1. **State the problem:** We have a discrete random variable $X$ taking values $0,1,2,3$. Given: $P(X \leq 2) = 0.9$, $P(X \leq 1) = 0.5$, and $E(X) = 1.4$. We need to find the variance $\mathrm{Var}(X)$. 2. **Recall formulas:** - Variance formula: $$\mathrm{Var}(X) = E(X^2) - [E(X)]^2$$ - We know $E(X) = 1.4$. 3. **Find probabilities:** - $P(X \leq 1) = P(X=0) + P(X=1) = 0.5$ - $P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) = 0.9$ From these: - $P(X=2) = P(X \leq 2) - P(X \leq 1) = 0.9 - 0.5 = 0.4$ - Since total probability is 1, $P(X=3) = 1 - P(X \leq 2) = 1 - 0.9 = 0.1$ 4. **Find $P(X=0)$ and $P(X=1)$:** Let $p_0 = P(X=0)$ and $p_1 = P(X=1)$. We have $p_0 + p_1 = 0.5$. 5. **Use expected value to find $p_0$ and $p_1$:** $$E(X) = 0 \cdot p_0 + 1 \cdot p_1 + 2 \cdot 0.4 + 3 \cdot 0.1 = p_1 + 0.8 + 0.3 = p_1 + 1.1$$ Given $E(X) = 1.4$, so: $$p_1 + 1.1 = 1.4 \implies p_1 = 0.3$$ Then: $$p_0 = 0.5 - p_1 = 0.5 - 0.3 = 0.2$$ 6. **Calculate $E(X^2)$:** $$E(X^2) = 0^2 \cdot 0.2 + 1^2 \cdot 0.3 + 2^2 \cdot 0.4 + 3^2 \cdot 0.1 = 0 + 0.3 + 4 \cdot 0.4 + 9 \cdot 0.1 = 0.3 + 1.6 + 0.9 = 2.8$$ 7. **Calculate variance:** $$\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = 2.8 - (1.4)^2 = 2.8 - 1.96 = 0.84$$ **Final answer:** The variance of $X$ is $0.84$.