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Question

Mathematics Question on Conditional Probability

A problem in mathematics is given to three students A, B, C and their respective probability of solving the problem is 12,13\frac{1}{2}, \frac{1}{3} and 14\frac{1}{4}. Probability that the problem is solved is

A

34\frac{3}{4}

B

12\frac{1}{2}

C

23\frac{2}{3}

D

13\frac{1}{3}

Answer

34\frac{3}{4}

Explanation

Solution

P(E1)=12,P(E2)=13P(E3)=14;P\left(E_{1}\right) = \frac{1}{2} ,P\left(E_{2} \right) = \frac{1}{3} P\left(E_{3}\right) = \frac{1}{4} ; P(E1UE2UE3)=1−P(Eˉ1)P(Eˉ2)P(Eˉ3)P\left(E_{1}UE_{2}UE_{3} \right) = 1- P \left(\bar{E}_{1}\right)P \left(\bar{E}_{2} \right)P \left(\bar{E}_{3}\right) =1−(1−12)(1−13)(1−14) = 1- \left(1- \frac{1}{2}\right) \left(1- \frac{1}{3}\right)\left(1- \frac{1}{4}\right) =1−12×23×34=34= 1- \frac{1}{2} \times\frac{2}{3}\times\frac{3}{4} = \frac{3}{4}