Three dice are thrown. The probability of getting a sum of numbers which is a perfect cube is:

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Accepted Answer

#### Step 1: Total Outcomes
When three dice are thrown, the total number of outcomes is:  
$$
6 	imes 6 	imes 6 = 216.
$$
#### Step 2: Identify Perfect Cube Sums
The possible sums when three dice are thrown range from **3** (minimum: $$1+1+1$$) to **18** (maximum: $$6+6+6$$). The perfect cubes within this range are:  
$$
1^3 = 1, \quad 2^3 = 8, \quad 3^3 = 27.
$$  
Only **8** is valid within the range of sums (3 to 18).
#### Step 3: Count Favorable Outcomes for Sum = 8
We need to find all combinations of dice rolls that sum to **8**. These combinations are:  
- (1, 1, 6)  
- (1, 2, 5)  
- (1, 3, 4)  
- (2, 2, 4)  
- (2, 3, 3)
Accounting for permutations of these combinations:
1. **(1, 1, 6):** $$ \frac{3!}{2!} = 3 $$ permutations.  
2. **(1, 2, 5):** $$ 3! = 6 $$ permutations.  
3. **(1, 3, 4):** $$ 3! = 6 $$ permutations.  
4. **(2, 2, 4):** $$ \frac{3!}{2!} = 3 $$ permutations.  
5. **(2, 3, 3):** $$ \frac{3!}{2!} = 3 $$ permutations.
Total favorable outcomes:  
$$
3 + 6 + 6 + 3 + 3 = 21.
$$
#### Step 4: Calculate Probability
The probability is given by:  
$$
P = \frac{	ext{Favorable Outcomes}}{	ext{Total Outcomes}} = \frac{21}{216} = \frac{7}{72}.
$$
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Thus, the correct answer is **(c) $$ \frac{5}{36} $$**.

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