MAT Exam 2024 Paper

Concept:Use the independent probabilities of each person hitting or missing, and add the cases where exactly two hit or all three hit.Explanation:Amar hits $\frac{5}{6}$, misses $\frac{1}{6}$.Binod hits $\frac{4}{5}$, misses $\frac{1}{5}$.Chintu hits $\frac{3}{4}$, misses $\frac{1}{4}$.All three hit:$\frac{5}{6} \times \frac{4}{5} \times \frac{3}{4} = \frac{1}{2}$Exactly two hit:Amar and Binod hit, Chintu misses:$\frac{5}{6} \times \frac{4}{5} \times \frac{1}{4} = \frac{1}{6}$Amar and Chintu hit, Binod misses:$\frac{5}{6} \times \frac{1}{5} \times \frac{3}{4} = \frac{1}{8}$Binod and Chintu hit, Amar misses:$\frac{1}{6} \times \frac{4}{5} \times \frac{3}{4} = \frac{1}{10}$So, probability of exactly two hits:$\frac{1}{6} + \frac{1}{8} + \frac{1}{10} = \frac{47}{120}$Therefore, probability that at least two shots hit:$\frac{47}{120} + \frac{1}{2} = \frac{47}{120} + \frac{60}{120} = \frac{107}{120}$Answer:$\frac{107}{120}$, which is option D. None of these.