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In a multiple choice question, there are five different answers, of which only one is correct. The probability that a student will know the correct answer is 0.6. If a student does not know the answer, he guesses an answer at random.

a) What is the probability that the student gives the correct answer?

b) If the student gives the correct answer, what is the probability that he guessed?

Let A$_1$ be: student knows the answer, A$_2$: student doesn't know the answer, B: student gives the correct answer

P(A$_1$) = 0.6 and P(A$_2$) = 0.4. How do I find P(B|A$_1$) and P(B|A$_2$)?

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    $\begingroup$ If there are five choices, what is the probability that the student obtains the correct answer by guessing? $\endgroup$ – N. F. Taussig Jun 8 at 22:18
  • $\begingroup$ @N.F.Taussig 0.2? $\endgroup$ – bmal Jun 8 at 22:26
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    $\begingroup$ Yes, that is correct. $\endgroup$ – N. F. Taussig Jun 8 at 22:29
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$P(B|A_2)$ is the probability that the student gives the correct answer, while guessing (i.e. not knowing), so that is $\frac{1}{5} = 0.2$

Clearly, by definition almost, $P(B|A_1)=1$, if the student knows the answer he/she gives it (basic assumption of the problem, I suppose).

Now complete your plan by applying the law of total probability:

$$P(B)=P(B|A_1)P(A_1) + P(B|A_2)P(A_2)$$

now that all values are known.

b) Is applying Bayes' rule, asking for $P(A_1|B)$ etc.

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Slightly different notation. Let $A$ be the event that the student knows the answer and $B$ is as you defined. $$ P(B)=P(B|A)P(A)+P(B|\neg A)P(\neg A)=1\times 0.6+0.2\times0.4=0.68. $$ Next, $$ P(\neg A|B)=\frac{P(\neg A\text{ and }B)}{P(B)}=\frac{P(B|\neg A)P(\neg A)}{P(B)}=\frac{0.08}{0.68}=\frac{2}{17}. $$

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