<s>[INST] Probability urn-envelope question Urn 1 contains 14 envelopes--6 with \$1 bills and 8 with \$5 bills. Urn 2 has 8 envelopes--3 with \$1 bills and 5 with \$5 bills. 3 bills are randomly transferred from urn 1 to urn 2. What is the probability of drawing a\$1 bill from urn 2?
Attempted answer:
Let $O_x$ be the event where $x$ \$1 bills are transferred from urn 1 to urn 2.
Let $A$ be the event that a \$1 bill is drawn from urn 2.
We either transfered 0 \$1 bills, 1 \$1 bills, 2 \$1 bills or 3 \$1 bills.
By law of total probability we have:
$$P(A) = P(O_3)P(A|O_3) + P(O_2)P(A|O_2) + P(O_1)P(A|O_1) + P(O_0)P(A|O_0)$$
$$P(A) = \left( \frac{6}{14}*\frac{5}{13}*\frac{4}{12} \right) \left( \frac{6}{11} \right) + \left( \frac{6}{14}*\frac{5}{13}*\frac{8}{12} \right) \left( \frac{5}{11} \right) + \left( \frac{6}{14}*\frac{8}{13}*\frac{7}{12} \right) \left( \frac{4}{11} \right) + \left( \frac{8}{14}*\frac{7}{13}*\frac{6}{12} \right) \left( \frac{3}{11} \right)$$
$$P(A) = \frac{178}{1001}$$
The answer is $\frac{30}{77}$
 [/INST] Hint:
(1,1,1), (1,1,5), (5,5,1), (5,5,5) are the possible bills transferred 
Similar to your approach:
P(USD1/Urn2) $$=\frac{{6\choose3}}{{14\choose3}}.\frac{6}{11}+\frac{{6\choose2}.{8\choose1}}{{14\choose3}}.\frac{5}{11}+\frac{{6\choose1}.{8\choose2}}{{14\choose3}}.\frac{4}{11}+\frac{{8\choose3}}{{14\choose3}}.\frac{3}{11}$$
$$ = \frac{720+3600+4032+1008}{24024} $$
$$= \frac{9360}{24024} $$
$$= \frac{30}{77}$$
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