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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 [tex]$2, 4, 5$[/tex], or [tex]$6$[/tex].

What is the probability that both events will occur?

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

Calculate [tex]$P(A \text{ and } B)$[/tex].

Give your answer in simplest form.
[tex]$\boxed{\phantom{?}}$[/tex] Enter


Sagot :

To determine the probability that both events A and B will occur, we first need to determine the probability of each event individually and then use the rule for independent events.

### Step 1: Determine the probability of Event A.

Event A is the coin landing on heads. A fair coin has two outcomes: heads or tails. The probability of the coin landing on heads is therefore:

[tex]\[ P(A) = \frac{1}{2} = 0.5 \][/tex]

### Step 2: Determine the probability of Event B.

Event B is the die landing on either 2, 4, 5, or 6. A standard six-sided die has six outcomes: 1, 2, 3, 4, 5, and 6. There are 4 favorable outcomes (2, 4, 5, or 6) out of the 6 possible outcomes. Thus, the probability of the die landing on one of these numbers is:

[tex]\[ P(B) = \frac{4}{6} = \frac{2}{3} \approx 0.6666666666666666 \][/tex]

### Step 3: Calculate the probability of both events occurring (Event A and Event B).

For two independent events, the probability that both events occur is the product of their individual probabilities. Therefore:

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

Substituting the values we found:

[tex]\[ P(A \text{ and } B) = 0.5 \times \frac{2}{3} \][/tex]

Multiplying these together:

[tex]\[ P(A \text{ and } B) = 0.5 \times 0.6666666666666666 = 0.3333333333333333 = \frac{1}{3} \][/tex]

Thus, the probability that both events will occur is:

[tex]\[ \boxed{\frac{1}{3}} \][/tex]

In simplest form, the probability [tex]\(P(A \text{ and } B)\)[/tex] is [tex]\(\frac{1}{3}\)[/tex].