Question

A promblem based on statistics 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,
i) find the probability that only one of them solves it correctly.
ii) find the probability that problem is solved.

Answer 1

$P$ ($A$ solving the question)=$\frac{1}{2}$

$P$($B$ solving the question)=$\frac{1}{3}$

$P$($C$ solving the question)=$\frac{1}{4}$

i)$P$ (only one at then solves the question)=$P$(only $A$ solves)+(only $B$ solves)+($P$ only $C$ solves)

$=P(A\cap B^0\cap C^0)+P(A^0\cap B\cap C^0)+P(A^0\cap B^0\cap C)$

$=\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}+\frac{1}{2}\times \frac{1}{3}\times \frac{3}{4}+\frac{1}{2}\times \frac{2}{3}\times \frac{1}{4}$

$=\frac{1}{4}+\frac{1}{8}+\frac{1}{12}=\frac{6+3+2+}{24}=\frac{11}{24}$

ii)$P$ (question is solved)=$P$(one solves it )+$P$(two of then solves it )+(all three solves)

$=\frac{11}{24}P(A\cap B\cap C^0)+P(A\cap B^0\cap C+P(A^0\cap B\cap C)+P(A\cap B\cap C)$

$=\frac{11}{24}+P\left(A\right).P\left(B\right).P\left(C^0\right)+P\left(A\right).P\left(B^0\right).P\left(C\right)+P\left(A^0\right).P\left(B\right).P\left(C\right)+P\left(A\right).P\left(B\right).P\left(C\right)$

$=\frac{11}{24}+\frac{1}{2}.\frac{1}{3}.\frac{3}{4}+\frac{1}{2}.\frac{2}{3}.\frac{1}{4}+\frac{1}{2}.\frac{1}{3}.\frac{1}{4}+\frac{1}{2}.\frac{1}{3}.\frac{1}{4}$

$=\frac{11}{24}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}+\frac{1}{24}=\frac{11+3+2+2}{24}$

$=\frac{18}{24}$

$=\frac{3}{4}$