Two golds after one of each is gone

Question

A pouch holds 5 gold tokens and 5 silver tokens, 10 in total. So far one gold and one silver token have been drawn and set aside. You now draw 2 more tokens.

Given the two tokens already removed, what is the probability the next two draws are both gold?

Accepted Answer

The condition resets the pouch. After removing one gold and one silver, what remains is 4	ext{ gold}+4	ext{ silver}=8	ext{ tokens}. Now compute two draws on the reduced pouch. Next draw gold. 4 gold out of 8: \frac{4}{8}. Following draw gold. One gold gone, so 3 gold out of 7: \frac{3}{7}. Multiply: \frac{4}{8}	imes\frac{3}{7}=\frac{12}{56}=\frac{3}{14}\approx 0.214. Takeaway. When equal numbers of each color are removed, the balance stays even, so the reduced pouch behaves like a smaller version of the original. Recount, then multiply the fresh draws.

Two golds after one of each is gone

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