Two whites after one black is gone

Question

A box holds 6 white balls and 4 black balls, 10 in total. One black ball has already been drawn and set aside. You now draw 2 more balls.

Given the black ball already removed, what is the probability the next two draws are both white?

Accepted Answer

The condition resets the box. After removing 1 black, what remains is 6	ext{ white}+3	ext{ black}=9	ext{ balls}. Now compute the two draws on the reduced box. Next draw white. 6 white out of 9: \frac{6}{9}. Following draw white. One white gone, so 5 white out of 8: \frac{5}{8}. Multiply: \frac{6}{9}	imes\frac{5}{8}=\frac{30}{72}=\frac{5}{12}\approx 0.417. Takeaway. Removing a black ball actually helps the white streak, since it leaves whites a larger share of the box. Always recount before computing the conditional draws.

Two whites after one black is gone

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