw, that leaves 10 bulbs, 6 of which work so the probability of drawing a working light bulb on the third draw is 6/10= 3/5.  The probability of three working bulbs in three draws is (2/3)(7/11)(3/5)= 14/55.
