Skip to main contentSkip to solution

A problem is given to 3 students A,B and C, whose probabilities of solving are 12,13\frac{1}{2}, \frac{1}{3} and 14\frac{1}{4} respectively, then

(A). P(Problem solved) = 34\frac{3}{4}

(B). P(Problem not solved) = 14\frac{1}{4}

(C). P(Problem solved by exactly one student) = 724\frac{7}{24}

(D). P(Problem solved by at least 2 students) = 1724\frac{17}{24}

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 3

P(not solved)=12⋅23⋅34=14P(\text{not solved})=\frac12\cdot\frac23\cdot\frac34=\frac14 and P(solved)=34P(\text{solved})=\frac34, so (A) and (B) hold. Exactly one solver gives 14+18+112=1124\frac14+\frac18+\frac{1}{12}=\frac{11}{24}, not 724\frac{7}{24}, so (C) fails. At least two is 34−1124=724\frac34-\frac{11}{24}=\frac{7}{24}, not 1724\frac{17}{24}, so (D) fails.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question