1. **Problem Statement:** You flip four coins. Let $X$ be the number of heads obtained. 2. **Sample Space:** Each coin can be Head (H) or Tail (T). The sample space for four coins is all sequences of length 4 with H or T: $$\{HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT\}$$ 3. **Possible values of $X$:** $X$ counts the number of heads, so possible values are: $$0, 1, 2, 3, 4$$ 4. **Type of Random Variable:** $X$ is discrete because it takes countable values. 5. **Probability Distribution:** The probability of each number of heads is given by the binomial distribution with $n=4$ and $p=0.5$: $$P(X = k) = \binom{4}{k} (0.5)^k (0.5)^{4-k} = \binom{4}{k} (0.5)^4$$ Calculate each: - $P(0) = \binom{4}{0} (0.5)^4 = 1 \times \frac{1}{16} = \frac{1}{16}$ - $P(1) = \binom{4}{1} (0.5)^4 = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4}$ - $P(2) = \binom{4}{2} (0.5)^4 = 6 \times \frac{1}{16} = \frac{6}{16} = \frac{3}{8}$ - $P(3) = \binom{4}{3} (0.5)^4 = 4 \times \frac{1}{16} = \frac{4}{16} = \frac{1}{4}$ - $P(4) = \binom{4}{4} (0.5)^4 = 1 \times \frac{1}{16} = \frac{1}{16}$ 6. **Summary Table:** | $X$ | 0 | 1 | 2 | 3 | 4 | |-----|---|---|---|---|---| | $P(X)$ | $\frac{1}{16}$ | $\frac{1}{4}$ | $\frac{3}{8}$ | $\frac{1}{4}$ | $\frac{1}{16}$ | This distribution sums to 1: $$\frac{1}{16} + \frac{4}{16} + \frac{6}{16} + \frac{4}{16} + \frac{1}{16} = \frac{16}{16} = 1$$ 7. **Histogram:** The histogram bars would have heights corresponding to these probabilities at each $X$ value. Final answer: The probability distribution for the number of heads when flipping four coins is as shown in the table above.