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?

Updated On: Feb 26, 2025
  • $\frac{1}{3}$
  • $\frac{1}{6}$
  • $\frac{1}{9}$
  • $\frac{1}{18}$
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The Correct Option is D

Solution and Explanation

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 \( \frac{1}{18} \).

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