In this problem, $X$ is the random variable representing the number of heads when flipping four coins. To analyze this, we can think of $X$ as having possible values corresponding to the number of heads obtained in the four flips. The random variable $X$ can take the values $\{0, 1, 2, 3, 4\}$, since we can get anywhere from 0 to 4 heads.

1. Sample Space:

The sample space for flipping four coins consists of all possible outcomes. Each flip can be either heads (H) or tails (T), so there are $2^4 = 16$ possible outcomes.

2. Values of $X$:

$X$ can take the following values:

• $X = 0$: No heads (all tails, $TTTT$)
• $X = 1$: One head (e.g., $HTTT$, $THTT$, etc.)
• $X = 2$: Two heads (e.g., $HHTT$, $HTHT$, etc.)
• $X = 3$: Three heads (e.g., $HHHT$, $THHH$, etc.)
• $X = 4$: Four heads (all heads, $HHHH$)

3. Probability Distribution:

The probability of getting a specific number of heads follows a binomial distribution because each coin flip is an independent Bernoulli trial with two outcomes (heads or tails). The binomial distribution formula is:

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

• $n = 4$ (number of trials, or coin flips),
• $p = 0.5$ (probability of getting heads on each flip),
• $k$ is the number of heads.

Let's calculate the probabilities for each possible value of $X$:

• $P(X = 0) = \binom{4}{0} (0.5)^0 (0.5)^4 = 1 \times 1 \times \frac{1}{16} = \frac{1}{16}$
• $P(X = 1) = \binom{4}{1} (0.5)^1 (0.5)^3 = 4 \times 0.5 \times \frac{1}{8} = \frac{4}{16} = \frac{1}{4}$
• $P(X = 2) = \binom{4}{2} (0.5)^2 (0.5)^2 = 6 \times \frac{1}{4} \times \frac{1}{4} = \frac{6}{16} = \frac{3}{8}$
• $P(X = 3) = \binom{4}{3} (0.5)^3 (0.5)^1 = 4 \times \frac{1}{8} \times 0.5 = \frac{4}{16} = \frac{1}{4}$
• $P(X = 4) = \binom{4}{4} (0.5)^4 (0.5)^0 = 1 \times \frac{1}{16} = \frac{1}{16}$

4. Probability Distribution Table:

$X$ (Number of Heads)	Probability $P(X)$
0	$\frac{1}{16}$
1	$\frac{4}{16} = \frac{1}{4}$
2	$\frac{6}{16} = \frac{3}{8}$
3	$\frac{4}{16} = \frac{1}{4}$
4	$\frac{1}{16}$

5. Expected Value of $X$:

The expected value $E(X)$ of a binomial distribution is calculated as: $E(X) = n \cdot p$ For this case: $E(X) = 4 \cdot 0.5 = 2$ So, the expected number of heads is 2.

Would you like further details or clarification on any part of this?

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Related Questions:

1. What is the variance and standard deviation for the number of heads in this scenario?
2. How would the probability distribution change if the probability of heads is not 0.5 but 0.7?
3. Can you explain how the binomial coefficient $\binom{n}{k}$ is calculated?
4. What is the cumulative probability of getting at most 2 heads?
5. What is the probability of getting an even number of heads?

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Tip:

When dealing with binomial distributions, remember that the sum of all probabilities for possible values of $X$ should always equal 1, ensuring that the probabilities cover all possible outcomes.