Question

A problem in Mathematics is given to three students P, Q and R whose chances of solving it are \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\) respectively. Arrange the following probabilities in decreasing order: A. The probability that P cannot solve the problem
B. The probability that Q cannot solve the problem
C. The probability that P, Q and R cannot solve the problem
D. The probability that the problem will be solved by at least one student Choose the correct answer:

• A. A, B, C, D
• B. B, C, D, A
• C. D, A, B, C
• D. D, B, A, C

Hint:

“At least one” = 1 − “none” → fastest way to solve such problems.

Answer 1

The Correct Option is D

Concept: Use probability rules:
• Complement: \(P(\text{not A}) = 1 - P(A)\)
• Independent events multiplication
• At least one = complement of none

Step 1: Given probabilities.
\[ P(P \text{ solves}) = \frac{1}{2}, \quad P(Q \text{ solves}) = \frac{1}{3}, \quad P(R \text{ solves}) = \frac{1}{4} \]

Step 2: Compute individual “cannot solve” probabilities.
A: \[ P(P \text{ cannot}) = 1 - \frac{1}{2} = \frac{1}{2} \] B: \[ P(Q \text{ cannot}) = 1 - \frac{1}{3} = \frac{2}{3} \] R: \[ P(R \text{ cannot}) = 1 - \frac{1}{4} = \frac{3}{4} \]

Step 3: Compute probability that all cannot solve (C).
\[ P(C) = \frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \] \[ = \frac{1}{4} \]

Step 4: Compute probability that at least one solves (D).
\[ P(D) = 1 - P(\text{none solve}) \] \[ = 1 - \frac{1}{4} = \frac{3}{4} \]

Step 5: List values.
\[ D = \frac{3}{4}, \quad B = \frac{2}{3}, \quad A = \frac{1}{2}, \quad C = \frac{1}{4} \]

Step 6: Arrange in decreasing order.
\[ \frac{3}{4} > \frac{2}{3} > \frac{1}{2} > \frac{1}{4} \] \[ D > B > A > C \] Final Answer: \[ \boxed{D, B, A, C} \]