There are $3$ red and $5$ black balls in bag$A$and $2$ red and $3$ black balls in bag $B$ . One ball is drawn from bag $A$ and two from bag $B$ . Find the probability that out of the three balls drawn one is red and $2$ are black.
My solution goes as follows:
Considering two events $A$ and $B$. We can select one red and two black balls by : first selecting a red ball from $A$ and then two black balls from $B$ ,we can select a black ball from $A$ and a red and a black ball from $B$ . These are the two ways in which we can choose the balls of rquired color. Now , let $A$ denote the event of first selecting a red ball from $A$ and then two black balls from $B$ let $B$ be the event of selecting a black ball from $A$ and a red and a black ball from $B$ . So, $P(A)=\frac{3.3.2}{8.5.4}$ and $P(B)=\frac{5.2.3}{8.5.4}$. However, in case of $P(B)$ the black ball and red ball can be chosen in any order. So there are $2$ possible ways to do this. Hence ,$P(B)=\frac{5.2.3.2}{8.5.4}$ . So the required probability is $P(A)+P(B)=P(A)=\frac{3.3.2}{8.5.4}+P(B)=\frac{5.2.3.2}{8.5.4}=\frac{39}{80}$.
However,we did not consider the ordering of choice of two black balls in case of $P(A)$ as the two black balls are indistinguishable objects. This is the
valid reason, right?
Best Answer
Your Math is accurate and your intuition is good, including your explanation of why you had to apply the factor $(2)$ to one of the terms when computing $p(B)$ and not apply the factor of $(2)$ when computing any of the $p(A)$ terms.
However, this type of analysis/intuition becomes convoluted, and does not generalize well for this type of problem. Therefore, although your Math was good and your intuition was valid, the approach that you took is not recommended.
Instead, I recommend a Combinatorics approach where the probability is expressed as
$$\frac{N\text{(umerator)}}{D\text{(enominator)}},$$
where
$~\displaystyle D = \binom{8}{1} \times \binom{5}{2}~ = 80$
which reflects the number of ways of selecting $(1)$ item from Bag-A and $(2)$ items from Bag-B.
Here, for convenience, the computation of $D$ assumes that the order that the items are taken from Bag-B is not relevant. Therefore, the computation of $N$ must be done in a consistent manner.
$N$ represents the union of two events, drawing a red ball from Bag-A and drawing a black ball from Bag-A.
Consequently,
$\displaystyle ~N = \left[\binom{3}{1} \binom{3}{2}\right] + \left[\binom{5}{1} \binom{2}{1}\binom{3}{1}\right] = 39.$