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A coin and a six-sided die are tossed.

Event A: The coin lands on heads.
Event B: The die lands on 1, 3, or 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) = \underline{[?]}[/tex]

Give your answer in simplest form.


Sagot :

To solve this problem, let's break it down step by step:

1. Event A: The coin lands on heads.

The probability of a coin landing on heads (event A) can be calculated since a fair coin has two possible outcomes: heads or tails. Therefore, the probability of the coin landing on heads is:
[tex]\[ P(A) = \frac{1}{2} \][/tex]

2. Event B: The die lands on 1, 3, or 6.

A standard six-sided die has six faces, each numbered from 1 to 6. To find the probability of the die landing on 1, 3, or 6, note that these are three of the six possible outcomes.
[tex]\[ P(B) = \frac{3}{6} = \frac{1}{2} \][/tex]

3. Both Events Occurring: The product of their probabilities.

Since tossing a coin and rolling a die are independent events, the probability that both events A and B will occur can be calculated by multiplying the probability of each event:
[tex]\[ P(A \text{ and } B) = P(A) \cdot P(B) \][/tex]

4. Substitute the probabilities into the formula:

[tex]\[ P(A \text{ and } B) = \left( \frac{1}{2} \right) \cdot \left( \frac{1}{2} \right) \][/tex]

Multiplying these fractions:
[tex]\[ P(A \text{ and } B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1 \times 1}{2 \times 2} = \frac{1}{4} \][/tex]

So, the probability that both events will occur is:
[tex]\[ P(A \text{ and } B) = \frac{1}{4} \][/tex]