Question

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

• A. $\frac{13}{28}$
• B. $\frac{11}{28}$
• C. $\frac{9}{35}$
• D. $\frac{26}{35}$

Answer 1

Answer: (A)

$\frac{13}{28}$

We have
When $A$ and $B$ plays , probability of winning of $A$ is $P\left(A\right)=\frac{1}{4}$ and probability of winning of $B$ is $P\left(B\right)=\frac{3}{4}$
When $A$ and $C$ plays, probability of winning of $A$ $A=\frac{3}{5}$ and probability of winning of $C$ is $\frac{2}{5}$.
When $A$ and $D$ plays, probability of winning of $A$ is $A=\frac{4}{7}$ and probability of winning of $D=\frac{3}{7}.$
In $3$ trials, $A$ has to succeed at least twice. So following four cases are possible :
1) $A$ may either win with all three
So, the chance in this case is $\frac{1}{4}.\frac{3}{5}.\frac{4}{7}$
2) $A$ fails with $B$ and wins with $C$ and $D$.
So, the chance in this case is $\frac{3}{4}.\frac{3}{5}.\frac{4}{7}$
3) $A$ fails with $C$ and wins with $B$ and $D$.
So, the chance in this case is $\frac{1}{4}.\frac{2}{5}.\frac{4}{7}$
4) $A$ fails with $D$ and wins with $B$ and $C$.
So, the chance in this case is $\frac{1}{4}.\frac{3}{5}.\frac{3}{7}$
The sum of these four gives the reqd. chance $\frac{12+36+8+9}{140}=\frac{55}{140}=\frac{11}{28}.$