Both gold coins out before any silver

Question

A pouch holds 2 gold coins and 5 silver coins, 7 in total. You draw them one at a time until the pouch is empty. What is the probability that both gold coins come out before any silver coin?

Accepted Answer

Look at the full color sequence at once. Picture the 7 coins in a random row of 7 positions; only which 2 positions hold the gold coins matters. Every choice of "which 2 positions are gold" is equally likely, and there are \binom{7}{2}=21 of them. "Both golds before any silver" is exactly one pattern: the two golds take the first two positions. So P=\frac{1}{\binom{7}{2}}=\frac{1}{21}\approx 0.048. Draw-by-draw check: \frac{2}{7}	imes\frac{1}{6}=\frac{2}{42}=\frac{1}{21}. Takeaway. Only one of the \binom{7}{2} equally likely gold-position patterns puts both golds first, so the probability is…

Both gold coins out before any silver

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