An urn contains five red and seven blue balls. Suppose that two balls are selected at random and without replacement.

Question

B is the event that the second ball is red, and A is the event that the first ball drawn is red.

I understand why P(B/A) is 4/11 in this question, and why P(B) = 5/12 by the law of total probability, but this intuitive explanation below does not make sense to me.

"If no information is given on the outcome of the first draw, there is no reason for the probability of second ball being red to differ from 5/12."

What does this mean? In my thought process, if we know that a ball has been drawn without replacement, then doesn't that mean that the denominator of the probability of the second ball being red has already been reduced to 11? How does the 5/12 make sense in a without replacement context?

Answer 1

When you are not given any information about the first ball drawn, then you should expect each individual ball to have an identical chance to be the second ball drawn.

Five of the twelve are red.

Answer 2

If I steal five balls from the urn and destroy them without ever looking at them, what is your best estimate of the probability that a ball drawn from this impoverished urn is red?

The answer is that the draw of that ball is indistinguishable from the draw of any other ball in the absence of conditioning on the colors of prior balls.

Equivalently, draw all the balls one at a time. We cannot predict the color of any one ball a priori, but we are absolutely certain that five out of twelve of the drawn balls are red. Meaning, if we blindly (meaning without observation) draw all twelve balls, keeping track of the drawing order. Then pick a draw index (a number from 1 to 12) to inspect. What is the probability that the ball with that draw index is red? It is 5/12.

In the absence of additional data (for instance, conditioning on the color of previous balls), every ball is 5/12 likely to be red. As soon as you have additional data, the probability may change (to be consistent with that additional data).

Answer 3

Suppose that you continue drawing at random until you’ve drawn all $12$ balls. In effect you’ve simply produced a random ordering of the balls, all possible orderings being equally likely. If you pick an ordering at random, what is the probability that the first ball in line is red? What about the second ball? The tenth ball? It should be intuitively clear that the position in line has no bearing on the color of the ball, so that the answer in each case is $\frac5{12}$.

If you have any doubt, it’s easy enough to confirm this. If the ball in position $k$ in the ordering is red, there are $\binom{11}4$ possible arrangements of the remaining $11$ balls, and if it is blue, there are $\binom{11}5$ possible arrangements of the remaining $11$ balls. Thus, $\binom{11}4$ of the $\binom{11}4+\binom{11}5=\binom{12}5$ possible orderings have a red ball in position $k$. The orderings are equally likely, so the probability that the ball in position $k$ is red is

$$\frac{\binom{11}4}{\binom{12}5}=\frac{11!5!7!}{4!7!12!}=\frac5{12}\;.$$