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: Form the equation for the outcomes.
Let the outcomes of the four dice rolls be
\[ x_1,x_2,x_3,x_4 \] where each
\[ 1\leq x_i\leq 6 \] We need the number of solutions of
\[ x_1+x_2+x_3+x_4=14 \]

Step 2: Remove the lower bound.
Define
\[ y_i=x_i-1 \] Then
\[ y_i\geq 0 \] and the equation becomes
\[ y_1+y_2+y_3+y_4=10 \] without considering the upper bound initially.
The number of non-negative integer solutions is
\[ \binom{10+4-1}{4-1} = \binom{13}{3} = 286 \]

Step 3: Apply the upper bound restriction.
Since
\[ x_i\leq 6, \] we have
\[ y_i\leq 5 \] Now subtract the invalid solutions where some
\[ y_i\geq 6 \] Suppose
\[ y_1\geq 6 \] Let
\[ z_1=y_1-6 \] Then
\[ z_1+y_2+y_3+y_4=4 \] The number of solutions is
\[ \binom{4+4-1}{3} = \binom{7}{3} = 35 \] Similarly for each variable. Hence total invalid solutions are
\[ 4\times 35=140 \] No two variables can simultaneously exceed \(5\), because then the sum would exceed \(10\).
Therefore, the valid number of outcomes is
\[ 286-140=146 \]

Step 4: Compute the probability.
Total possible outcomes are
\[ 6^4=1296 \] Hence,
\[ p=\frac{146}{1296} \] Therefore,
\[ 10p=\frac{1460}{1296} \] \[ =1.126543\ldots \] Rounded off to three decimal places,
\[ 1.127 \]

Step 5: Final conclusion.
Hence,
\[ \boxed{1.127} \]