Nathan will roll two six-sided dice. What is the probability that he will roll a number less than three on the first die and a number greater than three on the second die? Express your answer as a common fraction.
For the first die to be less than three, it must be a 1 or a 2, which occurs with probability $\frac{1}{3}$. For the second die to be greater than 3, it must be a 4 or a 5 or a 6, which occurs with probability $\frac{1}{2}$. The probability of both of these events occuring, as they are independent, is $\frac{1}{3} \cdot \frac{1}{2} = \boxed{\frac{1}{6}}$.
\frac{1}{6}
Level 2
Counting & Probability
Problem: Nathan will roll two six-sided dice. What is the probability that he will roll a number less than three on the first die and a number greater than three on the second die? Express your answer as a common fraction.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the probability of rolling a number less than three on the first die, we identify the possible outcomes for the first die. The numbers less than three are 1 and 2, giving us 2 favorable outcomes out of 6 total outcomes. Thus, the probability for the first die is 2/6, which simplifies to 1/3.

Next, we consider the second die, where we need to roll a number greater than three. The numbers greater than three are 4, 5, and 6, resulting in 3 favorable outcomes out of 6 total outcomes. Therefore, the probability for the second die is 3/6, which simplifies to 1/2.

Since the rolls of the two dice are independent events, we multiply the probabilities of the two events to find the overall probability. The calculation is as follows:

(1/3) * (1/2) = 1/6.

Thus, the final answer is 1/6.<|END_OF_LATENT|> The final answer is $\frac{1}{6}$.