1. **Problem Statement:** Create the probability distribution for the experiment: "Four coins are tossed. Let X be the number of heads." 2. **Formula and Rules:** The number of heads in four coin tosses follows a binomial distribution with parameters $n=4$ (number of trials) and $p=0.5$ (probability of heads in each toss). The probability mass function (PMF) for a binomial distribution is: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $k=0,1,2,\ldots,n$. 3. **Calculate probabilities for each possible value of $X$:** - For $k=0$ heads: $$P(X=0) = \binom{4}{0} (0.5)^0 (0.5)^4 = 1 \times 1 \times 0.0625 = 0.0625$$ - For $k=1$ head: $$P(X=1) = \binom{4}{1} (0.5)^1 (0.5)^3 = 4 \times 0.5 \times 0.125 = 0.25$$ - For $k=2$ heads: $$P(X=2) = \binom{4}{2} (0.5)^2 (0.5)^2 = 6 \times 0.25 \times 0.25 = 0.375$$ - For $k=3$ heads: $$P(X=3) = \binom{4}{3} (0.5)^3 (0.5)^1 = 4 \times 0.125 \times 0.5 = 0.25$$ - For $k=4$ heads: $$P(X=4) = \binom{4}{4} (0.5)^4 (0.5)^0 = 1 \times 0.0625 \times 1 = 0.0625$$ 4. **Summary of the probability distribution:** | Number of Heads ($X$) | Probability $P(X)$ | |-----------------------|-------------------| | 0 | 0.0625 | | 1 | 0.25 | | 2 | 0.375 | | 3 | 0.25 | | 4 | 0.0625 | 5. **Explanation:** Each probability corresponds to the chance of getting exactly $k$ heads in 4 tosses. The probabilities sum to 1, confirming a valid distribution. Final answer: $$P(X=0)=0.0625, P(X=1)=0.25, P(X=2)=0.375, P(X=3)=0.25, P(X=4)=0.0625$$