1. **Problem Statement:** We are given that 30% of students offered English, 20% offered Hindi, and 10% offered both English and Hindi. We need to find the probability that a randomly selected student offered English or Hindi. 2. **Formula Used:** The probability of the union of two events $A$ and $B$ is given by: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ where $P(A)$ is the probability of event $A$, $P(B)$ is the probability of event $B$, and $P(A \cap B)$ is the probability of both events occurring. 3. **Assigning Values:** - Let $E$ be the event that a student offered English. - Let $H$ be the event that a student offered Hindi. Given: $$P(E) = 0.30$$ $$P(H) = 0.20$$ $$P(E \cap H) = 0.10$$ 4. **Calculate $P(E \cup H)$:** $$P(E \cup H) = P(E) + P(H) - P(E \cap H) = 0.30 + 0.20 - 0.10 = 0.40$$ 5. **Interpretation:** The probability that a randomly selected student offered English or Hindi is $0.40$, which as a fraction is $\frac{2}{5}$. **Final Answer:** $$\boxed{\frac{2}{5}}$$