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Two six-sided dice are tossed.

Event A: The first die lands on 1, 2, 3, or 4.

Event B: The second die lands on 6.

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) = \frac{1}{?}[/tex]

Give your answer in simplest form.


Sagot :

Let's solve the problem step by step:

1. Identify the total number of possible outcomes:
- Since we have two six-sided dice, and each die has 6 faces, the total number of possible outcomes when two dice are tossed is given by [tex]\(6 \times 6 = 36\)[/tex].

2. Determine the number of favorable outcomes for Event A:
- Event A is that the first die lands on 1, 2, 3, or 4. There are 4 favorable outcomes for Event A out of the 6 possible outcomes for the first die.

3. Determine the number of favorable outcomes for Event B:
- Event B is that the second die lands on 6. There is 1 favorable outcome for Event B out of the 6 possible outcomes for the second die.

4. Calculate the combined favorable outcomes when both events A and B occur together:
- To find the number of favorable outcomes where both Event A (first die lands on 1, 2, 3, or 4) and Event B (second die lands on 6) occur simultaneously, we multiply the number of favorable outcomes for Event A by the number of favorable outcomes for Event B. That gives us:
[tex]\[ \text{Number of favorable outcomes} = 4 \times 1 = 4 \][/tex]

5. Calculate the probability:
- The probability [tex]\(P(A \text{ and } B)\)[/tex] is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. Thus:
[tex]\[ P(A \text{ and } B) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{4}{36} \][/tex]
- Simplify this fraction to its simplest form:
[tex]\[ \frac{4}{36} = \frac{1}{9} \][/tex]

Therefore, the probability that both Event A and Event B will occur is [tex]\( \frac{1}{9} \)[/tex].
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