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

The correct option is 1/18.

Let us break down the problem step-by-step to understand how to arrive at this probability.

When a standard six-sided die is rolled, the possible outcomes are: 1, 2, 3, 4, 5, and 6. Thus, the total number of possible outcomes for any single roll is 6.

Let us analyze the required conditions for each of the three throws:

First Throw: We need to get a number greater than 4.
The numbers greater than 4 on a die are 5 and 6. Therefore, there are 2 favorable outcomes (5 and 6).
The probability of getting a number greater than 4 in the first throw is:

$$P_1=\frac{2}{6}=\frac{1}{3}$$

Second Throw: We also need to get a number greater than 4.
Similarly, the favorable outcomes are 5 and 6.
The probability of getting a number greater than 4 in the second throw is:

$$P_2=\frac{2}{6}=\frac{1}{3}$$

Third Throw: We need to get a number less than 4.
The numbers less than 4 on a die are 1, 2, and 3. Therefore, there are 3 favorable outcomes (1, 2, and 3).
The probability of getting a number less than 4 in the third throw is:

$$P_3=\frac{3}{6}=\frac{1}{2}$$

Since the three rolls of the die are independent events, the combined probability of all three events occurring in sequence is the product of their individual probabilities:

$$P=P_1\times P_2\times P_3$$

Substituting the calculated values:

$$P=\frac{1}{3}\times \frac{1}{3}\times \frac{1}{2}=\frac{1}{18}$$

Thus, the overall probability of satisfying all three conditions is 1/18.