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An urn initially contains 5 white balls and 7 black balls. If a ball is extracted at random and then returned to the urn with two more balls of the same color. What is the probability that the first two extracted balls are black and the last two are white?

I'm having a lot of trouble with this problem, I have done the following but I'm not convinced that my method is correct...

Let $N_i = \{\text{the $i$ th extraction is a black ball}\}$ and $B_i = \{\text{the $i$ th extraction is a white ball}\}$.

I'm searching for $\mathbf{P}(N_1 \cap N_2 \cap B_3 \cap B_4)$:

\begin{align} \mathbf{P}(N_1)& = \frac{7}{12} & \mathbf{P}(N_2|N_1)& = \frac{9}{14}\\ \mathbf{P}(B_3|N_2 \cap N_1)& = \frac{5}{16} & \mathbf{P}(B_4|B_3 \cap N_2 \cap N_1)& = \frac{7}{18} \end{align}

Then the solution would be: $\frac{(7)(9)(5)(7)}{(12)(18)(14)(16)} = .045$

Any help is appreciated!

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You did everything right, i dont see any mistakes $\mathbf{P}(N_1 \cap N_2 \cap B_3 \cap B_4)$ = $\mathbf{P}(B_4|B_3 \cap N_2 \cap N_1)$$P(B_3 \cap N_2 \cap N_1)$= $\mathbf{P}(B_4|B_3 \cap N_2 \cap N_1)$$\mathbf{P}(B_3|N_2 \cap N_1)$$P(N_2 \cap N_1)$=$\mathbf{P}(B_4|B_3 \cap N_2 \cap N_1)$$\mathbf{P}(B_3|N_2 \cap N_1)$$P(N_2|N_1)*P(N_1)$

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  • $\begingroup$ Thanks I guess I'm confused by the fact that that I don't see clearly that when I do the product of all the probabilities the events are independent. Can you give me some input on this? $\endgroup$ – migueldva Dec 14 '16 at 18:59

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