46. Dev can hit a target 3 times in 6 shorts Pawan can hit the target 2 times in 6 shorts and Lakhan can hit the target 4 times in 4 shorts. What is the probability that at least 2 shorts hit the target

### Concept & Probability of Independent Events When multiple independent events occur, the probability of "at least" a certain number of successes can often be found by calculating the probability of the exact required scenarios or by subtracting the unwanted scenarios from 1. $$P(\text{at least 2}) = P(\text{exactly 2}) + P(\text{all 3})$$ ### Step-by-Step Solution * Let's find the probability of each person hitting the target in a single shot: * Probability Dev hits, $P(D) = \frac{3}{6} = \frac{1}{2}$. Probability he misses, $P(D') = \frac{1}{2}$. * Probability Pawan hits, $P(P) = \frac{2}{6} = \frac{1}{3}$. Probability he misses, $P(P') = \frac{2}{3}$. * Probability Lakhan hits, $P(L) = \frac{4}{4} = 1$. Probability he misses, $P(L') = 0$. * We need the probability of at least 2 hits. Since Lakhan always hits ($P(L)=1$), we only need to ensure at least one of the other two (Dev or Pawan) hits to get a total of $\geq 2$ hits. * **Method 1 (Direct Calculation):** * **Case 1: Exactly 2 hits.** This means Lakhan hits, one other hits, and one misses. $P(\text{Exactly 2}) = [P(D) \times P(P') \times P(L)] + [P(D') \times P(P) \times P(L)]$ $= (\frac{1}{2} \times \frac{2}{3} \times 1) + (\frac{1}{2} \times \frac{1}{3} \times 1) = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$. * **Case 2: All 3 hit.** $P(\text{All 3}) = P(D) \times P(P) \times P(L) = \frac{1}{2} \times \frac{1}{3} \times 1 = \frac{1}{6}$. * Total Probability $= \frac{1}{2} + \frac{1}{6} = \frac{3}{6} + \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$. * **Method 2 (Complementary Probability):** * $P(\text{at least 2}) = 1 - P(< 2 \text{ hits}) = 1 - [P(0 \text{ hits}) + P(1 \text{ hit})]$ * Since Lakhan always hits, 0 hits is impossible. * $P(1 \text{ hit})$ means only Lakhan hits, so Dev and Pawan must both miss. $P(1 \text{ hit}) = P(L) \times P(D') \times P(P') = 1 \times \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}$. * $P(\text{at least 2}) = 1 - 0 - \frac{1}{3} = \frac{2}{3}$. ### Exam Strategy & Shortcut Recognize that since Lakhan's probability of hitting is 1, he guarantees one hit. Therefore, the problem simplifies to: "What is the probability that at least one of Dev or Pawan hits?" $P(\text{at least one hits}) = 1 - P(\text{both miss}) = 1 - (P(D') \times P(P')) = 1 - (\frac{1}{2} \times \frac{2}{3}) = 1 - \frac{1}{3} = \frac{2}{3}$. ### Common Pitfall Ignoring the fact that Lakhan has a 100% hit rate can lead to unnecessarily complex calculations involving cases where Lakhan might miss, which wastes time and invites errors. ### Final Answer Therefore, the correct answer is **$\frac{2}{3}$**.