1. **State the problem:** We have a probability distribution for the number of games $X$ each child played: $X = \{2,3,4,5,6,7\}$ with probabilities $p(x) = \{0.05, 0.15, 0.15, 0.25, 0.20, 0.10, 0.10\}$. We need to find: a) The probability that a child will play exactly two games. b) The probability that a child will play more than four games. 2. **Recall the rules:** - The probability of an event $A$ is $P(A)$. - For discrete random variables, $P(X = x)$ is the probability that $X$ takes the value $x$. - The probability of $X$ being greater than a value $a$ is $P(X > a) = \sum_{x > a} P(X = x)$. 3. **Solve part (a):** The probability that a child plays exactly two games is given directly by $p(2)$. From the table, $p(2) = 0.05$. 4. **Solve part (b):** We want $P(X > 4) = P(X=5) + P(X=6) + P(X=7)$. From the table: $P(X=5) = 0.25$ $P(X=6) = 0.20$ $P(X=7) = 0.10$ So, $$P(X > 4) = 0.25 + 0.20 + 0.10 = 0.55$$ 5. **Final answers:** a) $P(X=2) = 0.05$ b) $P(X > 4) = 0.55$