Sure, let’s walk through this problem step-by-step to find the probability that both events will occur:
### Step 1: Determine the probabilities of each individual event.
Event A: The probability that the first die lands on [tex]\(1, 2, 3\)[/tex], or [tex]\(4\)[/tex]:
- Total possible outcomes for the first die: [tex]\(6\)[/tex] (since it's a standard six-sided die).
- Number of favorable outcomes for Event A: [tex]\(4\)[/tex] (since [tex]\(1, 2, 3\)[/tex], and [tex]\(4\)[/tex] are considered favorable).
The probability of Event A occurring, [tex]\(P(A)\)[/tex], is given by the ratio of favorable outcomes to total possible outcomes:
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes for Event A}}{\text{Total possible outcomes for the first die}} = \frac{4}{6} = \frac{2}{3} \][/tex]
Event B: The probability that the second die lands on [tex]\(5\)[/tex] or [tex]\(6\)[/tex]:
- Total possible outcomes for the second die: [tex]\(6\)[/tex] (since it’s a standard six-sided die).
- Number of favorable outcomes for Event B: [tex]\(2\)[/tex] (since [tex]\(5\)[/tex] and [tex]\(6\)[/tex] are considered favorable).
The probability of Event B occurring, [tex]\(P(B)\)[/tex], is given by the ratio of favorable outcomes to total possible outcomes:
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes for Event B}}{\text{Total possible outcomes for the second die}} = \frac{2}{6} = \frac{1}{3} \][/tex]
### Step 2: Calculate the probability of both events occurring.
For independent events, the probability that both events [tex]\(A\)[/tex] and [tex]\(B\)[/tex] occur, denoted as [tex]\(P(A \text{ and } B)\)[/tex], is the product of their individual probabilities:
[tex]\[ P(A \text{ and } B) = P(A) \cdot P(B) \][/tex]
Substituting the probabilities we found:
[tex]\[ P(A \text{ and } B) = \left(\frac{2}{3}\right) \cdot \left(\frac{1}{3}\right) = \frac{2}{9} \][/tex]
Thus, the probability that both Event A and Event B will occur is:
[tex]\[ \boxed{\frac{2}{9}} \][/tex]
