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. $\frac{1}{18}$

Answer 1

Answer: (D)

$\frac{1}{18}$

Answer 2

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

A number greater than 4 includes {5, 6}. The probability of this event is:

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

A number less than 4 includes {1, 2, 3}. The probability of this event is:

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

The probability of the required outcome (a number greater than 4 on the first and second throws, and a number less than 4 on the third throw) is the product of the probabilities:

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

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

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

Thus, the probability of the required outcome is $\frac{1}{18}$.