Question

A fair six-sided die is rolled \(4\) times independently. If \(p\) is the probability that the sum of the outcomes is \(14\), then \(10p\) equals (rounded off to three decimal places).

Hint:

For dice-sum problems, use the stars-and-bars method together with inclusion-exclusion to account for upper bounds.

Answer 1

Correct Answer: 1.127

Step 1: Translate the sum.
Need solutions of $x_1+x_2+x_3+x_4=14$ with each $1\le x_i\le6$. Put $y_i=x_i-1$ to get $y_1+\cdots+y_4=10$ with $0\le y_i\le5$.

Step 2: Count without the cap.
Non-negative solutions number $\binom{13}{3}=286$.

Step 3: Remove the overflow.
If one $y_i\ge6$, set $z=y_i-6$ to get a sum of $4$, with $\binom73=35$ solutions; four variables give $140$. No two can overflow at once. So valid count is $286-140=146$.

Step 4: Form $10p$.
Total outcomes are $6^4=1296$, so $p=\frac{146}{1296}$ and $10p=\frac{1460}{1296}=1.1265\ldots$

Step 5: Round.
\[ \boxed{1.127} \]