# Conditional Probability Inconsistency

Let $D$ be a person who answers a question whose answer is true, correctly with $0.4$ probability. $D$ also answers a question whose answer is false, correctly with $0.9$ probability. Moreover, the probability of the question having answers as true $=0.3$

Now, I denoted the event $D=T,\, D=F$ ,as the events where $D$ answers $T,F$respectively, and the event $A=T ,\,A=F$ as the events where the answer to a question is true or false.

Writing as an equation:

$$P(A=T)=0.3\quad \& \quad P(A=F)=0.7$$

$$P(D=T|A=T)=0.4\quad \&\quad P(D=F|A=F)=0.9$$

It follows that $P(D=F|A=T)=0.6$ and $P(D=T|A=F)=0.1$

We compute:

$$P(D=T)= P(D=T|A=T)(P(A=T) + P(D=T|A=F)(P(A=F) = 0.4*0.3 + 0.1*0.7$$ $$\implies P(D=T) =0.19$$

Now$$P(A=T|D=T) = \frac {P(D=T|A=T)\times P(A=T)}{P(D=T)} =\frac {0.4\times 0.3}{0.19} =\frac {12}{19}$$

Now consider another person D1 who is identical to D(in terms of probability) ,but whose decisions are independent of D.

Now $$P(A=T|D=T,D1=T)= \frac{P(D=T,D1=T|A=T)\times P(A=T)}{P(D=T,D1=T)}$$ $$=\frac{P(D=T|A=T)\times P(D1=T|A=T)\times 0.3}{P(D1=T)*P(D=T)}$$ (as D and D1 are independent.)

But this is $$= \frac{0.4*0.4*0.3}{0.19*0.19} = 1.329 >1$$

Clearly this is wrong. What is the reason for this inconsistency?

• This is hard to read. It looks, however, as if you are assuming that the events "the true answer is $T$" and "$D$ answers $T$" are independent, but they are not. If the correct answer is $T$ then $D$ answers $T$ a mere $40\%$ of the time, however if the correct answer is $F$ then $D$ answers $T$ $10\%$ of the time.
– lulu
Aug 28 '17 at 13:57
• I know that the two events mentioned are independent and as far as I cannot see where I interpreted as otherwise. It will be helpful if you can point to the exact step which is incorrect. Note: My formatting is not great as this is my first answer. Any tips on improving this. Aug 28 '17 at 14:16
• As I said, they are not independent.
– lulu
Aug 28 '17 at 14:17
• You appear to compute $P(A=T\,|\,D=T,D_1=T)$ by using the definition of conditional probability $P(X\,|\,Y)=\frac {P(X\cap Y)}{P(Y)}$. Yes? But then you evaluate $P(X\cap Y)=P(X)\times P(Y)$ which is only true for independent events.
– lulu
Aug 28 '17 at 14:20
• Note: I reformatted your question, I suggest you review it to make sure I didn't introduce any errors. If you click on "edit" you can see the syntax I used.
– lulu
Aug 28 '17 at 14:28

Now $$P(A=T|D=T,D1=T)= \frac{P(D=T,D1=T|A=T)\times P(A=T)}{P(D=T,D1=T)}$$ $$=\frac{P(D=T|A=T)\times P(D1=T|A=T)\times 0.3}{P(D1=T)*P(D=T)}$$ (as D and D1 are independent.)
No.   $D$ and $D_1$ are conditionally independent for a given answer.   The numerator expansion is okay, but the denominator is bad. You should use that:
$${\mathsf P(D{=}T,D_1{=}T)~}{=~{{\mathsf P(D{=}T\mid A{=}T)\,\mathsf P(D_1{=}T\mid A{=}T)\,\mathsf P(A{=}T)}\\+{\mathsf P(D{=}T\mid A{=}F)\,\mathsf P(D_1{=}T\mid A{=}F)\,\mathsf P(A{=}F)}}\\ ={0.4\cdot 0.4\cdot 0.3+0.1\cdot 0.1\cdot 0.7}}$$
Thus $\mathsf P(A{=}T\mid D{=}T, D_1{=}T)~=~\dfrac{0.4\cdot 0.4\cdot 0.3}{0.4\cdot 0.4\cdot 0.3+0.1\cdot 0.1\cdot 0.7}=\dfrac{48}{55}$