A coin and a six-sided die are tossed.

Event A: The coin lands on tails.

Event B: The die does NOT land on a 1.

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.



Answer :

Let's find the probability that both Event A and Event B will occur step by step.

### Step 1: Determine the Probability of Event A
Event A is that the coin lands on tails. For a fair coin, there are two equally likely outcomes: heads and tails.
- The probability of the coin landing on tails is:
[tex]\[ P(A) = \frac{1}{2} = 0.5 \][/tex]

### Step 2: Determine the Probability of Event B
Event B is that the six-sided die does not land on a 1. There are six possible outcomes when rolling a die: 1, 2, 3, 4, 5, and 6. Out of these six outcomes, five are favorable outcomes (2, 3, 4, 5, 6).
- The probability of the die not landing on a 1 is:
[tex]\[ P(B) = \frac{5}{6} \approx 0.8333 \][/tex]

### Step 3: Use the Multiplication Rule for Independent Events
Since the events are independent, the probability of both events A and B occurring is the product of their individual probabilities:
[tex]\[ P(A \text{ and } B) = P(A) \cdot P(B) \][/tex]

Substituting the probabilities we obtained:
[tex]\[ P(A \text{ and } B) = 0.5 \cdot 0.8333 \][/tex]

### Step 4: Calculate the Result
Multiplying these probabilities together gives:
[tex]\[ P(A \text{ and } B) = 0.5 \cdot 0.8333 = 0.4167 \][/tex]

So, the probability that both the coin lands on tails and the die does not land on a 1 is:
[tex]\[ P(A \text{ and } B) \approx 0.4167 \][/tex]

In simplest fractional form, this probability is:
[tex]\[ P(A \text{ and } B) = \frac{5}{12} \][/tex]

Therefore, the probability that both events will occur is:
[tex]\[ P(A \text{ and } B) = \boxed{\frac{5}{12}} \][/tex]