Browse Questions

# In answering a question on a multiple choice test, a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$ . What is the probability that the student knows the answer given that he answered it correctly?

$\begin{array}{1 1}12/13 \\ 1/13 \\ 6/13 \\ 8/13 \end{array}$

Toolbox:
• Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: $\;P(E_i/A)$=$\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}$
Let $E_1$: student knows the answer $\rightarrow$ P($E_1$) = $\large\frac{3}{4}$
Let $E_2$: student guesses the answer $\rightarrow$ P($E_2$) = $\large\frac{1}{4}$
Probability that a student answered correctly knowing the answer = P (A|$E_1$) = 1.
Probability that a student answered by guessing = P (A|$E_2$) = $\large\frac{1}{4}$
We need to find the probability that a student who is chosen from random knows the answer given that he answered correctly:
We can use Baye's theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)+P(A|E_2)}$
Using Baye's theorem, P ($E_1$|A) =$\Large\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{4}.\frac{1}{4}}$ = $\large\frac{3 \times 16 }{(12+1) \times 4} = \frac{12}{13}$