
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
162k+ views
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 $\overline{P(E)}:P(E)$.
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 $\overline{P(E)}:P(E)$.
Recently Updated Pages
Crack JEE Advanced 2025 with Vedantu's Live Classes

JEE Advanced Maths Revision Notes

JEE Advanced Chemistry Revision Notes

Download Free JEE Advanced Revision Notes PDF Online for 2025

The students S1 S2 S10 are to be divided into 3 groups class 11 maths JEE_Advanced

Solutions Class 12 Notes JEE Advanced Chemistry [PDF]

Trending doubts
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

IIT CSE Cutoff: Category‐Wise Opening and Closing Ranks

JEE Advanced Cut Off 2024

JEE Advanced Exam Pattern 2025

Other Pages
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
