Two six-sided dice are tossed. Event \( A \) : The first die does NOT land on 1 . Event \( B \) : The second die does NOT land on 6 . What is the probability that both events will occur? For Independent Events: \( P(A \) and \( B)=P(A) \cdot P(B) \) \( P(A \) and \( B)=[?] \) Give your answer in simplest form.

To find the probability of both events \( A \) and \( B \) occurring, first we need to determine the individual probabilities of each event. Event \( A \): The first die does NOT land on 1. The total outcomes are 6 (1 through 6). The favorable outcomes for \( A \) are 2, 3, 4, 5, and 6, which gives us 5 favorable outcomes. Thus, the probability \( P(A) \) is: \[ P(A) = \frac{5}{6} \] Event \( B \): The second die does NOT land on 6. Similar to event A, the favorable outcomes are 1, 2, 3, 4, and 5, which again gives us 5 favorable outcomes. Therefore, the probability \( P(B) \) is: \[ P(B) = \frac{5}{6} \] Now, since the two events are independent, we can find the combined probability \( P(A \text{ and } B) \): \[ P(A \text{ and } B) = P(A) \cdot P(B) = \frac{5}{6} \cdot \frac{5}{6} = \frac{25}{36} \] Thus, the probability that both events will occur is: \[ P(A \text{ and } B) = \frac{25}{36} \]