A coin is tossed thrice. Let a random variable X count the number of heads. What's P(X=2)? (Report your answer in decimals.)

To solve the problem, we need to determine the probability \( P(X = 2) \), where \( X \) represents the number of heads in three coin tosses. Each coin toss is independent, and there are two possible outcomes (head or tail) for each toss.

### Steps:

1. **Number of possible outcomes**: 
   Since the coin is tossed three times, the total number of possible outcomes is \( 2^3 = 8 \).

2. **Favorable outcomes (X = 2)**: 
   We want to count the number of outcomes where exactly 2 heads occur. The number of ways to choose 2 heads from 3 tosses is given by the binomial coefficient:
   \[
   \binom{3}{2} = \frac{3!}{2!(3-2)!} = 3
   \]
   The favorable outcomes are: HHT, HTH, and THH.

3. **Probability**: 
   The probability of exactly 2 heads is the ratio of favorable outcomes to the total outcomes:
   \[
   P(X = 2) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{3}{8}
   \]

Converting to decimal:
\[
P(X = 2) = 0.375
\]

Thus, the probability \( P(X = 2) \) is \( \boxed{0.375} \).

Would you like more details or have any questions?

Here are 5 related questions:
1. What is the probability of getting exactly 1 head in three coin tosses?
2. What is the probability of getting 3 heads in three coin tosses?
3. How would the probability change if the coin is biased?
4. What is the expected value of \( X \) in this experiment?
5. What is the probability of getting at least 2 heads?

**Tip**: The binomial distribution is often useful when dealing with independent trials, such as multiple coin tosses.

Learn how to calculate the probability of getting exactly 2 heads when tossing a coin three times. This problem involves binomial distribution, combinatorics, and key probability concepts, making it ideal for students in Grades 9-11.

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