0 votes
in Probability by (17.9k points)
closed by

A problem in mathematics is given to 3 students whose chances of solving it are \(\frac{1}{2},\frac{1}{3},\frac{1}{4},\) What is the probability that the problem is solved?

1 Answer

+1 vote
by (17.1k points)
selected by
Best answer

Let the respective events of solving the problem be denoted by A, B, C. 

Then P(A) = \(\frac{1}{2}\) , P(B) = \(\frac{1}{3} \) , P(C) = \(\frac{1}{4}\)

Clearly, A, B, C are independent events and the problem will be considered to have been solved if at least one student solves it.

 ∴  Required probability = P(A or B or C) = \(P(A\,\cup\,B\,\cup\,C)\) = \(1- P(\overline{A})P(\overline{B})P(\overline{C})\)

\(\overline{A}, \overline{B},\overline{C}\) are the respective events of not solving the problem.


 \(P(\overline{A}) =1-P(A) =1- \frac{1}{2} =\frac{1}{2}\) ,

 \(P(\overline{B}) =1-P(B) =1- \frac{1}{3} =\frac{2}{3}\) 

 \(P(\overline{C}) =1-P(C) =1- \frac{1}{4} =\frac{3}{4}\)

 Required probability \(1- P(\overline{A})P(\overline{B})P(\overline{C}) =1- \frac{1}{2}\times \frac{2}{3}\times \frac{3}{4} =1-\frac{1}{4} \)


Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.