To solve this problem, we need to find the probability that both event A (the coin does NOT land on heads) and event B (the die lands on 4) occur.
### Step-by-Step Solution:
1. Determine the probability of Event A (The coin does NOT land on heads):
- A fair coin has two sides: heads (H) and tails (T).
- The total probability for any side is 1, and each side has an equal probability of 1/2.
- Since we're looking for the event where the coin does NOT land on heads, it must land on tails.
- So, the probability of Event A occurring is:
[tex]\[
P(A) = \frac{1}{2} = 0.5
\][/tex]
2. Determine the probability of Event B (The die lands on 4):
- A fair six-sided die has six faces numbered from 1 to 6.
- Each face has an equal probability of 1/6.
- So, the probability of Event B occurring is:
[tex]\[
P(B) = \frac{1}{6} \approx 0.1667
\][/tex]
3. Multiply the probabilities of the two independent events:
- Since the events are independent, the probability of both events occurring (Event A and Event B) is the product of their individual probabilities.
- Using the formula for independent events:
[tex]\[
P(A \text{ and } B) = P(A) \cdot P(B)
\][/tex]
- Substitute the values we found:
[tex]\[
P(A \text{ and } B) = 0.5 \cdot 0.1667
\][/tex]
4. Calculate the result:
[tex]\[
P(A \text{ and } B) = 0.0833
\][/tex]
Thus, the probability that both the coin does not land on heads and the die lands on 4 is approximately [tex]\( 0.0833 \)[/tex]. In simplest fraction form, this is:
[tex]\[
P(A \text{ and } B) = \frac{1}{12}
\][/tex]
