Question

A fair six-faced dice, with the faces labelled ‘1’, ‘2’, ‘3’, ‘4’, ‘5’, and ‘6’, is rolled thrice. What is the probability of rolling ‘6’ exactly once?

• A. \( \frac{75}{216} \)
• B. \( \frac{1}{6} \)
• C. \( \frac{1}{18} \)
• D. \( \frac{25}{216} \)

Hint:

Use the binomial probability formula for independent events to calculate probabilities in dice or coin tossing problems.

Answer 1

The Correct Option is A

The problem is asking for the probability of rolling exactly one '6' when rolling a fair dice three times. This is a binomial probability problem. The probability of rolling a '6' on a fair die is \( \frac{1}{6} \), and the probability of not rolling a '6' is \( \frac{5}{6} \). We are rolling the die three times, and we want exactly one of those rolls to be a '6'. The binomial probability formula is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where:
\( n \) is the number of trials (3 rolls),
\( k \) is the number of successful outcomes (exactly one '6'),
\( p \) is the probability of success on a single trial (\( \frac{1}{6} \)). Substituting the values: \[ P({exactly one '6'}) = \binom{3}{1} \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^2 \] \[ = 3 \times \frac{1}{6} \times \frac{25}{36} = 3 \times \frac{25}{216} = \frac{75}{216}. \] Thus, the probability of rolling exactly one '6' is \( \frac{75}{216} \). Therefore, the correct answer is \( \boxed{A} \).