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Probability question

2011 · Shift 1 · Q37
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Probability question

2011 · Shift 1 · Q37

JEE AdvancedMathematicsProbabilityMCQ+4 / −1
Let U1{U_1}U1​ and U2{U_2}U2​ be two urns such that U1{U_1}U1​ contains 333 white and 222 red balls, and U2{U_2}U2​ contains only 111 white ball. A fair coin is tossed. If head appears then 111 ball is drawn at random from U1{U_1}U1​ and put into U2{U_2}U2​. However, if tail appears then 222 balls are drawn at random from U1{U_1}U1​ and put into U2{U_2}U2​. Now 111 ball is drawn at random from U2{U_2}U2​ being white isGiven that the drawn ball from U2{U_2}U2​ is white, the probability that head appeared on the coin is
  1. A
    1723{{17} \over {23}}2317​
  2. B
    1123{{11} \over {23}}2311​
  3. C
    1523{{15} \over {23}}2315​
  4. D
    1223{{12} \over {23}}2312​
View written solutionFree

Correct answer: D

  1. Define events

Let:

  • HHH = head appears
  • TTT = tail appears
  • WWW = the ball finally drawn from U2U_2U2​ is white

Since the coin is fair, P(H)=P(T)=12.P(H)=P(T)=\frac12.P(H)=P(T)=21​.

We need: P(H∣W)=P(H∩W)P(W).P(H\mid W)=\frac{P(H\cap W)}{P(W)}.P(H∣W)=P(W)P(H∩W)​.

  1. Case 1: Head appears

If head appears, exactly 111 ball is transferred from U1U_1U1​ to U2U_2U2​.

Initially:

  • U1U_1U1​: 3W,2R3W, 2R3W,2R
  • U2U_2U2​: 1W1W1W

When one ball is transferred from U1U_1U1​:

  • transferred white with probability 35\frac3553​
  • transferred red with probability 25\frac2552​

Now compute probability that a random ball drawn from U2U_2U2​ is white.

If white is transferred:

Then U2U_2U2​ has 222 white balls, so P(W∣H,white transferred)=1.P(W\mid H, \text{white transferred})=1.P(W∣H,white transferred)=1.

If red is transferred:

Then U2U_2U2​ has 1W,1R1W,1R1W,1R, so P(W∣H,red transferred)=12.P(W\mid H, \text{red transferred})=\frac12.P(W∣H,red transferred)=21​.

Hence,

=\frac35+\frac15 =\frac45.$$ Thus, $$P(H\cap W)=P(H)P(W\mid H)=\frac12\cdot \frac45=\frac25.$$ --- 3. **Case 2: Tail appears** If tail appears, exactly $2$ balls are transferred from $U_1$ to $U_2$. We find the composition of the two transferred balls. Total ways to choose $2$ balls from $5$: $$\binom52=10.$$ ### (i) Two whites transferred Ways: $$\binom32=3$$ So probability is $$\frac{3}{10}.$$ Then $U_2$ has $3$ white balls, hence $$P(W\mid T, 2W)=1.$$ ### (ii) One white and one red transferred Ways: $$\binom31\binom21=3\cdot 2=6$$ So probability is $$\frac{6}{10}=\frac35.$$ Then $U_2$ has $2W,1R$, hence $$P(W\mid T, 1W1R)=\frac23.$$ ### (iii) Two reds transferred Ways: $$\binom22=1$$ So probability is $$\frac{1}{10}.$$ Then $U_2$ has $1W,2R$, hence $$P(W\mid T, 2R)=\frac13.$$ Therefore, $$P(W\mid T)=\frac{3}{10}\cdot 1+\frac35\cdot \frac23+\frac{1}{10}\cdot \frac13.$$ Now simplify: $$P(W\mid T)=\frac{3}{10}+\frac{2}{5}+\frac{1}{30}.$$ Taking LCM $30$, $$P(W\mid T)=\frac{9}{30}+\frac{12}{30}+\frac{1}{30}=\frac{22}{30}=\frac{11}{15}.$$ Thus, $$P(T\cap W)=P(T)P(W\mid T)=\frac12\cdot \frac{11}{15}=\frac{11}{30}.$$ --- 4. **Compute $P(W)$** $$P(W)=P(H\cap W)+P(T\cap W)=\frac25+\frac{11}{30}.$$ Taking LCM $30$, $$P(W)=\frac{12}{30}+\frac{11}{30}=\frac{23}{30}.$$ --- 5. **Apply Bayes' theorem** $$P(H\mid W)=\frac{P(H\cap W)}{P(W)} =\frac{\frac25}{\frac{23}{30}}.$$ So, $$P(H\mid W)=\frac25\cdot \frac{30}{23}=\frac{12}{23}.$$ --- 6. **Match with options** $$\boxed{\frac{12}{23}}$$ This corresponds to **Option D**.
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