Question

If three dice are thrown at a time, then the probability of getting the sum of the numbers on them as a prime number is

• A. $3/8$
• B. $73/216$
• C. $4/27$
• D. $5/54$

Hint:

To find the number of ways to get a certain sum with multiple dice, it's systematic to list the partitions of the sum into parts no larger than 6. Then, for each partition, calculate the number of permutations.

Answer 1

The Correct Option is B

When three dice are thrown, the total number of possible outcomes is $6 \times 6 \times 6 = 216$.
The minimum possible sum is $1+1+1=3$, and the maximum is $6+6+6=18$.
The prime numbers between 3 and 18 are 3, 5, 7, 11, 13, 17.
We need to find the number of ways to obtain each of these sums.
Sum = 3: (1,1,1) - $\frac{3!}{3!} = 1$ way.
Sum = 5: (1,1,3) - $\frac{3!}{2!} = 3$ ways; (1,2,2) - $\frac{3!}{2!} = 3$ ways. Total = 6 ways.
Sum = 7: (1,1,5) - 3 ways; (1,2,4) - $3! = 6$ ways; (1,3,3) - 3 ways; (2,2,3) - 3 ways. Total = 15 ways.
Sum = 11: (1,4,6) - 6 ways; (1,5,5) - 3 ways; (2,3,6) - 6 ways; (2,4,5) - 6 ways; (3,3,5) - 3 ways; (3,4,4) - 3 ways. Total = 27 ways.
Sum = 13: (1,6,6) - 3 ways; (2,5,6) - 6 ways; (3,4,6) - 6 ways; (3,5,5) - 3 ways; (4,4,5) - 3 ways. Total = 21 ways.
Sum = 17: (5,6,6) - 3 ways.
Total number of favorable outcomes is the sum of ways for all prime sums:
Favorable outcomes = $1 + 6 + 15 + 27 + 21 + 3 = 73$.
The probability is $\frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{73}{216}$.