Question

Four students of class XII are given a problem to solve independently. Their respective chances of solving the problem are: \[ \frac{1}{2},\quad \frac{1}{3},\quad \frac{2}{3},\quad \frac{1}{5} \] Find the probability that at most one of them will solve the problem.

Answer 1

Let the students be \( A_1, A_2, A_3, A_4 \).

• \( P(A_1) = \frac{1}{2} \Rightarrow P(A_1') = \frac{1}{2} \)
• \( P(A_2) = \frac{1}{3} \Rightarrow P(A_2') = \frac{2}{3} \)
• \( P(A_3) = \frac{2}{3} \Rightarrow P(A_3') = \frac{1}{3} \)
• \( P(A_4) = \frac{1}{5} \Rightarrow P(A_4') = \frac{4}{5} \)

Case 1: None solves the problem

\[ P_0 = P(A_1') \cdot P(A_2') \cdot P(A_3') \cdot P(A_4') = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{3} \cdot \frac{4}{5} = \frac{8}{90} = \frac{4}{45} \]

Case 2: Exactly one solves the problem

Case 2.1: Only \( A_1 \) solves

\[ P = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{3} \cdot \frac{4}{5} = \frac{4}{45} \]

Case 2.2: Only \( A_2 \) solves

\[ P = \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{4}{5} = \frac{2}{45} \]

Case 2.3: Only \( A_3 \) solves

\[ P = \frac{2}{3} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{4}{5} = \frac{16}{45} \]

Case 2.4: Only \( A_4 \) solves

\[ P = \frac{1}{5} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{3} = \frac{2}{45} \]

Total for exactly one solves:

\[ P_1 = \frac{4}{45} + \frac{2}{45} + \frac{16}{45} + \frac{2}{45} = \frac{24}{45} \]

Final Answer:

\[ P(\text{At most one solves}) = P_0 + P_1 = \frac{4}{45} + \frac{24}{45} = \boxed{\frac{28}{45}} \]