Two blue after two reds are gone

Question

An urn holds 7 red marbles and 5 blue marbles, 12 in total. You have already drawn 2 red marbles and set them aside (they are not going back). You now draw 2 more marbles.

Given the two reds already removed, what is the probability the next two draws are both blue?

Accepted Answer

The condition simply changes the starting urn. After removing 2 reds, what remains is 5	ext{ red}+5	ext{ blue}=10	ext{ marbles}. Now find the chance the next two draws are both blue, using the shrunken urn and multiplying draw by draw. Next draw blue. 5 blue out of 10: \frac{5}{10}. Following draw blue. One blue gone, so 4 blue out of 9: \frac{4}{9}. Multiply: \frac{5}{10}	imes\frac{4}{9}=\frac{20}{90}=\frac{2}{9}\approx 0.222. Takeaway. Conditioning on what is already gone just resets the counts. Update the urn to its current contents, then compute the fresh draws as usual.

Two blue after two reds are gone

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