An urn contains 5 red balls and 4 black balls. A sample of two balls is selected at random. What is the probability that the sample contains one red and one white ball?
The answer as I have it is: $\frac{{5\choose 1} \times {4\choose 1}}{9 \choose 2} = \frac{5}{9}$
On the other hand, If I use the formula: $P(A \cap B) = P(A) . P(B|A)$ gives $\frac{5}{9}.\frac{4}{8} = \frac{5}{18}$. Not the Answer!!
Here, Am I taking the question in a wrong way? When I do $\frac{5}{9}.\frac{4}{8} = \frac{5}{18}$, is it that it only calculates the part where the first ball is Red and the second is Black and leaves out the part that the first ball is Black and the second ball is Red? If so, then I do get the right answer $\frac{5}{9}.\frac{4}{8} + \frac{4}{9}.\frac{5}{8} = 2 \times \frac{5}{9}.\frac{4}{8} = \frac{5}{9}$
Which means the right question I should ask is "What is the probability that the two balls are different colors?"
Somehow using combinatorial formula captures the whole thing in a single step that I don't need to worry about each case separately, although on the numerator we considered the red ball first, then the black ball. As it is a combination and the order does not matter it seems that the numerator should be divided by 2: $\displaystyle{\frac{\frac{5 \times 4}{2}}{9 \choose 2}}$ Which is very confusing.
My question is the whole thing seems a little confusing to me. Can anyone please address:
- if I'm right about using the $P(A\cap B) formula$
- What exactly is wrong about the $\displaystyle{\frac{\frac{5 \times 4}{2}}{9 \choose 2}}$ part?
- Can you please explain the whole thing probably with an example in a generic that also works for other similar questions.
Best Answer
Let's put numbers on the balls so we can tell exactly which ones are chosen. So the balls are $(r_1, r_2, r_3, r_4, r_5)$ and $(b_1, b_2, b_3, b_4).$
Then all $\binom 92 = 36$ possible pairs of balls that can be selected are listed in the following table:
\begin{array}{ccccc} \{b_1,r_1\} & \{b_1,r_2\} & \{b_1,r_3\} & \{b_1,r_4\} & \{b_1,r_5\} \\ \{b_2,r_1\} & \{b_2,r_2\} & \{b_2,r_3\} & \{b_2,r_4\} & \{b_2,r_5\} \\ \{b_3,r_1\} & \{b_3,r_2\} & \{b_3,r_3\} & \{b_3,r_4\} & \{b_3,r_5\} \\ \{b_4,r_1\} & \{b_4,r_2\} & \{b_4,r_3\} & \{b_4,r_4\} & \{b_4,r_5\} \\ \{r_1,r_2\} & \{r_1,r_3\} & \{r_1,r_4\} & \{r_1,r_5\} & \{r_2,r_3\} \\ \{r_2,r_4\} & \{r_2,r_5\} & \{r_3,r_4\} & \{r_3,r_5\} & \{r_4,r_5\} \\ \{b_1,b_2\} & \{b_1,b_3\} & \{b_1,b_4\} & \{b_2,b_3\} & \{b_2,b_4\} \\ \{b_3,b_4\}. \end{array}
The first four rows of five pairs (containing $4 \times 5$ pairs!), you may observe, includes every pair that has one red ball and one black ball. The remaining $16$ pairs are either both red or both black.
If you pay attention to which ball is drawn first, of course, you get ordered pairs instead of the unordered sets listed above, and for each sequence of balls in the table above there is another opposite sequence; for example both $(b_1,r_1)$ and $(r_1,b_1).$ And you can very well solve the problem by counting those two sequences separately; but if you count both $(b_1,r_1)$ and $(r_1,b_1)$ you must also count both $(r_1,r_2)$ and $(r_2,r_1),$ you must count both $(b_1,b_2)$ and $(b_2,b_1),$ and so forth. Instead of $36$ pairs in the denominator you will have $72$ pairs, and the probability $N/D$ will come out the same:
$$ \frac{\binom41 \binom51}{\binom92} = \frac{4\times5}{36} = \frac{5}{9}; $$
$$ \frac{\binom41 \binom51 + \binom51 \binom41}{2\binom92} = \frac{4\times5 + 5\times4}{2\times36} = \frac{40}{72} = \frac{5}{9}. $$