probability - A box with $3$ types of colored balls.

In a box there are $15$ white balls, $8$ black balls, and $12$ red balls. We extract $6$ balls, without putting them back.

$(a)$ What is the probability that the first ball is red, the second and third are black, and last three are white ?

$p=1\frac{12}{35}+2\frac{8}{35}+3\frac{15}{35}$

$(b)$ Knowing the second ball was black, which is the probability that the first ball was red?

$p=1\frac{8}{35}+1\frac{12}{35}+4\frac{15}{35}$

I'm terrible at this problems. Can someone help me and explain if my answers are correct. Thank you for your help!

Answer

I see you're fairly new in probability mathematics, so I try to proceed through the idea of things slowly. I hope this doesn't make my answer too long for you.

a. In general, $P(A \land B) = P(A|B) \times P(B)$. The probability of both $A$ and $B$ happening is the probability of $B$ happening multiplied by the probability of $A$ happening in case $B$ happened. This simple rule goes a long way in many probability problems and can be extended to longer chains as well.

In this case, we denote colours as $R, W, B$ and different outcomes as $R_1$ "first ball is red", $B_2$ "second ball is black", etc. The question in part a is finding $P(R_1 \land B_2 \land B_3 \land W_4 \land W_5 \land W_6)$.

The probability of $R_1$ is simple: we have 35 balls in total and 12 are red, hence $P(R_1) = \frac{12}{35}$. But now things start to get interesting. We need to compute $P(B_2 |R_1)$ next: "the probability of getting a black ball second if the first ball was red". If the first ball was red, there are only 34 balls left and 8 of them are black, so $P(B_2 |R_1) = \frac{8}{34}$.

Next, we want $P(B_3 | B_2 \land R_1)$, "the probability of getting a black ball third if first ball was red and second was black". Now we have 33 balls left, 7 of them black, so $P(B_3 | B_2 \land R_1) = \frac{8}{33}$. Notice how the probability changes?

Using the same logic we can get the final three probabilities:

$P(W_4 | B_3 \land B_2 \land R_1) = \frac{15}{32}$

$P(W_5 | W_4 \land B_3 \land B_2 \land R_1) = \frac{14}{31}$

$P(W_6 | W_5 \land W_4 \land B_3 \land B_2 \land R_1) = \frac{13}{30}$

Multiply the six probabilities together to get the result.

On to part b.

Here we want $P(R_1 | B_2)$. Here it's natural to use a well-known probability equation known as Bayes' Theorem: $P(X|Y) = \frac{P(Y|X) \times P(X)}{P(Y)}$. So our problem becomes:

$P(R_1 | B_2) = \frac{P(B_2 | R_1) \times P(R_1)}{P(B_2)}$

Now we just need to fill in the correct values:

$P(R_1) = \frac{12}{35}$

$P(B_2) = P(B_2 | B_1) \times P(B_1) + P(B_2 | \lnot B_1) \times (1 - P(B_1))$

$P(B_2|R_1)$ you can probably figure out from the a section.

Now all you need to do is solve for the given values.