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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
A and B are two students. Their chances of solving a problem correctly are
$\dfrac{1}{3}$ and $\dfrac{1}{4}$ respectively.
If the probability of their making a **common error** is $\dfrac{1}{20}$,
and they obtain the same answer, then the probability that their answer is correct is:
Solution
**Solution:** Let $A_1 =$ A correct, $A_2 =$ A wrong $B_1 =$ B correct, $B_2 =$ B wrong Then, $P(A_1) = \dfrac{1}{3}, \quad P(A_2) = \dfrac{2}{3}$ $P(B_1) = \dfrac{1}{4}, \quad P(B_2) = \dfrac{3}{4}$ They give the **same answer** if both are correct or both are wrong. \[ P(\text{same}) = P(A_1,B_1) + P(A_2,B_2) \] Assuming independence except for the given *common error*, \[ P(A_1,B_1) = \dfrac{1}{3} \cdot \dfrac{1}{4} = \dfrac{1}{12}, \qquad P(A_2,B_2) = \dfrac{1}{20} \] Then total \[ P(\text{same}) = \dfrac{1}{12} + \dfrac{1}{20} = \dfrac{8}{60} = \dfrac{2}{15} \] Hence, \[ P(\text{correct | same}) = \dfrac{P(A_1,B_1)}{P(\text{same})} = \dfrac{\tfrac{1}{12}}{\tfrac{2}{15}} = \dfrac{15}{24} = \dfrac{5}{8} \]Online Test Series, Information About Examination,
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