# How to find the probability that the first ball was red given that the second ball was black

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?

• You need to multiply the probabilities instead of adding them. The probability that two independent events occur is the product of the probabilities. As to part b), do you know Bayes' rule? Sep 3 '19 at 13:25
• Yeah, sorry, my mistake, I wrote + instead of *, but it's ok now. I know it, but I'm also confused. Sep 3 '19 at 13:40
• If F1 is "the second ball is black" and F2 is " the first ball is red" the the probability is P(F2/F1)= (P(F2)*P(F1))/P(F1), right? So it's (12/35 * 8/34 )/ P(F1). Now there is where I get confused. What is the value of P(F1)? I saw an exercise similar to this one, and according to that it should 8/35 but I don't understand why. Sep 3 '19 at 13:46
• That's the right way to do it. The probability that the second ball is black is just the probability that a ball drawn at random is black. Imagine that you draw the balls one at a time with you eyes closed, and place them in a line as you draw them. Now, before you open your eyes, what is the probability that the second ball in the line is black? Sep 3 '19 at 13:49
• Ah, that makes sense, I understand now. Thank you! Sep 3 '19 at 14:04

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:

Supposing the first ball is black, we have $$15+7+12$$ balls remaining and $$12$$ of them are red, so the probability the second ball is red given the first is black will simply be $$\dfrac{12}{15+7+12}$$.

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.