An urn contains one black ball and one green ball. A second urn contains one white and one green ball. One ball is drawn at random from each urn.
What is the probability of getting at least one green ball?

Options

A. A. `1//2`
B. B. `1//3`
C. C. `1//4`
D. D. `3//4`

Answer

To solve the problem of finding the probability of getting at least one green ball when drawing one ball from each of the two urns, we can follow these steps:

### Step 1: Identify the contents of the urns
- **Urn 1** contains: 1 black ball (B) and 1 green ball (G).
- **Urn 2** contains: 1 white ball (W) and 1 green ball (G).

### Step 2: Determine the total outcomes
When drawing one ball from each urn, we can list the possible outcomes:
1. (B from Urn 1, W from Urn 2)
2. (B from Urn 1, G from Urn 2)
3. (G from Urn 1, W from Urn 2)
4. (G from Urn 1, G from Urn 2)

### Step 3: Count the favorable outcomes for at least one green ball
We want to find the outcomes where at least one green ball is drawn. The favorable outcomes are:
- (B, G)
- (G, W)
- (G, G)
- (G, G)

Thus, all outcomes except (B, W) contain at least one green ball.

### Step 4: Calculate the probability of getting no green balls
The only outcome where no green ball is drawn is (B, W):
- Probability of drawing B from Urn 1 = 1/2
- Probability of drawing W from Urn 2 = 1/2

Therefore, the probability of getting no green balls is:
\[ P(\text{no green}) = P(B) \times P(W) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

### Step 5: Calculate the probability of getting at least one green ball
Using the complement rule:
\[ P(\text{at least one green}) = 1 - P(\text{no green}) \]
\[ P(\text{at least one green}) = 1 - \frac{1}{4} = \frac{3}{4} \]

### Conclusion
The probability of getting at least one green ball is:
\[ \frac{3}{4} \]

### Final Answer
Thus, the answer is \( \frac{3}{4} \).

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