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There are two urns. In urn I there are 5 white balls and 6 black balls, in urn II there are 4 white balls and 3 black balls. 2 balls from urn I are picked at random and transfered in urn II. Then, two balls are chosen at random from urn II.

1) What is the probability that both balls drawn from urn II are black?

2) What is the probability that the balls drawn form urn II are of different color if the balls selected from urn I were white?

3) What is the probability that the balls selected initially from urn I are of the same color, given that the balls drawn from urn II were white?

1) I tried to use the law of Bayes. Here's what I've got:

P(H1) = 2/11 (2 white balls), P(H2)=6/11 (a white and a black ball), P(H3)=3/11 (two black balls). If H1 happens, we get 6 white and 3 black balls in Urn II, if H2 - 5 white and 4 black, if H3 - 4 white and 5 black.

P(A|H1)=3/9=1/3. P(A|H2)=4/9. P(A|H3)=5/9.

P(A)=P(H1)*P(A|H1)+P(H2)*P(A|H2)+P(H3)*P(A|H3)=45/99=5/11. Is it correct? I have a feeling that I should have shown that there are 2 balls, not just one, but I don't know how to do it.

2) There are 6 white balls and 3 black balls in Urn II by condition. So I'm using combinations: P(A)=(6*3)/(9!/2!*7!)=3/6=1/2.

3) I'm also not sure about this one, but it seems that I have to construct a probability tree.

I would greatly appreciate it if you could help me solve it or give me a piece of advice. Thank you in advance!

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  • $\begingroup$ $P(A\mid H_1)=\frac39\frac28$ (two balls must be black, not one). Likewise for the other $P(A\mid H_i)$ $\endgroup$ – drhab Oct 23 '17 at 12:08
  • $\begingroup$ Thank you so much! I don't know why I didn't think about it earlier... $\endgroup$ – Christian Oct 23 '17 at 12:14
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Let $B$ denote the event that the balls taken from urn I are both black.

Let $D$ denote the event that the balls taken from urn I have different color.

Let $W$ denote the event that the balls taken from urn I are both white.

2) Let $E$ denote the event that the balls taken from urn II have different color.

Then: $$\mathsf P(E\mid W)=\frac69\frac38+\frac39\frac68=\frac12$$

3) Let $G$ denote the event that the balls taken from urn II are white.

Then: $$\mathsf P(B\cup W\mid G)=\frac{\mathsf P((B\cup W)\cap G)}{\mathsf P(G)}=\frac{\mathsf P(B\cap G)+\mathsf P(W\cap G)}{\mathsf P(G)}=\frac{\mathsf P(G\mid B)P(B)+\mathsf P(G\mid W)P(W)}{\mathsf P(G\mid B)P(B)+\mathsf P(G\mid W)P(W)+P(G\mid D)P(D)}$$

Try to work this out yourself.

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