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A student is taking a multiple-choice test. Each question on the test has five possible answers and only one choice is correct. The probability this student knows the correct answer is 70%. If the student does not know the answer, they select an answer at random with each answer having an equal probability of being picked. Calculate the conditional probability that the student knew the answer given they answered the question correctly.

I started off like this: Let B denote the event that the student knew the answer. Let A denote the event that the student answered the question correctly.

I was able to work out that $P(A)=\frac{3}{10}*\frac{1}{5} + \frac{7}{10} = \frac{19}{25}$

And I know the formula $P(B|A)=\frac{P(A \cap B)}{P(A)}$

I am unsure on how to work out $P(A \cap B)$. Any advice would be greatly apprectiated.

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  • $\begingroup$ Hint: The event $B$ is included in the event $A$ (since knowing the correct answer implies that the question was answered correctly). $\endgroup$ – lulu Feb 16 at 1:55
  • $\begingroup$ If the student knows the answer, we assume he will answer correctly, so the probability that he knows the answer and answers correctly is the same as the probability that he knows the answer. $\endgroup$ – saulspatz Feb 16 at 1:56
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It is easiest to draw a tree for these.


know        70%   -   correct   1       .7

                  /   correct   1/5     .06
don't know  30%   
                  \   incorrect 4/5     .24

Thus $P(A\cap B)=.7$ and your desired probability is $\frac{P(A\cap B)}{P(A)}=\frac{.7}{.76}= \frac{35}{38}$. It is very likely that the student knew the answer if they got it correct.

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    $\begingroup$ For simple pictorial explanation ,+1 $\endgroup$ – true blue anil Feb 16 at 7:20
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Let us denote the events: $A$: answered correctly, $B$: knew the answer.

We are provided that: $\mathsf P(B)=0.70$, $\mathsf P(A\mid B')=0.20$, and $\mathsf P(A\mid B)=1.00$.

NB: Thus ${\mathsf P(A\cap B)~{=\mathsf P(A\mid B)\,\mathsf P(B) \\= \mathsf P(B)\\=0.70}\\\mathsf P(A)\qquad~{=\mathsf P(B)+\mathsf P(A\mid B')\;\mathsf P(B')\\=0.70+0.20\cdot0.30\\=0.76}}$

We seek $\mathsf P(B\mid A)$.   Just use Bayes' Rule and the Law of Total Probability. $$\begin{align}\mathsf P(B\mid A) &=\dfrac{\mathsf P(A\mid B)\,\mathsf P(B)}{\mathsf P(A)} \\[1ex]&=\dfrac{\mathsf P(A\mid B)\,\mathsf P(B)}{\mathsf P(A\mid B)\;\mathsf P(B)+\mathsf P(A\mid B')\;\mathsf P(B')}\\&=\dfrac{35}{38}\end{align}$$

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We are given that the student answered the question correctly, and we must find the probability that they actually knew the answer. Thus, we have two cases, one being that the student knew the answer and got it right, or the student did not know, and guessed correctly. The probability of the first case is $(7/10)$. The probability of the $2$nd case is $(3/10)(1/5)= 3/50$. Adding these up, we get that the total probability that the student answered the question correctly is $19/25$. Since we need to find the probability that the student KNEW the answer and got it correct, we do the probability he knew the answer and divide it by the total probability of getting the question right, so we get $(7/10)/(19/25)$ or $35/38$.

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    $\begingroup$ Sorry, I don't understand the fancy notation so I just used my intuition $\endgroup$ – Some Guy Feb 16 at 3:41

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