Question

A fair dice is thrown 3 times. The probability that the product of the three outcomes is a prime number, is

• A. $\frac{1}{24}$
• B. $\frac{1}{36}$
• C. $\frac{1}{32}$
• D. $\frac{1}{8}$

Answer 1

Answer: (A)

$\frac{1}{24}$

Answer 2

To find the probability that the product of the three outcomes is a prime number, we need to consider the possible outcomes when a fair six-sided die is thrown three times.

Understanding Prime Numbers

A prime number is a number greater than 1 that has only two divisors: 1 and itself. The prime numbers that can appear on a standard die are 2, 3, and 5.

Possible Outcomes

For the product of the three outcomes to be prime, two of the outcomes must be 1, and the third outcome must be a prime number (2, 3, or 5).

Listing Favorable Outcomes

The possible sets of outcomes are {1, 1, 2}, {1, 1, 3}, and {1, 1, 5}. We need to consider all the permutations (orderings) of these sets:

• For {1, 1, 2}: (1, 1, 2), (1, 2, 1), (2, 1, 1) - 3 permutations
• For {1, 1, 3}: (1, 1, 3), (1, 3, 1), (3, 1, 1) - 3 permutations
• For {1, 1, 5}: (1, 1, 5), (1, 5, 1), (5, 1, 1) - 3 permutations

So, there are a total of 3 + 3 + 3 = 9 favorable outcomes.

Calculating Total Possible Outcomes

Each throw of the die has 6 possible outcomes. Since we throw the die three times, the total number of possible outcomes is $6 \times 6 \times 6 = 216$.

Calculating Probability

The probability is the ratio of favorable outcomes to total possible outcomes:

$P(\text{product is prime}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{9}{216} = \frac{1}{24}$

Therefore, the probability that the product of the three outcomes is a prime number is $\frac{1}{24}$.