An urn contains $15$ white balls, $8$ black balls and $12$ red balls. We extract $6$ balls without reintroducing them back.
a) Which is the probability that the first ball is red, the second and the third are black and the last three are white?
b) Knowing that the second ball was black, which is the probability that the first ball was red?
I did the first one like this: $12/35 \cdot 8/34 \cdot 7/33 \cdot 15/32 \cdot 14/31 \cdot 13/30$,
but what should I do at b?
Best Answer
Hint:
For (b), the probability that the first ball is red given that the second ball is black is going to be exactly the same as the probability that the second ball is red given that the first ball is black.
That these should be equal should be clear since pulling out two balls and labeling one as first and the other as second yields the same results as having pulled out two balls and labeled one as second and the other as first. In many problems like this, we may choose to perform the actions in different orders without affecting the probabilities at all, for example choosing who comes in last place in a race in the beginning of the calculation rather than having to choose every other person's position in the entire race before deciding the person who came in last.
We have then:
If you remain unconvinced, you may do this more directly the tedious way via the definitions. Let $R_1,B_2$ be the events that the first ball is red and the second ball is black respectively. We would have then:
$$Pr(R_1\mid B_2) = \dfrac{Pr(R_1\cap B_2)}{Pr(B_2)} = \dfrac{Pr(R_1)Pr(B_2\mid R_1)}{Pr(R_1)Pr(B_2\mid R_1)+Pr(W_1)Pr(B_2\mid W_1)+Pr(B_1)Pr(B_2\mid B_1)}$$
This is incredibly tedious however, and you can see how if we were talking about the probability that the first ball is red given that the last ball is black will be horrendous to calculate, which is why noting what we did above is so helpful.