A’s skill is to B’s 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.

Hint: In this question we can assume that the skills are proportional to the chance of winning. Therefore, in a trial, the probability that A will succeed depends on the ratio of the skills of A to the skills of the other person that A is facing. Thus, having found the probability of succeeding in each trial, we can find the probability of the events given in the question.

Complete step-by-step answer:

We can assume that the probability of success of each person is proportional to its skills………….(1.1)
We can find out the probability of succeeding of A in each trial as follows:
Trial of A with B
Let P1(A) be the probability of succeeding of A and P1(B) be the probability of succeeding of A in a trial with B. Form (1.1), we find that
$\dfrac{{{P}_{1}}(A)}{{{P}_{1}}(B)}=\dfrac{1}{3}\Rightarrow {{P}_{1}}(B)=3{{P}_{1}}(A)..............(1.2)$
However, the total probability must be 1. Therefore,
$\begin{align}
  & {{P}_{1}}(A)+{{P}_{1}}(B)=1\Rightarrow {{P}_{1}}(A)+3{{P}_{1}}(A)=1(\text{from equation 1}\text{.2}) \\
 & \Rightarrow 4{{P}_{1}}(A)=1\Rightarrow {{P}_{1}}(A)=\dfrac{1}{4}.....................(1.3) \\
\end{align}$
Trial of A with C
Let P2(A) be the probability of succeeding of A and P2(C) be the probability of succeeding of A in a trial with B. Form (1.1), we find that
$\dfrac{{{P}_{2}}(A)}{{{P}_{2}}(C)}=\dfrac{3}{2}\Rightarrow {{P}_{2}}(C)=\dfrac{2}{3}{{P}_{2}}(A)..............(1.4)$
However, the total probability must be 1. Therefore,
$\begin{align}
  & {{P}_{2}}(A)+{{P}_{2}}(C)=1\Rightarrow {{P}_{2}}(A)+\dfrac{2}{3}{{P}_{2}}(A)=1(\text{from equation 1}\text{.4}) \\
 & \Rightarrow 5{{P}_{2}}(A)=3\Rightarrow {{P}_{2}}(A)=\dfrac{3}{5}..................(1.5) \\
\end{align}$
Trial of A with D
Let P3(A) be the probability of succeeding of A and P3(B) be the probability of succeeding of A in a trial with B. Form (1.1), we find that
$\dfrac{{{P}_{3}}(A)}{{{P}_{3}}(D)}=\dfrac{4}{3}\Rightarrow {{P}_{3}}(D)=\dfrac{3}{4}{{P}_{3}}(A)..............(1.6)$
However, the total probability must be 1. Therefore,
$\begin{align}
  & {{P}_{3}}(A)+{{P}_{3}}(D)=1\Rightarrow {{P}_{3}}(A)+\dfrac{3}{4}{{P}_{3}}(A)=1(\text{from equation 1}\text{.6}) \\
 & \Rightarrow 7{{P}_{3}}(A)=4\Rightarrow {{P}_{3}}(A)=\dfrac{4}{7}...............(1.7) \\
\end{align}$

Now, the events corresponding to at least two successes are WWW, WWL, WLW, LWW where W represents win and L represents loss in a trial. As the trials do not depend on each other, the probability of the composite events will be the product of the probability of individual events.
However, in the individual event I, the probability that A succeeds is \[{{P}_{i}}\left( A \right)\]and that he loses if\[\left( 1-{{P}_{i}}\left( A \right) \right)\]. Thus,
\[\begin{align}
  & P(\text{A succeeds at least twice})=P(WWW)+P(WWL)+P(WLW)+P(LWW) \\
 & ={{P}_{1}}(A)\times {{P}_{3}}(A)\times {{P}_{2}}(A)+{{P}_{1}}(A)\times {{P}_{2}}(A)\times \left( 1-{{P}_{3}}(A) \right)+{{P}_{1}}(A)\times \left( 1-{{P}_{2}}(A) \right){{P}_{3}}(A)+\left( 1-{{P}_{1}}(A) \right){{P}_{2}}(A)\times {{P}_{3}}(A) \\
 & =\dfrac{1}{4}\times \dfrac{3}{5}\times \dfrac{4}{7}+\dfrac{1}{4}\times \dfrac{3}{5}\times \dfrac{3}{7}+\dfrac{1}{4}\times \dfrac{2}{5}\times \dfrac{4}{7}+\dfrac{3}{4}\times \dfrac{3}{5}\times \dfrac{4}{7} \\
 & =\dfrac{12+9+8+36}{4\times 5\times 7}=\dfrac{65}{4\times 5\times 7}=0.4628 \\
\end{align}\]
Thus, the probability of A succeeding twice at least is 0.4628.

Note: In this question, we should be careful about the events to be summed and remember that succeeding at least twice should also include the event of succeeding all three times.