Answer :
To find the probability that both events A and B will occur when two six-sided dice are tossed, we need to determine the probabilities of each individual event and then multiply them. Let’s break it down step-by-step.
1. Event A: The first die does NOT land on 5.
There are 6 possible outcomes for a six-sided die: 1, 2, 3, 4, 5, or 6. The event that the first die does not land on 5 excludes one outcome, so there are 5 favorable outcomes (1, 2, 3, 4, or 6).
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{5}{6} \][/tex]
2. Event B: The second die lands on 4.
Similarly, there are 6 possible outcomes for the second die: 1, 2, 3, 4, 5, or 6. Only one of these outcomes is favorable for event B (landing on 4).
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{1}{6} \][/tex]
3. Finding the Probability of Both Events Occurring:
Since events A and B are independent (the outcome of the first die does not affect the outcome of the second die), 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]
Substituting the probabilities we found:
[tex]\[ P(A \text{ and } B) = \left( \frac{5}{6} \right) \cdot \left( \frac{1}{6} \right) \][/tex]
4. Simplifying the Expression:
[tex]\[ P(A \text{ and } B) = \frac{5}{6} \times \frac{1}{6} = \frac{5 \cdot 1}{6 \cdot 6} = \frac{5}{36} \][/tex]
Hence, the probability that both events will occur is:
[tex]\[ P(A \text{ and } B) = \frac{5}{36} \][/tex]
This is the simplest form of the probability.
1. Event A: The first die does NOT land on 5.
There are 6 possible outcomes for a six-sided die: 1, 2, 3, 4, 5, or 6. The event that the first die does not land on 5 excludes one outcome, so there are 5 favorable outcomes (1, 2, 3, 4, or 6).
[tex]\[ P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{5}{6} \][/tex]
2. Event B: The second die lands on 4.
Similarly, there are 6 possible outcomes for the second die: 1, 2, 3, 4, 5, or 6. Only one of these outcomes is favorable for event B (landing on 4).
[tex]\[ P(B) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{1}{6} \][/tex]
3. Finding the Probability of Both Events Occurring:
Since events A and B are independent (the outcome of the first die does not affect the outcome of the second die), 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]
Substituting the probabilities we found:
[tex]\[ P(A \text{ and } B) = \left( \frac{5}{6} \right) \cdot \left( \frac{1}{6} \right) \][/tex]
4. Simplifying the Expression:
[tex]\[ P(A \text{ and } B) = \frac{5}{6} \times \frac{1}{6} = \frac{5 \cdot 1}{6 \cdot 6} = \frac{5}{36} \][/tex]
Hence, the probability that both events will occur is:
[tex]\[ P(A \text{ and } B) = \frac{5}{36} \][/tex]
This is the simplest form of the probability.
