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:

• A. \( \frac{5}{216} \)
• B. \( \frac{4}{27} \)
• C. \( \frac{35}{216} \)
• D. \( \frac{20}{216} \)

Hint:

List all valid cases systematically to compute probability.

Answer 1

The Correct Option is C

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).