Two urns contains 12 balls each. in the 1st, 4 balls are white, while in the 2nd there are 6 white balls. An urn is selected at random and 3 balls are drawn with replacement. what is the probability that 1st urn is selected and EXACTLY two balls have been drawn?

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    $\begingroup$ Do you mean exactly two white balls are drawn? $\endgroup$ – Aerinmund Fagelson Jan 23 '16 at 19:26

The probability that the first urn is chosen is just 1/2. Supposing this urn is chosen, there are three ways in which two white balls can be chosen, each with probability (1/3)x(1/3)x(2/3). So the final answer is

$\frac{1}{2}(\frac{1}{3}\frac{1}{3}\frac{2}{3}+\frac{1}{3}\frac{2}{3}\frac{1}{3}+\frac{2}{3}\frac{1}{3}\frac{1}{3}) =\frac{1}{2}(\frac{2}{3}\frac{1}{3})=\frac{1}{9}$


  • $\begingroup$ no it shouldn't... $\endgroup$ – Aerinmund Fagelson Jan 28 '16 at 22:31
  • $\begingroup$ ugh, of course I meant $\frac{1}{2}\left(3\cdot\frac{2}{27}\right)=\frac{1}{2}\cdot\frac{2}{9}= \frac{1}{9}$ $\endgroup$ – Logophobic Jan 28 '16 at 22:45

The probability of picking the first urn is $\frac{1}{2}.$ Now we have to compute the probability that we pick exactly $2$ white balls of the $4$ available, given that we draw $3$ times.

The probability of drawing exactly $2$ white balls from a draw of $3$ is simply $3(\frac{1}{3})(\frac{1}{3})(\frac{2}{3}) = \frac{2}{9}.$

Our final probability is thus $\frac{1}{2} \cdot \frac{2}{9} = \boxed{\frac{1}{9}}.$

  • $\begingroup$ Perfect! It was exactly how I thought. $\endgroup$ – L.F. Cavenaghi Jan 23 '16 at 19:41
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    $\begingroup$ It's truly exciting when I know I've got the same answers as others have. Thank you @Leonardo Francisco Cavenaghi $\endgroup$ – K. Jiang Jan 23 '16 at 19:43
  • $\begingroup$ @K.Jiang The question states that the balls are drawn with replacement $\endgroup$ – Logophobic Jan 23 '16 at 20:57
  • $\begingroup$ I see. Let me fix my answer @Logophobic $\endgroup$ – K. Jiang Jan 23 '16 at 20:58

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