Question:medium

Three urns contain 3 green and 2 white balls, 5 green and 6 white balls and 2 green and 4 white balls respectively. One ball is drawn at random from each of the urn. Then, the expected number of white balls drawn, is

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Linearity of Expectation (\(E[X+Y] = E[X] + E[Y]\)) is a powerful tool. For problems asking for the 'expected number' of occurrences, consider using indicator variables. It's often much faster than calculating the full probability distribution. Also, be aware that questions in exams can sometimes be flawed; in such cases, double-check your method and calculations.
Updated On: Feb 18, 2026
  • \( \frac{2}{55} \)
  • \( \frac{6}{330} \)
  • \( \frac{3}{330} \)
  • 1
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Concept Explanation:
The problem requires finding the expected number of white balls drawn when selecting one ball from each of the three urns. This is solved using the linearity of expectation: the expected value of a sum of random variables equals the sum of their individual expected values, regardless of independence.

Step 2: Formula and Approach:
Define indicator random variables \(X_1, X_2, X_3\) for drawing a white ball from Urn 1, Urn 2, and Urn 3, respectively. \(X_i = 1\) if a white ball is drawn from Urn \(i\), and \(X_i = 0\) otherwise. The total number of white balls drawn is \(X = X_1 + X_2 + X_3\). Applying the linearity of expectation: \[ E[X] = E[X_1 + X_2 + X_3] = E[X_1] + E[X_2] + E[X_3] \] For an indicator variable, \(E[X_i] = 1 . P(X_i=1) + 0 . P(X_i=0) = P(X_i=1)\). Therefore, \(E[X]\) is the sum of the probabilities of drawing a white ball from each urn.

Step 3: Detailed Calculation:
Calculate the probability of drawing a white ball from each urn:
- Urn 1: Contains 3 green and 2 white balls (Total: 5). \[ P(\text{White from Urn 1}) = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{2}{5} \] Thus, \(E[X_1] = \frac{2}{5}\).
- Urn 2: Contains 5 green and 6 white balls (Total: 11). \[ P(\text{White from Urn 2}) = \frac{6}{11} \] Thus, \(E[X_2] = \frac{6}{11}\).
- Urn 3: Contains 2 green and 4 white balls (Total: 6). \[ P(\text{White from Urn 3}) = \frac{4}{6} = \frac{2}{3} \] Thus, \(E[X_3] = \frac{2}{3}\).
Now, calculate the total expected number of white balls: \[ E[X] = E[X_1] + E[X_2] + E[X_3] = \frac{2}{5} + \frac{6}{11} + \frac{2}{3} \] Find a common denominator: \(5 \times 11 \times 3 = 165\). \[ E[X] = \frac{2 \times 33}{165} + \frac{6 \times 15}{165} + \frac{2 \times 55}{165} \] \[ E[X] = \frac{66 + 90 + 110}{165} = \frac{266}{165} \approx 1.612 \]
Step 4: Final Conclusion:
The calculated expected value is \(\frac{266}{165}\), approximately 1.612. None of the provided options match this result, indicating a potential error in the question or options. Selecting (D) is based on the assumption of a flawed question, where a simple integer might have been the intended answer.
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