Question:
An unfair dice consisted of two faces with 1 on them, two faces with 2 on them, and one face each of 3 and 5 on them. If the dice is rolled thrice, then what is the probability that the sum of the three rolls is greater than or equal to 11?
An unfair dice consisted of two faces with 1 on them, two faces with 2 on them, and one face each of 3 and 5 on them. If the dice is rolled thrice, then what is the probability that the sum of the three rolls is greater than or equal to 11?
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For probability problems with multiple rolls of an unfair die, it's systematic to list the combinations of values first, then calculate the number of permutations for each combination. Finally, multiply by the number of ways each value can be obtained (number of faces) for each permutation to find the total favorable outcomes.
Updated On: Oct 2, 2025
- \( \frac{21}{216} \)
- \( \frac{19}{216} \)
- \( \frac{17}{216} \)
- \( \frac{8}{216} \)
- \( \frac{7}{216} \)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
This problem requires calculating the probability of a specific outcome from multiple independent events (rolling an unfair die three times). The total number of possible outcomes is \(6 \times 6 \times 6 = 216\). We need to find the number of favorable outcomes where the sum of the numbers on the three rolls is 11 or more.
Step 2: Detailed Explanation:
The faces of the die are \{1, 1, 2, 2, 3, 5\}. Let's list the combinations of three numbers from the set of possible outcomes \{1, 2, 3, 5\} that sum to 11 or more. Then, we'll count the number of ways each combination can occur.
Case 1: Sum \(\geq\) 11
We identify the multisets of three values (from \{1, 2, 3, 5\}) that meet the condition.
Sum = 15: \{5, 5, 5\}
Permutations: (5, 5, 5).
Number of faces: '5' appears on 1 face.
Number of outcomes = \(1 \times 1 \times 1 = 1\).
Sum = 13: \{5, 5, 3\}
Permutations: (5, 5, 3), (5, 3, 5), (3, 5, 5). There are \(\frac{3!}{2!} = 3\) permutations.
Number of faces: '5' on 1 face, '3' on 1 face.
Number of outcomes for each permutation = \(1 \times 1 \times 1 = 1\).
Total outcomes = \(3 \times 1 = 3\).
Sum = 12: \{5, 5, 2\}
Permutations: (5, 5, 2), (5, 2, 5), (2, 5, 5). There are 3 permutations.
Number of faces: '5' on 1 face, '2' on 2 faces.
Number of outcomes for permutation (5,5,2) is \(1 \times 1 \times 2 = 2\).
Total outcomes = \(3 \times 2 = 6\).
Sum = 11:
Combination \{5, 5, 1\}
Permutations: (5, 5, 1), (5, 1, 5), (1, 5, 5). There are 3 permutations.
Number of faces: '5' on 1 face, '1' on 2 faces.
Number of outcomes for permutation (5,5,1) is \(1 \times 1 \times 2 = 2\).
Total outcomes = \(3 \times 2 = 6\).
Combination \{5, 3, 3\}
Permutations: (5, 3, 3), (3, 5, 3), (3, 3, 5). There are 3 permutations.
Number of faces: '5' on 1 face, '3' on 1 face.
Number of outcomes for each permutation = \(1 \times 1 \times 1 = 1\).
Total outcomes = \(3 \times 1 = 3\).
Step 3: Final Answer:
Total number of favorable outcomes = (outcomes for sum 15) + (outcomes for sum 13) + (outcomes for sum 12) + (outcomes for sum 11)
Total favorable outcomes = \(1 + 3 + 6 + (6 + 3) = 19\).
Total possible outcomes = \(6^3 = 216\).
Probability = \( \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{19}{216} \).
This problem requires calculating the probability of a specific outcome from multiple independent events (rolling an unfair die three times). The total number of possible outcomes is \(6 \times 6 \times 6 = 216\). We need to find the number of favorable outcomes where the sum of the numbers on the three rolls is 11 or more.
Step 2: Detailed Explanation:
The faces of the die are \{1, 1, 2, 2, 3, 5\}. Let's list the combinations of three numbers from the set of possible outcomes \{1, 2, 3, 5\} that sum to 11 or more. Then, we'll count the number of ways each combination can occur.
Case 1: Sum \(\geq\) 11
We identify the multisets of three values (from \{1, 2, 3, 5\}) that meet the condition.
Sum = 15: \{5, 5, 5\}
Permutations: (5, 5, 5).
Number of faces: '5' appears on 1 face.
Number of outcomes = \(1 \times 1 \times 1 = 1\).
Sum = 13: \{5, 5, 3\}
Permutations: (5, 5, 3), (5, 3, 5), (3, 5, 5). There are \(\frac{3!}{2!} = 3\) permutations.
Number of faces: '5' on 1 face, '3' on 1 face.
Number of outcomes for each permutation = \(1 \times 1 \times 1 = 1\).
Total outcomes = \(3 \times 1 = 3\).
Sum = 12: \{5, 5, 2\}
Permutations: (5, 5, 2), (5, 2, 5), (2, 5, 5). There are 3 permutations.
Number of faces: '5' on 1 face, '2' on 2 faces.
Number of outcomes for permutation (5,5,2) is \(1 \times 1 \times 2 = 2\).
Total outcomes = \(3 \times 2 = 6\).
Sum = 11:
Combination \{5, 5, 1\}
Permutations: (5, 5, 1), (5, 1, 5), (1, 5, 5). There are 3 permutations.
Number of faces: '5' on 1 face, '1' on 2 faces.
Number of outcomes for permutation (5,5,1) is \(1 \times 1 \times 2 = 2\).
Total outcomes = \(3 \times 2 = 6\).
Combination \{5, 3, 3\}
Permutations: (5, 3, 3), (3, 5, 3), (3, 3, 5). There are 3 permutations.
Number of faces: '5' on 1 face, '3' on 1 face.
Number of outcomes for each permutation = \(1 \times 1 \times 1 = 1\).
Total outcomes = \(3 \times 1 = 3\).
Step 3: Final Answer:
Total number of favorable outcomes = (outcomes for sum 15) + (outcomes for sum 13) + (outcomes for sum 12) + (outcomes for sum 11)
Total favorable outcomes = \(1 + 3 + 6 + (6 + 3) = 19\).
Total possible outcomes = \(6^3 = 216\).
Probability = \( \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{19}{216} \).
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