Throw $4$ red and $4$ black balls into $4$ pots (probability question)

Question

Suppose we have $4$ red and $4$ black balls. We throw these balls into $4$ pots. Find the probability:

(a) each pot contains $1$ red and $1$ black ball.

(b) each pot contains $2$ balls.

Attempt. (a) Let $R_i,~B_i$ be the events that pot $i$ contains exactly $1$ red, $1$ black ball, respectively. The desired event is $E=R_1B_1R_2B_2R_3B_3R_4B_4$ and so:

$$P(E)=P(R_1B_1)\,P(R_2B_2|R_1B_1)\,P(R_3B_3|R_1B_1R_2B_2)\,\,P(R_4B_4|R_1B_1R_2B_2R_3B_3)$$

where: $$P(R_1B_1)=P(R_1)P(B_1)=\frac{1}{4}\frac{1}{4},~ P(R_2B_2|R_1B_1)=P(R_2|R_1)P(B_2|B_1)=\frac{1}{3}\frac{1}{3}$$ etc so the desired probability is $1/(4!)^2.$

(b) I am not sure how to approach this.

Am I on the right path regarding (a)? How should I approach (b)?

Thanks in advance for helping.

Answer 1

Suppose we have $4$ red and $4$ black balls. We throws these balls into $4$ pots. Find the probability that each pot receives one red and one black ball.

Method 1: Assuming each ball lands in one of the pots, there are four places each of the eight balls could land. Therefore, there are $4^8$ ways to distribute the balls to the pots.

For the favorable cases, each red ball and lands in a different pot, as does each black ball. There are $4!$ ways to distribute each red ball to a different pot and $4!$ ways to distribute each black ball to a different pot. Hence, the number of favorable cases is $4!4!$.

Therefore, the probability that each pot receives one red and one black ball is $$\Pr(\text{each pot receives one red and one black ball}) = \frac{4!4!}{4^8} = \frac{9}{1024}$$

Method 2: Assume the first red ball lands in some pot. The probability that the second red ball lands in a different pot is $3/4$. The probability that the third red ball lands in a pot different from the first two is $2/4$. The probability that the fourth red ball lands in a pot different from the first three is $1/4$. Hence, the probability that each red ball lands in a different pot is $$\Pr(\text{each pot receives one red ball}) = 1 \cdot \frac{3}{4} \cdot \frac{2}{4} \cdot \frac{1}{4} = \frac{3}{32}$$ Since the probability that each pot receives one black ball is the same, the probability that one red and one black ball lands in each pot is $$\Pr(\text{each pot receives one red and one black ball}) = \frac{3}{32} \cdot \frac{3}{32} = \frac{9}{1024}$$

Suppose we have $4$ red and $4$ black balls. We throws these balls into $4$ pots. Find the probability that each pot receives one red and one black ball.

Line the balls up in some order to throw them in the pots. The number of favorable cases is the number of ways we can select two of the eight balls to be placed in the first pot, two of the remaining six balls to be placed in the second pot, two of the remaining four balls to be placed in the third pot, and both of the remaining two balls to be placed in the fourth pot. Hence, the number of favorable outcomes is $$\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$ Therefore, the probability that two balls land in each pot is $$\Pr(\text{each pot receives two balls}) = \frac{\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}}{4^8} = \frac{315}{8192}$$

Answer 2

Part A:

Supposing the balls were numbered so as to be distinct as well as coloured. There would be $4^8$ ways of the balls landing and I think $(4!)^2$ of those would have one of each colour in each pot because if we first distribute the four black balls there are $4!$ ways they can go which fulfil the requirement and then the same for the four red ones.

So the answer to part a is $\frac{9}{1024}$.

Part B is harder!

EDIT: I've deleted this part of my answer as it was wrong.