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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability1 Mark
MCQ
Choose the correct answer from the given four options. $A$ and $B$ are two students. Their chances of solving a problem correctly are $\frac{1}{3}$ and $\frac{1}{4},$respectively. If the probability of their making a common error is, $\frac{1}{20}$ and they obtain the same answer, then the probability of their answer to be correct is :
  • A
    $\frac{1}{12}$
  • B
    $\frac{1}{40}$
  • C
    $\frac{13}{120}$
  • ✓
    $\frac{10}{30}$
✓

Answer

Correct option: D.
$\frac{10}{30}$
Let $E_1= $ Event that both $A$ and $B$ solve the problem
$\therefore\text{P}(\text{E}_1)=\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$
Let $E_2 =$ Event that both $A$ and $B$ got incorrect solution of the problem
$\therefore\text{P}(\text{E}_2)=\frac{2}{3}\times\frac{3}{4}=\frac{1}{2}$
Here, $\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=1,\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{1}{20}$
$\therefore\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=\frac{\text{P}(\text{E}_1\cap\text{E})}{\text{P}(\text{E})}=\frac{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)}{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)} $
$=\frac{\frac{1}{12}\times1}{\frac{1}{12}\times1+\frac{1}{2}\times\frac{1}{20}}=\frac{10}{30}$

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