Calculating Probability of a Perfect Square Sum with Three Dice

The question asks for the probability of getting a sum that is a perfect square when three standard dice are thrown.

Understanding the Basics of Probability

Probability is calculated as:

$$ \text{Probability} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}} $$

Step 1: Determine Total Possible Outcomes

When a single standard die is thrown, there are 6 possible outcomes (1, 2, 3, 4, 5, 6). When three dice are thrown, the outcomes for each die are independent. Therefore, the total number of possible outcomes is the product of the number of outcomes for each die.

Total Possible Outcomes = $6 \times 6 \times 6 = 216$

Step 2: Identify Favorable Outcomes (Perfect Square Sums)

The sum of the outcomes of three dice can range from a minimum of $1 + 1 + 1 = 3$ to a maximum of $6 + 6 + 6 = 18$. We need to find which perfect squares fall within this range [3, 18].

Perfect squares are numbers obtained by squaring an integer:

• $1^2 = 1$ (less than 3, not possible sum)
• $2^2 = 4$ (within range)
• $3^2 = 9$ (within range)
• $4^2 = 16$ (within range)
• $5^2 = 25$ (greater than 18, not possible sum)

So, the possible perfect square sums are 4, 9, and 16. We need to find the number of ways to get each of these sums.

Step 3: Count Outcomes for Each Perfect Square Sum

Sum = 4

We need to find combinations of three numbers (from 1 to 6) that add up to 4.

The only combination is (1, 1, 2).

Let's list the permutations of (1, 1, 2):

• (1, 1, 2)
• (1, 2, 1)
• (2, 1, 1)

Number of outcomes for sum 4 = 3.

Sum = 9

We need to find combinations of three numbers (from 1 to 6) that add up to 9. Let's list the combinations in non-decreasing order:

• (1, 2, 6)
• (1, 3, 5)
• (1, 4, 4)
• (2, 2, 5)
• (2, 3, 4)
• (3, 3, 3)

Now, let's find the number of permutations for each combination type:

• For (1, 2, 6): These are three distinct numbers. The number of permutations is $3! = 3 \times 2 \times 1 = 6$.
• For (1, 3, 5): These are three distinct numbers. The number of permutations is $3! = 3 \times 2 \times 1 = 6$.
• For (1, 4, 4): Two numbers are the same. The number of permutations is $\frac{3!}{2!} = \frac{6}{2} = 3$.
• For (2, 2, 5): Two numbers are the same. The number of permutations is $\frac{3!}{2!} = \frac{6}{2} = 3$.
• For (2, 3, 4): These are three distinct numbers. The number of permutations is $3! = 3 \times 2 \times 1 = 6$.
• For (3, 3, 3): All numbers are the same. The number of permutations is $\frac{3!}{3!} = \frac{6}{6} = 1$.

Total number of outcomes for sum 9 = $6 + 6 + 3 + 3 + 6 + 1 = 25$.

Sum = 16

We need to find combinations of three numbers (from 1 to 6) that add up to 16. Let's list the combinations in non-decreasing order:

• (4, 6, 6)
• (5, 5, 6)

Let's find the number of permutations for each combination type:

• For (4, 6, 6): Two numbers are the same. The number of permutations is $\frac{3!}{2!} = \frac{6}{2} = 3$.
• For (5, 5, 6): Two numbers are the same. The number of permutations is $\frac{3!}{2!} = \frac{6}{2} = 3$.

Total number of outcomes for sum 16 = $3 + 3 = 6$.

Step 4: Calculate Total Favorable Outcomes

Total number of favorable outcomes is the sum of the outcomes for sums 4, 9, and 16.

Total Favorable Outcomes = (Outcomes for sum 4) + (Outcomes for sum 9) + (Outcomes for sum 16)

Total Favorable Outcomes = $3 + 25 + 6 = 34$

Step 5: Calculate the Probability

Now we can calculate the probability using the formula:

$$ \text{Probability} = \frac{\text{Total Favorable Outcomes}}{\text{Total Possible Outcomes}} $$

$$ \text{Probability} = \frac{34}{216} $$

This fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 2.

$$ \text{Probability} = \frac{34 \div 2}{216 \div 2} = \frac{17}{108} $$

Thus, the probability of getting a sum which is a perfect square when three dice are thrown is $\frac{17}{108}$.

The correct option is $\frac{17}{108}$.

Revision Table: Three Dice Probability

Additional Information: Dice Probability and Combinations

When calculating probability with multiple dice, it's crucial to correctly identify and count all possible outcomes. The total number of outcomes increases exponentially with the number of dice.

Counting favorable outcomes often involves listing combinations that meet the condition and then finding the number of permutations for each combination. The formula for permutations of $n$ objects where there are $n_1$ identical objects of type 1, $n_2$ identical objects of type 2, ..., up to $n_k$ identical objects of type k, is $\frac{n!}{n_1! n_2! \cdots n_k!}$. For three dice (n=3):

• If all three numbers are distinct (e.g., 1, 2, 6), $n_1=1, n_2=1, n_3=1$. Permutations = $\frac{3!}{1!1!1!} = 3! = 6$.
• If two numbers are identical (e.g., 1, 4, 4), $n_1=2, n_2=1$. Permutations = $\frac{3!}{2!1!} = \frac{6}{2} = 3$.
• If all three numbers are identical (e.g., 3, 3, 3), $n_1=3$. Permutations = $\frac{3!}{3!} = \frac{6}{6} = 1$.

Understanding combinations and permutations is key to solving many probability problems involving multiple events like rolling dice.