Draws from a bag until the first red ball

Question

A bag contains 10 balls, 3 of them red and 7 non-red. You draw balls one at a time without replacement in random order until you get the first red ball. On average, how many draws does that take (including the red ball)?

Accepted Answer

Sampling without replacement, so use the gap argument. The 3 red balls split the 7 non-red balls into 4 gaps, and each non-red ball is equally likely to land in any gap. So on average \frac{7}{4} non-red balls come before the first red, and the red is the next draw: \frac{7}{4} + 1 = \frac{11}{4} = 2.75 	ext{ draws.} This is \frac{N+1}{m+1} = \frac{11}{4} with N = 10, m = 3. Takeaway. First of m reds among N balls sits at expected position \frac{N+1}{m+1} = \frac{11}{4} = 2.75.

Draws from a bag until the first red ball

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