Question

A die is rolled thrice. What is the probability of getting a number greater than $4$ in the first and second throws and a number less than $4$ in the third throw?

• A. $\frac{1}{3}$
• B. $\frac{1}{6}$
• C. $\frac{1}{9}$
• D. $\frac{1}{18}$

Hint:

When calculating the probability of multiple independent events, you can multiply the individual probabilities together. In this case, the events are independent (each die roll does not affect the others), so the total probability is simply the product of the individual probabilities. Make sure to simplify fractions to their lowest terms to avoid errors in final calculations.

Answer 1

The Correct Option is D

When a die is rolled, the possible outcomes are 1, 2, 3, 4, 5, and 6. The probabilities for the specified events are calculated as follows:

The event "a number greater than 4" includes the outcomes {5, 6}. The probability of this event is:

\( P(\text{greater than 4}) = \frac{2}{6} = \frac{1}{3} \)

The event "a number less than 4" includes the outcomes {1, 2, 3}. The probability of this event is:

\( P(\text{less than 4}) = \frac{3}{6} = \frac{1}{2} \)

The probability of the specific sequence of events (a number greater than 4 on the first roll, a number greater than 4 on the second roll, and a number less than 4 on the third roll) is the product of the individual probabilities:

\( P(\text{required outcome}) = P(\text{greater than 4}) \cdot P(\text{greater than 4}) \cdot P(\text{less than 4}) \)

\( P(\text{required outcome}) = \frac{1}{3} \times \frac{1}{3} \times \frac{1}{2} \)

\( P(\text{required outcome}) = \frac{1}{18} \)

Therefore, the probability of the required sequence of outcomes is \( \frac{1}{18} \).