Probability 4 Question 24
24. In a test an examinee either guesses or copies of knows the answer to a multiple choice question with four choices. The probability that he make a guess is $\frac{1}{3}$ and the probability that he copies the answer is $\frac{1}{6}$. The probability that his answer is correct given that he copied it, is $\frac{1}{8}$. Find the probability that he knew the answer to the question given that he correctly answered it.
(1991, 4M)
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Solution:
- Let $E _1, E _2, E _3$ and $A$ be the events defined as
$E _1=$ the examinee guesses the answer
$E _2=$ the examinee copies the answer
$E _3=$ the examinee knows the answer and $A$ =the examinee answer correctly
We have,
$$ P\left(E _1\right)=\frac{1}{3}, P\left(E _2\right)=\frac{1}{6} $$
Since, $E _1, E _2, E _3$ are mutually exclusive and exhaustive events.
$\therefore \quad P\left(E _1\right)+P\left(E _2\right)+P\left(E _3\right)=1$
$$ \Rightarrow \quad P\left(E _3\right)=1-\frac{1}{3}-\frac{1}{6}=\frac{1}{2} $$
If $E _1$ has already occured, then the examinee guesses. Since, there are four choices out of which only one is correct, therefore the probability that he answer correctly given that he has made a guess is $1 / 4$.
i.e. $\quad P\left(A / E _1\right)=\frac{1}{4}$
It is given that, $\quad P\left(A / E _2\right)=\frac{1}{8}$
and $P\left(A / E _3\right)=$ probability that he answer correctly given that he know the answer $=1$
By Baye’s theorem, we have
$$ \begin{aligned} & P\left(E _3 / A\right)=\quad P\left(E _3\right) \cdot P\left(A / E _3\right) \\ & P\left(E _1\right) \cdot P\left(A / E _1\right)+P\left(E _2\right) \cdot P\left(A / E _2\right) \\ & \therefore P\left(E _3 / A\right)=\frac{\frac{1}{2} \times 1}{\frac{1}{3} \times \frac{1}{4}+\frac{1}{6} \times \frac{1}{8}+\frac{1}{2} \times 1}=\frac{24}{29} \end{aligned} $$
