Certainly! Let's start by breaking down the problem into smaller steps to find the probability that both events [tex]\( A \)[/tex] and [tex]\( B \)[/tex] will occur.
1. Determine the total number of outcomes for a die:
Each die has 6 faces, so the total number of possible outcomes when rolling one die is 6.
2. Calculate the probability of Event [tex]\( A \)[/tex]:
Event [tex]\( A \)[/tex] occurs when the first die lands on 1, 2, 3, or 4. These outcomes are favorable to Event [tex]\( A \)[/tex].
- Number of favorable outcomes for [tex]\( A \)[/tex]: [tex]\(1, 2, 3, 4\)[/tex]
- Total favorable outcomes: 4
Therefore, the probability of Event [tex]\( A \)[/tex] is:
[tex]\[
P(A) = \frac{\text{Number of favorable outcomes for } A}{\text{Total possible outcomes}} = \frac{4}{6} = \frac{2}{3}
\][/tex]
3. Calculate the probability of Event [tex]\( B \)[/tex]:
Event [tex]\( B \)[/tex] happens when the second die lands on 6.
- Number of favorable outcomes for [tex]\( B \)[/tex]: 1 (only one face of the die shows 6)
Therefore, the probability of Event [tex]\( B \)[/tex] is:
[tex]\[
P(B) = \frac{\text{Number of favorable outcomes for } B}{\text{Total possible outcomes}} = \frac{1}{6}
\][/tex]
4. Calculate the probability that both events will occur (Event [tex]\( A \)[/tex] and Event [tex]\( B \)[/tex]):
Since the events are independent, the probability of both events occurring is the product of their individual probabilities:
[tex]\[
P(A \text{ and } B) = P(A) \cdot P(B)
\][/tex]
Substitute the values of [tex]\( P(A) \)[/tex] and [tex]\( P(B) \)[/tex]:
[tex]\[
P(A \text{ and } B) = \left(\frac{2}{3}\right) \cdot \left(\frac{1}{6}\right)
\][/tex]
Multiply the fractions:
[tex]\[
P(A \text{ and } B) = \frac{2}{3} \cdot \frac{1}{6} = \frac{2 \times 1}{3 \times 6} = \frac{2}{18} = \frac{1}{9}
\][/tex]
So, 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]
In decimal form, this probability is approximately [tex]\( 0.1111 \)[/tex].
