Probability of 3rd ball being green while drawn without replacement, in a case where there are 5 blue 3 green and 7 yellow marbles in an urn. I'm having a hard time coming to a conclusion about the process of the math. I drew a probability tree to have a better understanding, the final branch being P(green) and P(green)', then I added the probabilities of the 3rd one being green and divided it by all possible outcomes of the third pick. My answer comes to 7/39 but I'm not sure if my process is correct.
2026-03-31 13:10:23.1774962623
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An urn has 5 blue, 3 green and 7 yellow marbles. two marbles are drawn without replacement. probability that the 3rd one will be a green marble?
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There are a few cases we have to consider, which are the orders that the ball are drawn. Denote $G$ and green, $B$ as blue, and $Y$ as yellow. A tuple $(a,b)$ represents that a ball of type $a$ is drawn first, followed by a ball of type $b$, and finally a green ball is drawn. The cases are as follows:
$(B,B)$: $5\cdot4\cdot3$
$(B,G)$ and $(G,B)$: $2\cdot5\cdot3\cdot2$
$(B,Y)$ and $(Y,B)$: $2\cdot5\cdot7\cdot3$
$(G,G)$: $3\cdot2\cdot1$
$(G,Y)$ and $(Y,G)$: $2\cdot3\cdot7\cdot2$
$(Y,Y)$: $7\cdot6\cdot3$
The sum of all cases is $546$. Because there are $15\cdot14\cdot13$ total ways to draw three balls, the probability is therefore: $$\frac{546}{2730}=\boxed{\frac15}$$
Probability of case I $= \frac{12}{15}\cdot \frac{11}{14} \cdot \frac{3}{13}=\frac{66}{455}$
Probability of case II $= \frac{12}{15}\cdot \frac{3}{14} \cdot \frac{2}{13}=\frac{12}{455}$
Probability of case III $= \frac{3}{15}\cdot \frac{2}{14} \cdot \frac{1}{13}=\frac{1}{455}$
Probability of case IV $= \frac{3}{15}\cdot \frac{12}{14} \cdot \frac{2}{13}=\frac{12}{455}$
Total Probability$=\frac{66}{455}+\frac{12}{455}+\frac{1}{455}+\frac{12}{455}=\frac{91}{455}=\boxed{\frac{1}{5}}$