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.
[tex]\(\square\)[/tex] Enter



Answer :

Sure, let's work through this step-by-step to determine the probability that both events will occur.

Step 1: Determine the probability of event A

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

Step 2: Determine the probability of event B

Event B is the die landing on either 1, 3, or 6. A fair six-sided die has six possible outcomes: 1, 2, 3, 4, 5, and 6. The favorable outcomes for event B (1, 3, or 6) are three in total. Thus, the probability of event B is:
[tex]\[ P(B) = \frac{3}{6} = \frac{1}{2} \][/tex]

Step 3: Determine the probability of both events occurring

Since events A and B are independent, the probability of both events occurring (A and B) is obtained by multiplying their individual probabilities:
[tex]\[ P(A \text{ and } B) = P(A) \cdot P(B) \][/tex]

Substituting the values we obtained:
[tex]\[ P(A \text{ and } B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \][/tex]

Therefore, the probability that both events will occur is:
[tex]\[ \boxed{\frac{1}{4}} \][/tex]