Question

If odds against solving a question by three students are $2:1,5:2$ and $5:3$ respectively, then the probability that the question is solved only by one student is

• A. $\frac{32}{56}$
• B. $\frac{24}{56}$
• C. $\frac{25}{56}$
• D. None of these

Answer 1

The correct option is C

$\frac{25}{56}$

Explanation for the correct option

Step 1: Solve for the individual probabilities

Let the three students be $A,B$ and $C$

Given that odds against solving a question by three students are $2:1,5:2$ and $5:3$ respectively

Odds are used to describe the chance of an event occurring.

Odds are defined as (Probability that the event will occur)$:$ (Probability that the event will not occur)

Here the odds are against solving a question

$\Rightarrow$ (Probability that the student will not solve the question)$:$ (Probability that the student will solve the question)

Classical probability is a simple form of probability that has equal odds of something happening.

$\therefore$Probability of an event$=\frac{\text{Favorable outcome}}{\text{Total outcomes}}$

Therefore,

The probability that $A$ will solve the question$P\left(A\right)\begin{array}{ccccccc}=& \frac{1}{\left(2+1\right)}& =& \frac{1}{3}& \Rightarrow & P\left(A\text{'}\right)& =1-\frac{1}{3}=\frac{2}{3}\end{array}$

The probability that $B$ will solve the question$P\left(B\right)\begin{array}{ccccccc}=& \frac{2}{\left(5+2\right)}& =& \frac{2}{7}& \Rightarrow & P\left(B\text{'}\right)& =1-\frac{2}{7}=\frac{5}{7}\end{array}$

The probability that $C$ will solve the question$P\left(C\right)\begin{array}{ccccccc}=& \frac{3}{\left(5+3\right)}& =& \frac{3}{8}& \Rightarrow & P\left(C\text{'}\right)& =1-\frac{3}{8}=\frac{5}{8}\end{array}$

Step 2: Solve for the probability that the question is solved only by one student

$\therefore$The probability that the question is solved only by one student$=P\left(AB\text{'}C\text{'}\right)+P\left(A\text{'}BC\text{'}\right)+P\left(A\text{'}B\text{'}C\right)$

$$\begin{gathered} =P\left(A\right)\times P\left(B\text{'}\right)\times P\left(C\text{'}\right)+P\left(A\text{'}\right)\times P\left(B\right)\times P\left(C\text{'}\right)+P\left(A\text{'}\right)\times P\left(B\text{'}\right)\times P\left(C\right) \\ =\frac{1}{3}\times \frac{5}{7}\times \frac{5}{8}+\frac{2}{3}\times \frac{2}{7}\times \frac{5}{8}+\frac{2}{3}\times \frac{5}{7}\times \frac{3}{8} \\ =\frac{25}{168}+\frac{20}{168}+\frac{30}{168} \\ =\frac{75}{168} \\ =\frac{25}{56} \end{gathered}$$

Hence, the correct option is option(C).