Let's solve the problem step by step.

Part 1: Probability of getting exactly 5 heads

Since the coin is fair, each toss has a probability of $\frac{1}{2}$ for heads or tails. The number of heads $X$ obtained after 8 tosses follows a binomial distribution.

The probability of getting exactly $k$ heads in $n$ tosses is given by the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Where:

• $n = 8$ (the number of tosses),
• $k = 5$ (the number of heads),
• $p = \frac{1}{2}$ (the probability of getting heads on a single toss).

So, the probability of getting exactly 5 heads is:

$P(X = 5) = \binom{8}{5} \left(\frac{1}{2}\right)^5 \left(\frac{1}{2}\right)^{8-5}$

Simplifying:

$P(X = 5) = \binom{8}{5} \left(\frac{1}{2}\right)^8$

First, calculate the binomial coefficient $\binom{8}{5}$:

$\binom{8}{5} = \frac{8!}{5!(8-5)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

Now, calculate the probability:

$P(X = 5) = 56 \times \left(\frac{1}{2}\right)^8 = 56 \times \frac{1}{256} = \frac{56}{256} = \frac{7}{32}$

So, the probability of getting exactly 5 heads is $\frac{7}{32}$.

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Part 2: Expected Value of $X$

The expected value $E(X)$ for a binomial distribution $X \sim \text{Binomial}(n, p)$ is given by:

$E(X) = n \times p$

For $n = 8$ and $p = \frac{1}{2}$:

$E(X) = 8 \times \frac{1}{2} = 4$

So, the expected value of $X$ is $4$.

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Part 3: Variance of $X$

The variance $\text{Var}(X)$ for a binomial distribution $X \sim \text{Binomial}(n, p)$ is given by:

$\text{Var}(X) = n \times p \times (1 - p)$

For $n = 8$ and $p = \frac{1}{2}$:

$\text{Var}(X) = 8 \times \frac{1}{2} \times \frac{1}{2} = 8 \times \frac{1}{4} = 2$

So, the variance of $X$ is $2$.

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Final Answers:

• Probability of getting exactly 5 heads: $\frac{7}{32}$
• Expected value: $4$
• Variance: $2$

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Would you like more details or have any questions about the steps?

Here are some follow-up questions for further exploration:

1. How would the probability change if we were looking for 6 heads instead of 5?
2. What happens to the expected value and variance if the coin is biased, say with a 70% chance of heads?
3. How would the probability of 5 heads be calculated if the coin were tossed 10 times instead of 8?
4. What is the cumulative probability of getting 5 or fewer heads in 8 tosses?
5. How do we interpret the expected value in the context of this coin tossing experiment?

Tip: The binomial distribution works well when there are two possible outcomes (like heads or tails), and the number of trials is fixed. Make sure the trials are independent!