Question:
Two persons A and B throw three unbiased dice one after the other. If A gets the sum 13, then the probability that B gets a higher sum is:
Two persons A and B throw three unbiased dice one after the other. If A gets the sum 13, then the probability that B gets a higher sum is:
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List all valid cases systematically to compute probability.
Updated On: May 16, 2025
- \( \frac{5}{216} \)
- \( \frac{4}{27} \)
- \( \frac{35}{216} \)
- \( \frac{20}{216} \)
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The Correct Option is C
Approach Solution - 1
Step 1: Understanding the Problem Each person throws three unbiased dice, and their sum can range from: \[ \text{Minimum sum} = 1+1+1 = 3, \quad \text{Maximum sum} = 6+6+6 = 18 \] We are given that A obtains a sum of 13, and we need to find the probability that B obtains a sum greater than 13.
Step 2: Ways to obtain a sum of 13 We determine how many ways three dice can sum up to 13: Possible triplets that give 13: \[ (6,6,1), (6,5,2), (6,4,3), (5,6,2), (5,5,3), (5,4,4), (4,6,3), (4,5,4) \] Total ways = 21
Step 3: Ways to obtain a sum greater than 13 We count the number of ways to get sums of 14, 15, 16, 17, and 18. - Ways to get 14: \[ (6,6,2), (6,5,3), (6,4,4), (5,6,3), (5,5,4), (4,6,4) \] Total = 15 - Ways to get 15: \[ (6,6,3), (6,5,4), (5,6,4), (5,5,5) \] Total = 10 - Ways to get 16: \[ (6,6,4), (6,5,5), (5,6,5) \] Total = 6 - Ways to get 17: \[ (6,6,5), (6,5,6), (5,6,6) \] Total = 3 - Ways to get 18: \[ (6,6,6) \] Total = 1
Step 4: Compute Probability Total ways to get a sum greater than 13: \[ 15 + 10 + 6 + 3 + 1 = 35 \] Since the total number of ways to roll three dice is: \[ 6^3 = 216 \] The probability is: \[ P = \frac{35}{216} \] Thus, the correct answer is option (3).
Step 2: Ways to obtain a sum of 13 We determine how many ways three dice can sum up to 13: Possible triplets that give 13: \[ (6,6,1), (6,5,2), (6,4,3), (5,6,2), (5,5,3), (5,4,4), (4,6,3), (4,5,4) \] Total ways = 21
Step 3: Ways to obtain a sum greater than 13 We count the number of ways to get sums of 14, 15, 16, 17, and 18. - Ways to get 14: \[ (6,6,2), (6,5,3), (6,4,4), (5,6,3), (5,5,4), (4,6,4) \] Total = 15 - Ways to get 15: \[ (6,6,3), (6,5,4), (5,6,4), (5,5,5) \] Total = 10 - Ways to get 16: \[ (6,6,4), (6,5,5), (5,6,5) \] Total = 6 - Ways to get 17: \[ (6,6,5), (6,5,6), (5,6,6) \] Total = 3 - Ways to get 18: \[ (6,6,6) \] Total = 1
Step 4: Compute Probability Total ways to get a sum greater than 13: \[ 15 + 10 + 6 + 3 + 1 = 35 \] Since the total number of ways to roll three dice is: \[ 6^3 = 216 \] The probability is: \[ P = \frac{35}{216} \] Thus, the correct answer is option (3).
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Approach Solution -2
To solve the problem, we need to determine the probability that person B, rolling three unbiased dice, gets a sum higher than 13 given that person A has achieved a sum of 13. Each die can roll a number between 1 and 6, and the total number of possible outcomes for a single roll of three dice is \(6^3=216\).
Step 1: Count the Outcomes for Sum 13
For sum 13 with three dice, the possible combinations are:
For sum 13 with three dice, the possible combinations are:
- (6,6,1)
- (6,5,2)
- (6,4,3)
- (5,6,2)
- (5,5,3)
- (5,4,4)
- (4,6,3)
- (4,5,4)
- (4,4,5)
- (3,6,4)
- (3,5,5)
- (2,6,5)
- (1,6,6)
Step 2: Determine B's Probabilities with Outcomes Greater than 13
To achieve a sum greater than 13, person B needs to get either 14, 15, 16, 17, or 18. Calculating for these:
To achieve a sum greater than 13, person B needs to get either 14, 15, 16, 17, or 18. Calculating for these:
- Sum 14: (6,6,2), (6,5,3), (6,4,4), (5,6,3), (5,5,4), (4,6,4), (4,5,5), (3,6,5), (2,6,6). 9 outcomes.
- Sum 15: (6,6,3), (6,5,4), (6,4,5), (6,3,6), (5,6,4), (5,5,5), (5,4,6), (4,6,5), (3,6,6). 10 outcomes.
- Sum 16: (6,6,4), (6,5,5), (6,4,6), (5,6,5), (5,5,6), (4,6,6). 6 outcomes.
- Sum 17: (6,6,5), (6,5,6), (5,6,6). 3 outcomes.
- Sum 18: (6,6,6). 1 outcome.
Step 3: Calculate the Probability
Since A already achieved a sum of 13 (15 ways) and B achieved a higher sum with 29 successful ways, the probability B wins is the number of these successful outcomes divided by the total possibilities:
\(P(B>A)=\frac{35}{216}\).
This simplifies to the probability \( \frac{35}{216} \).
Since A already achieved a sum of 13 (15 ways) and B achieved a higher sum with 29 successful ways, the probability B wins is the number of these successful outcomes divided by the total possibilities:
\(P(B>A)=\frac{35}{216}\).
This simplifies to the probability \( \frac{35}{216} \).
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