Question

A dice is rolled three times and the sum of the three numbers on the uppermost face is 15. What is the chance that the first roll was a four?

• A. \(\frac{2}{5}\)
• B. \(\frac{1}{5}\)
• C. \(\frac{1}{6}\)
• D. None of these

Hint:

First list all dice combinations that add to 15, then count which ones start with a 4.

Answer 1

The Correct Option is B

Step 1: List the ways three dice can add up to 15.
Each die shows a value from 1 to 6, so the biggest possible total is 18. A total of 15 is 3 less than that. The sets of three numbers (order not fixed yet) that add to 15 are \(5,5,5\), \(4,5,6\), and \(3,6,6\). No other combination of three numbers from 1 to 6 gives 15.

Step 2: Count the arrangements for each set.
The set \(5,5,5\) has all three dice equal, so it gives only 1 arrangement. The set \(4,5,6\) has three different numbers, so it gives \(3! = 6\) arrangements. The set \(3,6,6\) has two equal numbers, so it gives \(\frac{3!}{2!} = 3\) arrangements. Adding these up, \(1 + 6 + 3 = 10\) total arrangements give a sum of 15.

Step 3: Pick out the arrangements where the first roll is 4.
A 4 can only appear in the set \(4,5,6\), since the other two sets have no 4 in them. Within \(4,5,6\), the first die is 4 in exactly 2 of the 6 arrangements: \(4,5,6\) and \(4,6,5\).

Step 4: Check the wrong options.
Option A, \(\frac{2}{5}\), would come from an arithmetic slip such as counting 4 out of 10 instead of 2 out of 10, so it is wrong. Option C, \(\frac{1}{6}\), treats the first roll as independent of the condition that the total is 15, which ignores the given information, so it is wrong too.

Final Answer:
Out of 10 equally likely arrangements that sum to 15, 2 have a first roll of 4, so the chance is \[ \boxed{\frac{2}{10} = \frac{1}{5}} \]