Answer :
Let's solve the problem step-by-step to find the probability that both Event [tex]\(A\)[/tex] and Event [tex]\(B\)[/tex] will occur when two six-sided dice are tossed.
### Step 1: Determine the probability of Event [tex]\(A\)[/tex]
Event [tex]\(A\)[/tex] is that the first die lands on 1, 2, 3, or 4.
- A six-sided die has outcomes: 1, 2, 3, 4, 5, 6.
- The favorable outcomes for Event [tex]\(A\)[/tex] are 1, 2, 3, and 4.
- Therefore, there are 4 favorable outcomes for Event [tex]\(A\)[/tex].
The probability of Event [tex]\(A\)[/tex] is given by:
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes for } A}{\text{Total number of possible outcomes}} = \frac{4}{6} = \frac{2}{3} \][/tex]
### Step 2: Determine the probability of Event [tex]\(B\)[/tex]
Event [tex]\(B\)[/tex] is that the second die lands on 6.
- A six-sided die has outcomes: 1, 2, 3, 4, 5, 6.
- The favorable outcome for Event [tex]\(B\)[/tex] is just 6.
- Therefore, there is 1 favorable outcome for Event [tex]\(B\)[/tex].
The probability of Event [tex]\(B\)[/tex] is given by:
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes for } B}{\text{Total number of possible outcomes}} = \frac{1}{6} \][/tex]
### Step 3: Calculate the probability of both events occurring (Event [tex]\(A\)[/tex] and Event [tex]\(B\)[/tex])
Given that the two events are independent, the probability of both events occurring is calculated by multiplying the probabilities of each event:
[tex]\[ P(A \text{ and } B) = P(A) \cdot P(B) \][/tex]
Substituting the probabilities calculated in previous steps:
[tex]\[ P(A) = \frac{2}{3} \][/tex]
[tex]\[ P(B) = \frac{1}{6} \][/tex]
Therefore:
[tex]\[ P(A \text{ and } B) = \frac{2}{3} \times \frac{1}{6} \][/tex]
To multiply these fractions:
[tex]\[ P(A \text{ and } B) = \frac{2 \times 1}{3 \times 6} = \frac{2}{18} = \frac{1}{9} \][/tex]
### Final Answer
The probability that both Event [tex]\(A\)[/tex] (the first die lands on 1, 2, 3, or 4) and Event [tex]\(B\)[/tex] (the second die lands on 6) will occur is:
[tex]\[ P(A \text{ and } B) = \frac{1}{9} \][/tex]
### Step 1: Determine the probability of Event [tex]\(A\)[/tex]
Event [tex]\(A\)[/tex] is that the first die lands on 1, 2, 3, or 4.
- A six-sided die has outcomes: 1, 2, 3, 4, 5, 6.
- The favorable outcomes for Event [tex]\(A\)[/tex] are 1, 2, 3, and 4.
- Therefore, there are 4 favorable outcomes for Event [tex]\(A\)[/tex].
The probability of Event [tex]\(A\)[/tex] is given by:
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes for } A}{\text{Total number of possible outcomes}} = \frac{4}{6} = \frac{2}{3} \][/tex]
### Step 2: Determine the probability of Event [tex]\(B\)[/tex]
Event [tex]\(B\)[/tex] is that the second die lands on 6.
- A six-sided die has outcomes: 1, 2, 3, 4, 5, 6.
- The favorable outcome for Event [tex]\(B\)[/tex] is just 6.
- Therefore, there is 1 favorable outcome for Event [tex]\(B\)[/tex].
The probability of Event [tex]\(B\)[/tex] is given by:
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes for } B}{\text{Total number of possible outcomes}} = \frac{1}{6} \][/tex]
### Step 3: Calculate the probability of both events occurring (Event [tex]\(A\)[/tex] and Event [tex]\(B\)[/tex])
Given that the two events are independent, the probability of both events occurring is calculated by multiplying the probabilities of each event:
[tex]\[ P(A \text{ and } B) = P(A) \cdot P(B) \][/tex]
Substituting the probabilities calculated in previous steps:
[tex]\[ P(A) = \frac{2}{3} \][/tex]
[tex]\[ P(B) = \frac{1}{6} \][/tex]
Therefore:
[tex]\[ P(A \text{ and } B) = \frac{2}{3} \times \frac{1}{6} \][/tex]
To multiply these fractions:
[tex]\[ P(A \text{ and } B) = \frac{2 \times 1}{3 \times 6} = \frac{2}{18} = \frac{1}{9} \][/tex]
### Final Answer
The probability that both Event [tex]\(A\)[/tex] (the first die lands on 1, 2, 3, or 4) and Event [tex]\(B\)[/tex] (the second die lands on 6) will occur is:
[tex]\[ P(A \text{ and } B) = \frac{1}{9} \][/tex]
