Question

One die has two faces marked 1, two faces marked 2, one face marked 3, and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3, and one face marked 4. The probability of getting the sum of numbers to be 4 or 5 when both the dice are thrown together is:

• A. \( \frac{1}{2} \)
• B. \( \frac{3}{5} \)
• C. \( \frac{2}{3} \)
• D. \( \frac{4}{9} \)

Hint:

When calculating probabilities for dice rolls, list all favorable outcomes and divide by the total number of outcomes (in this case, 36).

Answer 1

The Correct Option is A

To determine the probability that the sum of numbers on two special dice is either 4 or 5, we analyze the possible outcomes and their probabilities.

1. Probability Distributions:
For die \( D_1 \):
- \( P(D_1=1) = \frac{1}{3} \)
- \( P(D_1=2) = \frac{1}{3} \)
- \( P(D_1=3) = \frac{1}{6} \)
- \( P(D_1=4) = \frac{1}{6} \)

For die \( D_2 \):
- \( P(D_2=1) = \frac{1}{6} \)
- \( P(D_2=2) = \frac{1}{3} \)
- \( P(D_2=3) = \frac{1}{3} \)
- \( P(D_2=4) = \frac{1}{6} \)

2. Calculating Probability for Sum = 4:
Possible combinations:
- (1, 3): \( \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \)
- (2, 2): \( \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \)
- (3, 1): \( \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \)

Total probability: \( \frac{1}{9} + \frac{1}{9} + \frac{1}{36} = \frac{1}{4} \)

3. Calculating Probability for Sum = 5:
Possible combinations:
- (1, 4): \( \frac{1}{3} \times \frac{1}{6} = \frac{1}{18} \)
- (2, 3): \( \frac{1}{3} \times \frac{1}{3} = \frac{1}{9} \)
- (3, 2): \( \frac{1}{6} \times \frac{1}{3} = \frac{1}{18} \)
- (4, 1): \( \frac{1}{6} \times \frac{1}{6} = \frac{1}{36} \)

Total probability: \( \frac{1}{18} + \frac{1}{9} + \frac{1}{18} + \frac{1}{36} = \frac{1}{4} \)

4. Final Probability Calculation:
The combined probability for sums of 4 or 5 is:
\( \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \)

Final Answer:
The probability is \(\boxed{\dfrac{1}{2}}\).