you choose 2 balls at random, one at a time. Are you more likely to chose 2 white balls in a row when you replace the first ball or when you don't replace the first ball? Explain.

Answer 1

Suppose you have a total of n balls, out of which m balls are white.

For the question to make sense we need to assume m <= n and n >= 2.

First let me give you the answer:

If m = 0 then then both probabilities are equal 0.

If m = n then then both probabilities are equal 1.

Otherwise replacement gives you a higher probability for the following reason.

The first ball is white with probability m/n.

For the second ball the probability depends on whether you replace or not.

If you replace then the second ball is while with probability m/n.

If you do not replace then the second ball is while with probability (m-1)/(n-1).

So we need to prove

(m-1)/(n-1) < m/n

(m-1)n < m(n-1) (result of multiplying both sides by n(n-1)

mn - n < mn - m (result of expanding the expresions)

-n < -m (result of subtracting mn from both sides)

n > m (result of multiplying both sides by -1)

The last expression is true after having eliminated the case m = n.