Question

$A^{\prime }s$ skill is to $B^{\prime }s$ as $1:3$ ; to $C^{\prime }s$ as $3:2$ ; and to $D^{\prime }s$ as $4:3$: find the chance that A in three trials, one with each person, will succeed twice at least.

Answer 1

Chance of winning of $A$ with $B$ $=\frac{1}{4}$

$\Rightarrow$ chance of loosing with $B$ $=1-\frac{1}{4}=\frac{3}{4}$

Chance of winning of $A$ with $C$ $=\frac{3}{5}$

$\Rightarrow$ chance of loosing with $C$ $=1-\frac{3}{5}=\frac{2}{5}$

Chance of winning with $D$ $=\frac{4}{7}$

$\Rightarrow$ chance of loosing with $D$ =$1-\frac{4}{7}=\frac{3}{7}$

$A$ can succeed aleast twice if he win with all three or loose with one and win with the other two

$\Rightarrow P\left(E\right)=\frac{1}{4}.\frac{3}{5}.\frac{4}{7}+\frac{3}{4}.\frac{3}{5}.\frac{4}{7}+\frac{1}{4}.\frac{2}{5}.\frac{4}{7}+\frac{1}{4}.\frac{3}{5}.\frac{3}{7}\Rightarrow P\left(E\right)=\frac{13}{28}$