An unbiased die is tossed twice. Find the probability of getting a 1, 2, 3 or 4 on the first toss and a 4, 5 or 6 on the second toss.

Options

A. `1/3`
B. `2/3`
C. `5/6`
D. `1/6`

Answer

To solve the problem of finding the probability of getting a 1, 2, 3, or 4 on the first toss and a 4, 5, or 6 on the second toss of an unbiased die, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Events**:
- Let Event A be getting a 1, 2, 3, or 4 on the first toss.
- Let Event B be getting a 4, 5, or 6 on the second toss.

2. **Calculate the Probability of Event A**:
- The total outcomes when tossing a die are 6 (1, 2, 3, 4, 5, 6).
- The favorable outcomes for Event A (1, 2, 3, 4) are 4.
- Therefore, the probability of Event A, \( P(A) \), is given by:
\[ P(A) = \frac{\text{Number of favorable outcomes for A}}{\text{Total outcomes}} = \frac{4}{6} = \frac{2}{3} \]

3. **Calculate the Probability of Event B**:
- The total outcomes when tossing a die are still 6.
- The favorable outcomes for Event B (4, 5, 6) are 3.
- Therefore, the probability of Event B, \( P(B) \), is given by:
\[ P(B) = \frac{\text{Number of favorable outcomes for B}}{\text{Total outcomes}} = \frac{3}{6} = \frac{1}{2} \]

4. **Determine the Combined Probability**:
- Since the two events (tossing the die twice) are independent, the probability of both events occurring (Event A and Event B) is given by:
\[ P(A \cap B) = P(A) \times P(B) \]
- Substituting the probabilities we calculated:
\[ P(A \cap B) = \left(\frac{2}{3}\right) \times \left(\frac{1}{2}\right) = \frac{2}{6} = \frac{1}{3} \]

5. **Final Answer**:
- The probability of getting a 1, 2, 3, or 4 on the first toss and a 4, 5, or 6 on the second toss is \( \frac{1}{3} \).