Question

Three candidates solve a question. Odds in favor of the correct answer are $5:2$, $4:3$, $3:4$ respectively for the three candidates. What is the probability that at least two of them solve the question correctly?

Answer 1

Hint: We will first calculate the probabilities of solving the question by three students individually by using the given odds probability. From the individual probability of the students to solve the question, we will calculate the probability that at least two of them solve the question correctly. Here we will use the probability of the complementary events by using the formula $\text{P}\left( {\bar{A}} \right)=1-\text{P}\left( A \right)$, where $\text{P}\left( A \right)$ is the probability of the event $A$ and $\text{P}\left( {\bar{A}} \right)$ is the probability of the complementary event to the event $A$.

Complete step-by-step answer:
We have been given that the odds in favor of the three candidates solve a question correctly are $5:2$, $4:3$, $3:4$.
Let us consider that the three students are $x,y,z$, hence we can write that
The odds in favor probability of $x$ solving the problem is $5:2$.
The odds in favor probability of $y$ solving the problem is $4:3$.
The odds in favor probability of $y$ solving the problem is $3:4$.
We know that the odd probability of an event is the ratio of favorable outcomes of an event to unfavorable outcomes of that event.
Hence we have
The favorable outcomes for $x$ solving the problem is $5$ and unfavorable outcomes for $x$ solving the problem is $2$, so the probability of $x$ solving the problem is given by
$\begin{align}
  & \text{P}\left( x \right)=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}} \\
 & =\dfrac{\text{favorable outcomes}}{\text{favorable outcomes}+\text{unfavorable outcomes}} \\
 & =\dfrac{5}{5+2} \\
 & \text{P}\left( x \right)=\dfrac{5}{7} \\
\end{align}$
Now the problem is not solved by $x$ is the complementary event for $x$ solving the problem, hence we have the probability of $x$ not solving the problem given by
$\begin{align}
  & \text{P}\left( {\bar{x}} \right)=1-\text{P}\left( x \right) \\
 & =1-\dfrac{5}{7} \\
 & =\dfrac{7-5}{7} \\
 & =\dfrac{2}{7} \\
\end{align}$
The favorable outcomes for $y$ solving the problem is $4$ and unfavorable outcomes for $y$ solving the problem is $3$, so we can write the probability of $y$ solving the problem as
$\begin{align}
  & \text{P}\left( y \right)=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}} \\
 & =\dfrac{\text{favorable outcomes}}{\text{favorable outcomes}+\text{unfavorable outcomes}} \\
 & =\dfrac{4}{4+3} \\
 & \text{P}\left( y \right)=\dfrac{4}{7} \\
\end{align}$
Now the problem is not solved by $y$ is the complementary event for $y$ solving the problem, hence we have the probability of $y$ not solving the problem given by
$\begin{align}
  & \text{P}\left( {\bar{y}} \right)=1-\text{P}\left( y \right) \\
 & =1-\dfrac{4}{7} \\
 & =\dfrac{7-4}{7} \\
 & =\dfrac{3}{7} \\
\end{align}$
The favorable outcomes for $z$ solving the problem is $3$ and unfavorable outcomes for $z$ solving the problem is $4$, so we have the probability of $z$ solving the problem given by
$\begin{align}
  & \text{P}\left( z \right)=\dfrac{\text{favorable outcomes}}{\text{total number of outcomes}} \\
 & =\dfrac{\text{favorable outcomes}}{\text{favorable outcomes}+\text{unfavorable outcomes}} \\
 & =\dfrac{3}{3+4} \\
 & \text{P}\left( z \right)=\dfrac{3}{7} \\
\end{align}$
Now the problem is not solved by $z$ is the complementary event for $z$ solving the problem, hence we have the probability of $z$ not solving the problem given by
$\begin{align}
  & \text{P}\left( {\bar{z}} \right)=1-\text{P}\left( z \right) \\
 & =1-\dfrac{3}{7} \\
 & =\dfrac{7-3}{7} \\
 & =\dfrac{4}{7} \\
\end{align}$
Now from the given data we have calculate the probabilities as
$\text{P}\left( x \right)=\dfrac{5}{7}$, $\text{P}\left( y \right)=\dfrac{4}{7}$, $\text{P}\left( z \right)=\dfrac{3}{7}$
$\text{P}\left( {\bar{x}} \right)=\dfrac{2}{7}$, $\text{P}\left( {\bar{y}} \right)=\dfrac{3}{7}$, $\text{P}\left( {\bar{z}} \right)=\dfrac{4}{7}$
Now we have to calculate the probability that at least two of them solve the problem and that is given by
$\text{P}\left( r \right)=\text{P}\left( x\cap y\cap \bar{z} \right)+\text{P}\left( x\cap \bar{y}\cap z \right)+\text{P}\left( \bar{x}\cap y\cap z \right)+\text{P}\left( x\cap y\cap z \right)$
Here the events $x,y,z$ are independent events, so we can write $\text{P}\left( x\cap y\cap z \right)=\text{P}\left( x \right).\text{P}\left( y \right).\text{P}\left( z \right)$, now we will get
$\begin{align}
  & \text{P}\left( r \right)=\text{P}\left( x \right).\text{P}\left( y \right).\text{P}\left( {\bar{z}} \right)+\text{P}\left( x \right).\text{P}\left( {\bar{y}} \right).\text{P}\left( z \right)+\text{P}\left( {\bar{x}} \right).\text{P}\left( y \right).\text{P}\left( z \right)+\text{P}\left( x \right).\text{P}\left( y \right).\text{P}\left( z \right) \\
 & =\dfrac{5}{7}.\dfrac{4}{7}.\dfrac{4}{7}+\dfrac{5}{7}.\dfrac{3}{7}.\dfrac{3}{7}+\dfrac{2}{7}.\dfrac{4}{7}.\dfrac{3}{7}+\dfrac{5}{7}.\dfrac{4}{7}.\dfrac{3}{7} \\
 & =\dfrac{\left( 5\times 4\times 4 \right)+\left( 5\times 3\times 3 \right)+\left( 2\times 4\times 3 \right)+\left( 5\times 4\times 3 \right)}{7\times 7\times 7} \\
 & =\dfrac{80+45+24+60}{343} \\
 & =\dfrac{209}{343} \\
\end{align}$
Hence the probability of solving the problem by at least two of them is $\dfrac{209}{343}$.

Note: While calculating the probability of solving the problem by at least two of them some time students may forget to add the probability $\text{P}\left( x\cap y\cap z \right)$. We know that at least indicates greater than or equal to so at least two means we need to consider the probability that three students solve. For a simple calculation, we can also find the complementary event probability by the formula
$\text{P}\left( {\bar{A}} \right)=\dfrac{\text{unfavorable outcomes of event }A}{\text{total outcomes}}$
From this formula also we will get the same values for $\text{P}\left( {\bar{x}} \right),\text{P}\left( {\bar{y}} \right),\text{P}\left( {\bar{z}} \right)$.