Suppose there are 12 balls in an urn: 4 red, 4 yellow and 4 black. We draw at random without replacement.
What is the probability that the 6th draw will be a yellow ball, given that in the first 5 draws, there was exactly 1 more red ball drawn than yellow?
My intuitive attempt:
There are only 3 possible outcomes in terms of yellow balls drawn already during the first 5 draws, i.e.:
- 0 yellow balls drawn (and 1 red, 4 black): let this be event $A$
- 1 yellow balls drawn (and 2 red, 2 black): let this be event $B$
- 2 yellow balls drawn (and 3 red, 0 black): let this be event $C$
Given that one of the 3 events above happened, but we don't know which one, we could be in any of the 3 possible sample spaces going forward: even though the above events have different probabilities, ultimately it is given that one of them happened with certainty, therefore we need not be concerned with probabilities that we condition on. Therefore the answer can be computed as (i.e. equal probability that we are in any of the three possible sample spaces):
$$\frac{1}{3}\left(\frac{4}{7}+\frac{3}{7}+\frac{2}{7}\right)=\frac{3}{7}$$
Is the above the correct answer?
Best Answer
To elaborate on the discussion in the comments:
The events $A,B,C$ are not equally probable. Indeed, given that we are told that exactly one more red than yellow ball was drawn, we can be pretty confident that we are in case $B$, but $A,C$ are still possible (and not equally likely).
To be precise: With $i\in \{0,1,2\}$ let $p_i$ be the probability that $i$ yellow and $i+1$ red were drawn in the first $5$ draws. Then we have $$p_i=\binom 4i\times \binom 4{i+1}\times \binom 4{5-2i-1}\Big / \binom {12}5$$
Numerically, we have $$\{p_i\}=\{.005\overline {05}, \,.\overline {18},\,.\overline {03}\}$$
confirming the intuitive claim that $B$ was the most likely. The conditional probability you want is then given by $$\frac {p_0\times 4/7+p_1\times 3/7+p_2\times 2/7}{p_0+p_1+p_2}=.41196$$
which is (slightly) less than $\frac 37$ (reflecting the fact that case $A$ is considerably less probable than case $C$).