# Conditional probability and independent events.

In a test, an examinee either guesses or copies or knows the answer to a multiple-choice question with four choices, only one answer being correct. The probability that he makes 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 copies it, is $\frac{1}{8}$

What is the probability that he knew the answer to the question, given that he correctly answers it?

From the text I identified $3$ events regard to the same experience.So the sum of the $3$ probabilities must be $1$.It's known that:

$P(A)=\frac{1}{3}$

$P(B)=\frac{1}{6}$

So,

$P(C)+\frac{1}{3}+\frac{1}{6}=1$

$P(C)=\frac{1}{2}$, this is the probability of knowing the answer.

It's also known that $P(D|B)=\frac{1}{8}$. $P(D)$ is the probability of the question is correctly answered.

The problem ask about $P(C|D)$. From the knowledge that $P(D|B)=\frac{1}{8}$, $P(B)=\frac{1}{6}$ and by the definition of conditional probability, it's known that

$P(D \cap B)=\frac{1}{8} \cdot \frac{1}{6}$. This proves that $B$ and $D$ are indepentend events. And if the $P(B)$ it's known, the $P(D)$ must be $\frac{1}{8}$.

Now, using the conditional probability definition, one can find $P(C|D)$.But if $D$ and $B$ were independent, and $C$ and $B$ are events of the same experience, than $C$ and $D$ must also be independents. So $P(C|D)=P(C)=\frac{1}{2}$

Is my thought right?

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Writing a title in the imperative is usually frowned upon in MSE. –  user02138 Mar 24 '12 at 2:10
"$P(D \cap B)=\frac{1}{8} \cdot \frac{1}{6}$. This proves that $B$ and $D$ are indepentend events." Not quite. You correctly multiplied $P(D|B)$ and $P(B)$ to get $P(D \cap B)$; but independence needs $P(D \cap B)=P(D)P(B)$, not $P(D \cap B)=P(D|B)P(B)$. –  Dilip Sarwate Mar 24 '12 at 2:14
I didn't realize that was in imperative form.I already changed it. –  João Mar 24 '12 at 2:15
Also $C$ and $D$ are clearly dependent... The answer is $P(C\cap D)\over P(D)$. To find $P(D)$, condition on the three alternatives (guess, copy, knows the answer) $P(D)=P(A) P(D|A)+P(B)P(D|B)+P(C)P(D|C)$. –  David Mitra Mar 24 '12 at 2:17
Ok. I set a table, and found that $P(D \cap B)=\frac{1}{48}$.So they are not independent.But how can I find $P(A \cap D)$ and $P(C \cap D)$? –  João Mar 24 '12 at 10:53

First, some warm up:

Let's define our events at the start:

$\ \ \ D$ is the event that the student answers correctly.

$\ \ \ A$ is the event that the student guesses the answer.

$\ \ \ B$ is the event that the student copies the answer.

$\ \ \ C$ is the event that the student knows the answer.

Let's also write down what we know:

$$\textstyle P(A)={1\over3},\quad P(B)={1\over 6},\quad P(D\mid B)={1\over 8} .$$

Also note $$\textstyle P(D\mid C)=1, \quad P(D\mid A)={1\over 4},\quad P( C) =1-{1\over3}-{1\over6}={1\over2}.$$

Now on to the problem proper:

You want to find $P(C\mid D)$.

$C$ and $D$ are not independent. We have to use the basic formula defining conditional probabilities: $$\tag{1} P(C\mid D) ={P(C\cap D)\over P(D)}.$$

To find $P(C\cap D)$, we use the basic formula again (though it's usually called the multiplication principle when used this way): $$P(C\cap D) =P(C)P(D\mid C).$$ We know $P(C)={1\over2}$ (as you calculated); and, if we're given that the student knows the answer, it follows that in this case that the probability that the student answers correctly is 1. Thus $$\textstyle\tag{2}P(C\cap D) = {1\over2}\cdot 1={1\over 2}.$$

Now to find the term $P(D)$ in $(1)$, we first write $$\tag{3} P(D)=P(A\cap D)+P(B\cap D)+P(C\cap D)$$ this is allowed since $A$, $B$, and $C$ are mutually exclusive events and one of the three must occur; as sets, $D$ can be written as the disjoint union $D= (A\cap D)\cup (B\cap D)\cup (C\cap D)$.

On to calculating the terms in $(3)$:

We have already calculated $P(C\cap D)$.

To find $P(A\cap D)$: $$\tag{4}\textstyle P(A\cap D)=P(A)P(D\mid A)={1\over3}\cdot{1\over4}={1\over12}.$$

To find $P(B\cap D)$: $$\tag{5}\textstyle P(B\cap D)=P(B)P(D\mid B)={1\over6}\cdot{1\over8}={1\over48}.$$ So, substituting the information from $(2)$, $(4)$ and $(5)$ into equation $(3)$, we have $$\textstyle P(D)= {1\over12}+{1\over48}+{1\over2} ={29\over 48}.$$ Using this and $(1)$ and $(2)$ we finally obtain $$P(C\mid D) = {P(C\cap D)\over P(D)}={ 1/2\over 29/48}= {48\over 2\cdot 29}={24\over29}.$$

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I chosen this answer because the solution presented don't apply to baye's law.Thanks –  João Mar 24 '12 at 13:51
@João While the solution is excellent and fully deserving of your acceptance, it has used Bayes' law without saying so since it computed $P(C|D)$ starting from $P(D|C)$. –  Dilip Sarwate Mar 25 '12 at 2:39
@DilipSarwate It is Bayes' law of course; but I wouldn't say I "used" it, but rather derived it for this particular problem. –  David Mitra Mar 25 '12 at 2:46
@DavidMitra Yes, that you derived it is one way of looking at it. My students have often asked why Bayes' law deserves a special name since it is so "obvious" and claimed that maybe if they had been alive in the 18th century, they would have been famous instead of Bayes. They also wonder why it is so controversial.... –  Dilip Sarwate Mar 25 '12 at 12:01