Sure, let's find the probability that both events A and B will occur when two six-sided dice are tossed.
1. Define the Events:
- Event A: The first die does NOT land on 5.
- Event B: The second die lands on 4.
2. Probability of Event A:
- There are 6 faces on a die.
- The first die not landing on 5 means it can land on 1, 2, 3, 4, or 6.
- Therefore, there are 5 favorable outcomes.
- The probability [tex]\( P(A) \)[/tex] is given by the ratio of favorable outcomes to the total outcomes:
[tex]\[
P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{6} \approx 0.8333
\][/tex]
3. Probability of Event B:
- The second die landing on 4 means only 1 face is favorable.
- The probability [tex]\( P(B) \)[/tex] is:
[tex]\[
P(B) = \frac{1}{6} \approx 0.1667
\][/tex]
4. Combine the Probabilities:
- The events A and B are independent.
- For independent events, the probability of both events occurring [tex]\( P(A \text{ and } B) \)[/tex] is the product of their individual probabilities:
[tex]\[
P(A \text{ and } B) = P(A) \times P(B)
\][/tex]
- Plugging in the values:
[tex]\[
P(A \text{ and } B) = \left(\frac{5}{6}\right) \times \left(\frac{1}{6}\right) \approx 0.8333 \times 0.1667 = 0.1389
\][/tex]
Thus, the probability that both events will occur is approximately 0.1389.
