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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability1 Mark
Question
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 common error is $\frac{1}{20}$ and they obtain the same answer, then the probability of their answer to be correct is.
  1. $\frac{10}{13}$
  2. $\frac{13}{120}$
  3. $\frac{1}{40}$
  4. $\frac{1}{12}$
✓

Answer

  1. $\frac{10}{13}$

Solution:

Let E1 be the event that Both A and B solve the problem.

A and B are independent,

$\Rightarrow\ \text{P}(\text{E}_1)=\text{P(A)}\times\text{P(B)}$

$\Rightarrow\ \text{P}(\text{E}_1)=\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$

Let E2 both A and B got wrong solution.

$\text{P}(\text{E}_2)=\Big(1-\frac{1}{3}\Big)\times\Big(1-\frac{1}{4}\Big)=\frac{1}{2}$

Let E be the event getting same answer.

$\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=1,\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{1}{20}$

$\Rightarrow\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=\frac{\frac{1}{12}\times1}{\frac{1}{12}\times1+\frac{1}{2}\times\frac{1}{20}}=\frac{10}{13}$

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