There are two balls in an urn whose colours are not known (each ball c...
Let E
i (0
< i
< 2) denote the event that urn contains 'i' white and '(2 – i)' black balls.
Let A denote the event that a white ball is drawn from the urn.
We have P(E
i) = 1/3 for i = 0, 1, 2
P (A|E
1) = 1/3, P(A|E
2) = 2/3, P(A|E
3) = 1.
By the total probability rule,
P(A)= P(E
1)P(A|E
1) + P(E
2)P(A|E
2) + P(E
3)P(A|E
3)
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There are two balls in an urn whose colours are not known (each ball c...
Understanding the Problem
In the urn, we have two balls whose colors are unknown, and we add a white ball. We need to determine the probability of drawing a white ball.
Possible Scenarios
Initially, the two balls can be in any of the following combinations:
- Scenario 1: Both balls are white (WW)
- Scenario 2: One ball is white, and one is black (WB)
- Scenario 3: One ball is black, and one is white (BW)
- Scenario 4: Both balls are black (BB)
Calculating Probabilities
Since the color of the two initial balls is unknown, we assume each scenario is equally likely:
- Total Scenarios: 4 (WW, WB, BW, BB)
- Adding a white ball to the urn:
Now let's analyze the probability of drawing a white ball in each scenario:
- Scenario WW: 3 white balls (WW + W) → Probability of drawing white = 3/3 = 1
- Scenario WB or BW: 2 white balls and 1 black ball → Probability of drawing white = 2/3
- Scenario BB: 1 white ball and 2 black balls → Probability of drawing white = 1/3
Calculating Overall Probability
Now, to find the overall probability of drawing a white ball:
- WW: 1/4 * 1 = 1/4
- WB: 1/4 * 2/3 = 1/6
- BW: 1/4 * 2/3 = 1/6
- BB: 1/4 * 1/3 = 1/12
Adding these probabilities gives:
1/4 + 1/6 + 1/6 + 1/12 = 9/12 = 3/4
Now, since we are only interested in the scenarios with at least one white ball (WW, WB, BW):
- Total Probability of scenarios with white = (1/4 + 1/6 + 1/6) = 2/3
Final Answer
Therefore, the probability that the drawn ball is white is 2/3, which matches option 'C'.
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