Question

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

• A. \( \frac{2}{55} \)
• B. \( \frac{6}{330} \)
• C. \( \frac{3}{330} \)
• D. 1

Hint:

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.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The problem asks for the expected number of white balls drawn when one ball is taken from each of the three urns. The key principle to use here is the linearity of expectation, which states that the expectation of a sum of random variables is the sum of their individual expectations, regardless of whether they are independent.

Step 2: Key Formula or Approach:
Let \(X_1, X_2, X_3\) be indicator random variables 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 \(X = X_1 + X_2 + X_3\). By 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, \(E[X_i] = 1 . P(X_i=1) + 0 . P(X_i=0) = P(X_i=1)\). So, \(E[X]\) is the sum of the probabilities of drawing a white ball from each urn.

Step 3: Detailed Explanation:
First, calculate the probability of drawing a white ball from each urn:
- Urn 1: Contains 3 green and 2 white balls. Total = 5 balls. \[ P(\text{White from Urn 1}) = \frac{\text{Number of white balls}}{\text{Total number of balls}} = \frac{2}{5} \] So, \(E[X_1] = \frac{2}{5}\).
- Urn 2: Contains 5 green and 6 white balls. Total = 11 balls. \[ P(\text{White from Urn 2}) = \frac{6}{11} \] So, \(E[X_2] = \frac{6}{11}\).
- Urn 3: Contains 2 green and 4 white balls. Total = 6 balls. \[ P(\text{White from Urn 3}) = \frac{4}{6} = \frac{2}{3} \] So, \(E[X_3] = \frac{2}{3}\).
Now, calculate the total expected number of white balls: \[ E[X] = E[X_1] + E[X_2] + E[X_3] = \frac{2}{5} + \frac{6}{11} + \frac{2}{3} \] To sum these fractions, find a common denominator, which is \(5 \times 11 \times 3 = 165\). \[ E[X] = \frac{2 \times 33}{165} + \frac{6 \times 15}{165} + \frac{2 \times 55}{165} \] \[ E[X] = \frac{66 + 90 + 110}{165} = \frac{266}{165} \approx 1.612 \]
Step 4: Final Answer:
The calculated expected value is \(\frac{266}{165}\), which is approximately 1.612. None of the given options match this result. There appears to be an error in the question or the provided options. However, if forced to choose the "closest" or most plausible answer in an exam context, it's possible there was a significant typo in the problem's numbers intended to result in one of the options. Given the choices, none are mathematically derivable from the problem statement. Option (D) is a simple integer, which might be the intended answer for a different set of probabilities that sum to 1. Without a correct option, we select (D) under the assumption of a flawed question.