A question is given to three students A, B and C whose chances of solving it are \(\frac{1}{2},\frac{1}{3}\) and \(\frac{1}{4}\) respectively. What is the probability that the question will be solved?

Concept:

Complement of an event:

The complement of an event is the subset of outcomes in the sample space that are not in the event. A complement is itself an event.

The probability of the complement of an event is one minus the probability of the event.

P (A’) or P (A^c) or \({\rm{P\;}}\left( {{\rm{\bar A}}} \right)\) = 1 - P (A)

Where, P (A) be probability of Event A and P (A’) or P (A^c) or \({\rm{P\;}}\left( {{\rm{\bar A}}} \right)\)  be probability of the complement of Event A

Calculation:

Given:

Probability of solving problem by A = P (A) =1/2

Probability of solving problem by B = P (B) =1/3

Probability of solving problem by C = P (C) =1/4

P (problem will not solved by A) = 1 – (1/2) = 1/2

P (problem will not solved by B) = 1 – (1/3) = 2/3

P (problem will not solved by C) = 1 – (1/4) = 3/4

P (Problem will be solved) = 1 – P (problem will not solved by A, B and C)

\(= 1 - \left( {\frac{1}{2}\; \times \;\frac{2}{3}\; \times \;\frac{3}{4}} \right) = 1 - \;\frac{1}{4} = \;\frac{3}{4}\)