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

Give your answer in simplest form.



Answer :

GinnyAnswer
To find the probability that both events A and B occur, we first need to determine the individual probabilities of each event.

1. Probability of Event A (coin lands on heads):
- A standard coin has two sides: heads and tails.
- The probability of the coin landing on heads is therefore:
[tex]\[ P(A) = \frac{1}{2} \][/tex]

2. Probability of Event B (die lands on 1, 3, or 6):
- A standard six-sided die has six faces, numbered 1 through 6.
- The favorable outcomes for event B are the die landing on 1, 3, or 6.
- There are 3 favorable outcomes out of the 6 possible outcomes, so the probability of the die landing on 1, 3, or 6 is:
[tex]\[ P(B) = \frac{3}{6} = \frac{1}{2} \][/tex]

3. Probability of both events A and B occurring:
- Since events A and B are independent (the outcome of the coin toss does not affect the outcome of the die roll), we can use the formula for the probability of independent events both occurring:
[tex]\[ P(A \text{ and } B) = P(A) \cdot P(B) \][/tex]

Putting the calculated probabilities into the formula, we get:
[tex]\[ P(A \text{ and } B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \][/tex]

So, the probability that both events will occur is:
[tex]\[ P(A \text{ and } B) = \frac{1}{4} \][/tex]
In decimal form, this is 0.25.

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