Two six-sided dice are tossed.

Event A: The first die lands on 1 or 2.

Event B: The second die lands on 5.

What is the probability that both events will occur?

For independent events: [tex] P(A \text{ and } B) = P(A) \cdot P(B) [/tex]

[tex] P(A \text{ and } B) = \square [/tex]



Answer :

Let's determine the probability step-by-step.

1. Identify Event A:
- Event A is that the first die lands on either 1 or 2.
- A six-sided die has 6 possible outcomes: {1, 2, 3, 4, 5, 6}.
- The favorable outcomes for Event A are 1 and 2.

2. Calculate the Probability of Event A:
- There are 2 favorable outcomes (1 and 2) out of 6 possible outcomes.
- Thus, the probability of Event A, [tex]\( P(A) \)[/tex], is:
[tex]\[ P(A) = \frac{2}{6} = \frac{1}{3} \approx 0.3333 \][/tex]

3. Identify Event B:
- Event B is that the second die lands on 5.
- Again, a six-sided die has 6 possible outcomes: {1, 2, 3, 4, 5, 6}.
- The favorable outcome for Event B is 5.

4. Calculate the Probability of Event B:
- There is 1 favorable outcome (5) out of 6 possible outcomes.
- Thus, the probability of Event B, [tex]\( P(B) \)[/tex], is:
[tex]\[ P(B) = \frac{1}{6} \approx 0.1667 \][/tex]

5. Calculate the Combined Probability of Both Events A and B:
- Since these events are independent, we use the formula [tex]\( P(A \text{ and } B) = P(A) \cdot P(B) \)[/tex].
- Therefore, the probability [tex]\( P(A \text{ and } B) \)[/tex] is:
[tex]\[ P(A \text{ and } B) = P(A) \cdot P(B) = \left(\frac{1}{3}\right) \cdot \left(\frac{1}{6}\right) \approx 0.3333 \cdot 0.1667 = 0.0556 \][/tex]

Thus, [tex]\( P(A \text{ and } B) = 0.0556 \)[/tex]. So, the probability that both events will occur is approximately 0.0556.