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This question is already posted here ,but i want to know flaw in my approach.

Question

Let $U_1$ and $U_2$ be two urns such that $U_1$ contains $3$ white and $2$ red balls, and $U_2$ contains only $1$ white ball.

A fair coin is tossed:

  • If head appears then $1$ ball is drawn at random from $U_1$ and put into $U_2$.
  • If tail appears then $2$ balls are drawn at random from $U_1$ and put into $U_2$.

Now $1$ ball is drawn at random from $U_2$,What is the probability of the drawn ball from $U_2$ being white?

My Approach

There are $2$ cases.

  • $\text{Case 1(Head Appears)}$

If Head Appears,then only $1$ ball will be drawn from $U_1$(white/red) to $U_2$

Current possible status of $U_2$

$WW,RW$

i.e $$\frac{1}{2}\times( \frac{2}{2}+\frac{1}{2} )$$

  • $\text{Case 2(Tail Appears)}$

If Tail Appears,then $2$ balls will be drawn from $U_1$(WW,WR,RR) to $U_2$

Current possible status of $U_2$

$WWW,WRW,RRW$

i.e $$\frac{1}{2}\times( \frac{1}{1}+\frac{2}{3} +\frac{1}{3})$$

Total Probability =$\text{Case 1+case 2}$

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

But my answer is coming $>1$,So it is absolutely wrong.But i am not getting where i am doing wrong.

Please help

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  • $\begingroup$ You have not accounted for the probability of selecting a red/white ball from urn 1. $\endgroup$ – N. F. Taussig May 21 '18 at 11:58
  • $\begingroup$ @N.F.Taussig sir: so it would be better to use Bayes' theorem ..?? $\endgroup$ – laura May 21 '18 at 12:00
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    $\begingroup$ We need to use conditional probabilities, but this is not an application of Bayes' theorem since we not working with posterior probabilities. $\endgroup$ – N. F. Taussig May 21 '18 at 12:56
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Following your approach it should be:

Case 1 (Head Appears): $U_2=WW$ with probability $\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}$ and $U_2=RW$ with probability $\frac{1}{2}\cdot\frac{2}{5}=\frac{2}{10}$.

Case 2 (Tail Appears): $U_2=WWW$ with probability $\frac{1}{2}\cdot\frac{3}{10}=\frac{3}{20}$, $U_2=RWW$ with probability $\frac{1}{2}\cdot\frac{6}{10}=\frac{6}{20}$ and $U_2=RRW$ with probability $\frac{1}{2}\cdot\frac{1}{10}=\frac{1}{20}$.

Hence the probability of the drawn ball from $U_2$ being white is: $$\frac{3}{10}+\frac{1}{2}\cdot \frac{2}{10}+\frac{3}{20}+\frac{2}{3}\cdot \frac{6}{20}+\frac{1}{3}\cdot \frac{1}{20}=\frac{23}{30}.$$

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  • $\begingroup$ I am not getting how you got $\frac{3}{5}$ and $\frac{2}{5}$ in case $1$ $\endgroup$ – laura May 21 '18 at 12:19
  • $\begingroup$ @laura The probability of selecting a white ball from urn 1 is $3/5$; the probability of selecting a red ball from urn 1 is $2/5$. $\endgroup$ – N. F. Taussig May 21 '18 at 12:41
  • $\begingroup$ $U_1$ contains 3 white and 2 red balls. $\endgroup$ – Robert Z May 21 '18 at 12:41
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Refer to the probability tree diagram:

enter image description here

Hence, the required probability is: $$\frac12\cdot \frac35+\frac12\cdot\frac25\cdot\frac12+\frac12\cdot\frac35\cdot\frac24+\frac12\cdot\frac35\cdot\frac24\cdot\frac23+\frac12\cdot\frac35\cdot\frac24\cdot\frac23+\frac12\cdot\frac25\cdot\frac14\cdot\frac13=\\ \frac{18+6+9+6+6+1}{60}=\frac{46}{60}=\frac{23}{30}.$$

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