Question

Three fair dice are thrown. What is the probability of getting a total of 15 given that they exhibit three different numbers that are in arithmetic progression?

• A. \( \frac{1}{8} \)
• B. \( \frac{1}{6} \)
• C. \( \frac{1}{4} \)
• D. \( \frac{1}{2} \)

Hint:

In conditional probability, always restrict the sample space first based on the given condition, then count favorable cases.

Answer 1

The Correct Option is B

Step 1: Understand the given condition.
We are given that three dice show distinct numbers in arithmetic progression (AP). We need to find the probability that their sum is 15 under this condition.

Step 2: Find all possible AP triplets from {1,2,3,4,5,6\.}
Three distinct numbers in AP must have equal difference. Possible AP triplets are:
\[ (1,2,3),\ (2,3,4),\ (3,4,5),\ (4,5,6),\ (1,3,5),\ (2,4,6). \]

Step 3: Count total outcomes satisfying the condition.
Each triplet can appear in \(3! = 6\) different permutations (since dice are distinguishable).
So total favorable outcomes under condition:
\[ 6 \times 6 = 36. \]

Step 4: Find triplets whose sum is 15.
Check sums:
\[ (4,5,6) \Rightarrow 4+5+6=15. \]
This is the only AP triplet with sum 15.

Step 5: Count favorable outcomes.
This triplet also has \(3! = 6\) permutations.
So favorable outcomes = 6.

Step 6: Compute conditional probability.
\[ P = \frac{\text{favorable outcomes}}{\text{total AP outcomes}} = \frac{6}{36}. \]

Step 7: Simplify the result.
\[ P = \frac{1}{6}. \]
Final Answer:
\[ \boxed{\frac{1}{6}}. \]