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:
Show Hint
- \( \frac{1}{2} \)
- \( \frac{3}{5} \)
- \( \frac{2}{3} \)
- \( \frac{4}{9} \)
The Correct Option is A
Approach Solution - 1
To determine the probability of getting a sum of 4 or 5 when both dice are thrown, we must first understand the face distribution of the two dice:
- First Die: Two faces with 1, two faces with 2, one face with 3, and one face with 4.
- Second Die: One face with 1, two faces with 2, two faces with 3, and one face with 4.
Now, let's calculate the total number of possible outcomes when throwing both dice. Since each die has 6 faces, the total outcomes are:
\(6 \times 6 = 36\)
Next, we calculate the number of favorable outcomes for each sum.
Sum of 4 Outcomes:
- (1,3) - First die shows 1, second die shows 3.
- (2,2) - First die shows 2, second die shows 2.
- (3,1) - First die shows 3, second die shows 1.
Now count the probability for each combination:
- Probability of (1,3): \(\frac{2}{6} \times \frac{2}{6} = \frac{4}{36}\)
- Probability of (2,2): \(\frac{2}{6} \times \frac{2}{6} = \frac{4}{36}\)
- Probability of (3,1): \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)
Sum of 5 Outcomes:
- (1,4) - First die shows 1, second die shows 4.
- (2,3) - First die shows 2, second die shows 3.
- (3,2) - First die shows 3, second die shows 2.
- (4,1) - First die shows 4, second die shows 1.
Now count the probability for each combination:
- Probability of (1,4): \(\frac{2}{6} \times \frac{1}{6} = \frac{2}{36}\)
- Probability of (2,3): \(\frac{2}{6} \times \frac{2}{6} = \frac{4}{36}\)
- Probability of (3,2): \(\frac{1}{6} \times \frac{2}{6} = \frac{2}{36}\)
- Probability of (4,1): \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\)
Total Probability:
Add up all the probabilities of getting either a sum of 4 or 5:
\(\left(\frac{4}{36} + \frac{4}{36} + \frac{1}{36}\right) + \left(\frac{2}{36} + \frac{4}{36} + \frac{2}{36} + \frac{1}{36}\right) = \frac{18}{36} = \frac{1}{2}\)
Thus, the probability of getting a sum of 4 or 5 is \(\frac{1}{2}\). Therefore, the correct option is:
- \(\frac{1}{2}\)
Approach Solution -2
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 \(\boxed{\dfrac{1}{2}}\).
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