Sure! Let's solve this problem step-by-step to find the probability that both events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] will occur.
First, we need to determine the probability of each individual event:
1. Event A: The first die lands on [tex]\( 1, 2, 3, \)[/tex] or [tex]\( 4 \)[/tex].
- A six-sided die has 6 possible outcomes: [tex]\( 1, 2, 3, 4, 5, \)[/tex] and [tex]\( 6 \)[/tex].
- Event A specifies landing on [tex]\( 1, 2, 3, \)[/tex] or [tex]\( 4 \)[/tex], which includes 4 favorable outcomes.
- Therefore, the probability of Event A, [tex]\( P(A) \)[/tex]:
[tex]\[
P(A) = \frac{\text{Number of favorable outcomes for A}}{\text{Total possible outcomes on a six-sided die}} = \frac{4}{6} = \frac{2}{3}
\][/tex]
2. Event B: The second die lands on 6.
- There is only 1 favorable outcome (landing on 6) out of the 6 possible outcomes.
- Therefore, the probability of Event B, [tex]\( P(B) \)[/tex]:
[tex]\[
P(B) = \frac{1}{6}
\][/tex]
Since the events are independent, we use the formula for the probability of both events occurring:
[tex]\[
P(A \text{ and } B) = P(A) \cdot P(B)
\][/tex]
Substituting the known probabilities [tex]\( P(A) \)[/tex] and [tex]\( P(B) \)[/tex] into the formula, we get:
[tex]\[
P(A \text{ and } B) = \left(\frac{2}{3}\right) \cdot \left(\frac{1}{6}\right)
\][/tex]
Now, multiply the fractions:
[tex]\[
P(A \text{ and } B) = \frac{2}{3} \times \frac{1}{6} = \frac{2 \times 1}{3 \times 6} = \frac{2}{18} = \frac{1}{9}
\][/tex]
Therefore, the probability that both events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] will occur is:
[tex]\[
P(A \text{ and } B) = \frac{1}{9}
\][/tex]
