An unbiased coin is tossed four times. Find the probability of the given event.

The coin lands heads exactly once.

To find the probability that the coin lands heads exactly once in four tosses, we can follow these steps:

### Step 1: Total Possible Outcomes
Each toss of the coin has two possible outcomes: heads (H) or tails (T). Since the coin is tossed 4 times, the total number of possible outcomes is:

\[
2^4 = 16
\]

### Step 2: Favorable Outcomes (Exactly One Head)
We now need to count how many of these outcomes include exactly one head. The possible outcomes where exactly one head appears are:

- HTTT
- THTT
- TTHT
- TTTH

There are 4 ways to arrange one head among the four tosses. Hence, there are 4 favorable outcomes.

### Step 3: Probability Calculation
The probability of getting exactly one head is the number of favorable outcomes divided by the total possible outcomes:

\[
P(\text{exactly 1 head}) = \frac{4}{16} = \frac{1}{4}
\]

So, the probability that the coin lands heads exactly once in four tosses is \( \frac{1}{4} \).

Would you like further details on any part of the solution?

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Here are some related questions to expand your understanding:

1. What is the probability of getting exactly two heads in four tosses?
2. How would the probability change if the coin was tossed five times instead of four?
3. What is the probability of getting no heads in four tosses?
4. What is the probability of getting at least one head in four tosses?
5. How can we generalize the formula for exactly \( k \) heads in \( n \) tosses?

**Tip:** For problems involving repeated independent events, the binomial probability formula is often very helpful!

An unbiased coin is tossed four times. Find the probability of the given event: The coin lands heads exactly once.

Explore how to calculate the probability of getting exactly one head in four tosses of an unbiased coin. This problem involves key probability concepts and combinatorics, providing a step-by-step solution. Suitable for students in Grades 6-8.

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