An urn has 2 white, 3 red and 5 black balls. Problem An urn has 2 white, 3 red and 5 black balls. 3 balls are randomly drawn, one at a time and without replacement.
Calculate the probability of extracting the sequence of colors (white, black, red) knowing that you have extracted a black ball.
solution
Let E be the event that refers to the indicated color sequence. The required probability is $P(E|X= 1)$ from which it follows $P(E|X= 1) = \frac{P(E⋂(X=1))}{P(X=1)} = \frac{P(E)}{P(X=1)} \overset{(question1)}{=} \frac{\frac{2}{10}\frac{5}{9}\frac{3}{8}}{\frac{5}{12}} = \frac{\frac{1}{24}}{\frac{5}{12}} = \frac{1}{10}$
question 1
If the sequence is (white, black, red) in the numerator we should not have: $\frac{3}{10}\frac{5}{9}\frac{2}{8}$ in the end the result is the same but surely the reasoning to obtain it is different.What can be the reasoning done in the solution?
And in the denominator , why $\frac{5}{12}$. $P(X=1)$ indicates the probability that a black ball has been extracted. This shouldn't be $\frac{5}{10}$
 A: Start by computing the probability of drawing a black ball. The probability that no black ball is drawn is $\frac{1}{2} \frac{4}{9} \frac{3}{8} = \frac{1}{12}$. Therefore, the probability that a black ball is drawn is $1 - \frac{1}{12} = \frac{11}{12}$.
Now, compute the probability of observing the sequence (white, black, red). This is straightforward: $\frac{1}{5} \frac{5}{9} \frac{3}{8} = \frac{1}{24}$.
The desired conditional probability is thus $\frac{\frac{1}{24}}{\frac{11}{12}} = \boxed{\frac{1}{22}}$.
Edit: Suppose that instead of "a black ball", we are interested in the condition that "exactly one black ball" is drawn. For this, consider that the number of permutations of three selected balls (regardless of color, and assuming they are distinguishable) is $(10) (9) (8) = 720$. To extract exactly one black ball, we must select it among $5$ choices, and choose one position out of $3$ possible. Then, for the first unselected position, we choose one of the $5$ non-black balls, and for the last position, we choose one of the $4$ remaining non-black balls. The number of ways is $(5) (3) (5) (4) = 300$. The probability of extracting exactly one black ball is thus $\frac{300}{720} = \frac{5}{12}$. And the desired conditional probability would be $\frac{\frac{1}{24}}{\frac{5}{24}} = \boxed{\frac{1}{10}}$.
A: Total Universe is $10 \times 9 \times 8 =720$
Part of universe is excluded : we know that at least one ball is black. So all combinations based on only Red+White are excluded : $5 \times 4 \times 3 =60$ combinations are excluded.
So Universe is reduced to $720-60=660$
Number of combinations that match with sequence (White,Black,Red) is $2 \times 5 \times 3= 30$
So, probability is $\frac {30}{660}=\frac{1}{22}$
Conditional probability means that you will divide a fraction by another fraction. It is too complex. It is easier to divide an integer (count of success) by another integer (size of restricted universe).
