To solve this problem, we need to calculate the probability that both specified events will happen when two six-sided dice are tossed.
### Step 1: Calculate the Probability of Event A
Event A is that the first die does NOT land on 5.
- A six-sided die has 6 possible outcomes: {1, 2, 3, 4, 5, 6}.
- The favorable outcomes for Event A are: {1, 2, 3, 4, 6}, which are 5 out of the 6 possible outcomes.
Hence, the probability of Event A ([tex]\(P(A)\)[/tex]) is:
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes for A}}{\text{Total number of possible outcomes}} = \frac{5}{6} \][/tex]
### Step 2: Calculate the Probability of Event B
Event B is that the second die lands on 4.
- A six-sided die has 6 possible outcomes: {1, 2, 3, 4, 5, 6}.
- The favorable outcome for Event B is: {4}, which is 1 out of the 6 possible outcomes.
Hence, the probability of Event B ([tex]\(P(B)\)[/tex]) is:
[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
Since Event A and Event B are independent events, the probability that both A and B will occur is given by the product of their individual probabilities.
[tex]\[ P(A \text{ and } B) = P(A) \cdot P(B) \][/tex]
Substituting the values of [tex]\(P(A)\)[/tex] and [tex]\(P(B)\)[/tex]:
[tex]\[ P(A \text{ and } B) = \frac{5}{6} \cdot \frac{1}{6} \][/tex]
To multiply these fractions:
[tex]\[ P(A \text{ and } B) = \frac{5 \times 1}{6 \times 6} = \frac{5}{36} \][/tex]
Thus, the probability that both the first die does not land on 5 and the second die lands on 4 is [tex]\( \frac{5}{36} \)[/tex].
