Question

If odds against solving a question by three students are $2:1$, $5:2$ and $5:3$ respectively. So, the probability that the question is solved only by one student is
A. $\dfrac{31}{56}$
B. $\dfrac{24}{56}$
C. $\dfrac{25}{56}$
D. None of the above

Answer 1

Hint: In this question, we have to find the probability of the event occurring. Here we have the odds of the event, we can calculate the probability.

Formula Used: The probability of an event $E$ is calculated by,
\[P(E)=\dfrac{n(E)}{n(S)}\]; And \[\overline{P(E)}=1-P(E)\]
\[n(E)\] is the number of favourable outcomes of the event and \[n(S)\] is the total number of outcomes.

Complete step by step solution: Consider the three students as A, B and C.
Then, the odds against solving a question by three students are
\[\begin{align}
  & \begin{array}{*{35}{l}}
   A\text{ }\text{ }2:1 \\
   B\text{ }\text{ }5:2 \\
\end{array} \\
 & C\text{ }\text{ }5:3 \\
\end{align}\]
So,
The number of favourable outcomes for student (A) \[n(A)=1\]
The total number of outcomes for student (A) \[n(S)=1+2=3\]
The number of favourable outcomes for student (B) \[n(B)=2\]
The total number of outcomes for student (B) \[n(S)=2+5=7\]
The number of favourable outcomes for student (C) \[n(C)=3\]
The total number of outcomes for student (A) \[n(S)=3+5=8\]
Then, the probabilities are
\[\begin{align}
  & P(A)=\dfrac{n(A)}{n(S)} \\
 & \text{ }=\dfrac{1}{3} \\
\end{align}\]
\[\begin{align}
  & P(B)=\dfrac{n(B)}{n(S)} \\
 & \text{ }=\dfrac{2}{7} \\
\end{align}\]
\[\begin{align}
  & P(C)=\dfrac{n(C)}{n(S)} \\
 & \text{ }=\dfrac{3}{8} \\
\end{align}\]
And
\[\begin{align}
  & \overline{P(A)}=1-P(A) \\
 & \text{ }=1-\dfrac{1}{3} \\
 & \text{ }=\dfrac{2}{3} \\
\end{align}\]
\[\begin{align}
  & \overline{P(B)}=1-P(B) \\
 & \text{ }=1-\dfrac{2}{7} \\
 & \text{ }=\dfrac{5}{7} \\
\end{align}\]
\[\begin{align}
  & \overline{P(C)}=1-P(C) \\
 & \text{ }=1-\dfrac{3}{8} \\
 & \text{ }=\dfrac{5}{8} \\
\end{align}\]
So, the probability that only one student can solve the question is
\[P(A\overline{B}\overline{C}\cup \overline{A}B\overline{C}\cup \overline{A}\overline{B}C)=P(A)P(\overline{B})P(\overline{C})+P(\overline{A})P(B)P(\overline{C})+P(\overline{A})P(\overline{B})P(C)\]
On substituting,
\[\begin{align}
  & P(A\overline{B}\overline{C}\cup \overline{A}B\overline{C}\cup \overline{A}\overline{B}C)=\dfrac{1}{3}\times \dfrac{5}{7}\times \dfrac{5}{8}+\dfrac{2}{3}\times \dfrac{2}{7}\times \dfrac{5}{8}+\dfrac{2}{3}\times \dfrac{5}{7}\times \dfrac{3}{8} \\
 & \text{ }=\dfrac{25}{168}+\dfrac{5}{42}+\dfrac{5}{28} \\
 & \text{ }=\dfrac{25}{56} \\
\end{align}\]


Option ‘C’ is correct

Note: Here we may go wrong with the ratio of odds. Here the ratio is odds against the event. So, we can find the required probability by using this ratio $\overline{P(E)}:P(E)$.