Let U1 and U2 be two urns such that U1 contains 3 white and 2 red balls, and U2 contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U1 and put into U2. However, if tail appears then 2 balls are drawn at random from U1 and put into U2. Now 1 ball is drawn at random from U2.

Engineering

Mathematics

Total Probability Theorem and Bayes Theorem

Question

Let U₁ and U₂ be two urns such that U₁ contains 3 white and 2 red balls, and U₂ contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U₁ and put into U₂. However, if tail appears then 2 balls are drawn at random from U₁ and put into U₂. Now 1 ball is drawn at random from U₂.

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Linked Question 1

The probability of the drawn ball from U₂ being white is

$\frac{11}{30}$

$\frac{19}{30}$

$\frac{23}{30}$

$\frac{13}{30}$

Required probability = P(H)[P(W/H) × P(W₂) + P(R/H)P(W₂)] + P(T) $[P (\frac{bothW}{T}) P (W_{2}) + P (\frac{bothR}{T})$

$P (W_{2}) + P (\frac{R_{1} & W_{1}}{T}) P (W_{2})]$

$= \frac{1}{2} [\frac{3}{5} \times 1 + \frac{2}{5} \times \frac{1}{2}] + \frac{1}{2} [\frac{^{3} C_{2}}{^{5} C_{2}} \times 1 + \frac{^{2} C_{2}}{^{5} C_{2}} \times \frac{1}{3} + \frac{^{3} C_{2} \times^{3} C_{2}}{^{5} C_{2}} \times \frac{2}{3}]$

$= \frac{1}{2} [\frac{3}{5} + \frac{1}{5}] + \frac{1}{2} [\frac{3}{10} + \frac{1}{30} + \frac{2}{5}] = \frac{2}{5} + \frac{11}{30} = \frac{23}{30}$

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Linked Question 2

Given that the drawn ball from U₂ is white, the probability that head appeared on the coin is :

$\frac{17}{23}$

$\frac{12}{23}$

$\frac{11}{23}$

$\frac{15}{23}$

Required probability

$= \frac{p (H) [P (\frac{W_{1}}{H}) P (W_{2}) + P (\frac{R_{1}}{H}) P (W_{2})]}{p (H) [P (\frac{W_{1}}{H}) P (W_{2}) + P (\frac{R_{1}}{H}) P (W_{2})] + p (T) [P (\frac{bothW}{T}) P (W_{2}) + P (\frac{bothR}{T}) P (W_{2}) + P (\frac{R_{1} & W_{1}}{T}) P (W_{2})]}$

$= \frac{\frac{1}{2} [\frac{3}{5} \times 1 + \frac{2}{5} \times \frac{1}{2}]}{\frac{23}{30}} = \frac{12}{13}$

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