1. **State the problem:** We have a discrete random variable $X$ representing the number of games played per visit with the given probability distribution: $$\begin{array}{c|ccccccc} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(x) & 0.05 & 0.15 & 0.15 & 0.25 & 0.20 & 0.10 & 0.10 \end{array}$$ We want to find: - (a) The probability that a child will play more than four games, i.e., $P(X > 4)$. - (b) The probability that a child will play at least two games, i.e., $P(X \geq 2)$. 2. **Calculate $P(X > 4)$:** This means $X$ can be 5, 6, or 7. $$P(X > 4) = P(5) + P(6) + P(7) = 0.20 + 0.10 + 0.10 = 0.40$$ 3. **Calculate $P(X \geq 2)$:** This means $X$ can be 2, 3, 4, 5, 6, or 7. $$P(X \geq 2) = P(2) + P(3) + P(4) + P(5) + P(6) + P(7)$$ $$= 0.15 + 0.15 + 0.25 + 0.20 + 0.10 + 0.10 = 0.95$$ 4. **Summary of answers:** - $P(X > 4) = 0.40$ - $P(X \geq 2) = 0.95$ --- **Note:** The second question (7.23) about mean and variance is not solved here as per instructions to solve only the first question.