All reds out before any blue

Question

An urn holds 3 red balls and 4 blue balls, 7 in total. You draw them one at a time until the urn is empty, recording the color sequence.

What is the probability that every red ball is drawn before any blue ball, that is, the sequence is red, red, red, then all blues?

Accepted Answer

Instead of multiplying draw by draw, look at the whole color sequence at once. Think of the 7 balls landing in a random row of 7 positions. What matters is only which 3 of those 7 positions are occupied by the red balls; the rest are blue. Every choice of "which 3 positions are red" is equally likely, and the number of such choices is \binom{7}{3}=35. The event "all reds before any blue" happens in exactly one of these patterns: the reds must take the first 3 positions. So P=\frac{1}{\binom{7}{3}}=\frac{1}{35}\approx 0.029. A quick sanity check with the draw-by-draw view gives the same thing:…

All reds out before any blue

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