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A and B are independent witness in a case. The probablity that A speaks the truth is 'x' and that of B is 'y'.If A and B agree on a certain statement, how to find the probability that the statement is true ?

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$xy/(xy+(1-x)(1-y))$ – picakhu Nov 28 '11 at 16:15
@picakhu: why wouldn't you put this as an answer? – Ilya Nov 28 '11 at 16:58
@Ilya, I am too lazy to write it up and it is trivial. – picakhu Nov 28 '11 at 16:59
@picakhu: How is this trivial? Maybe you're a-naturally talented. – Gigili Nov 28 '11 at 17:03
@Gigili, please refer to answers below to see why. It should be trivial to everyone not just me I hope. – picakhu Nov 28 '11 at 19:07
up vote 0 down vote accepted


$A_t$ stand for "A says statement is true." and $A_f$ for "A says statement is false" and

$B_t$ stand for "B says statement is true." and $B_f$ for "B says statement is false" and

$S_t$ stand for "Statement is true" and $S_f$ for "Statement is false" and

Then, we know that:

$\text{Prob}(A_t | S_t) = \text{Prob}(A_f | S_f) = x$, and

$\text{Prob}(A_t | S_f) = \text{Prob}(A_f | S_t) = 1-x$, and

$\text{Prob}(B_t | S_t) = \text{Prob}(B_f | S_f) = y$, and

$\text{Prob}(B_t | S_f) = \text{Prob}(B_f | S_t) = 1-y$, and

We want to know:

$\text{Prob}(S_t | A_t \cap B_t)$

Using Bayes theorem, we have:

$$\text{Prob}(S_t | A_t \cap B_t) = \frac{\text{Prob}(A_t \cap B_t |S_t) \text{Prob}(S_t)}{\text{Prob}(A_t \cap B_t)}$$

But, we know that,

$\text{Prob}(A_t \cap B_t |S_t) = xy$ and

$\text{Prob}(A_t \cap B_t) = \text{Prob}(A_t \cap B_t |S_t) \text{Prob}(S_t) + \text{Prob}(A_t \cap B_t |S_f) \text{Prob}(S_f)$


$\text{Prob}(A_t \cap B_t) = xy \text{Prob}(S_t) + (1-x)(1-y) (1-\text{Prob}(S_t))$

Simplifying the above, we get:

$\text{Prob}(A_t \cap B_t) = (1+2xy-x-y) \text{Prob}(S_t)$

Thus, we have:

$$\text{Prob}(S_t | A_t \cap B_t) = \frac{xy}{1+2xy-x-y}$$

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$ P(A)=x $ and $ P(B)=y $,

$A$ and $B$ are independent, so $P(A \cap B)=P(A)P(B)$

Therefore, the probability that both speak the truth will be $P(A \cap B)=P(A)P(B)=xy$.

And then, the probability that they agree on a certain statement is, they both speak the truth or they both tell lie, which would be:

$$xy + (1-x)(1-y) $$

As a result, the probability that the statement is true is:

$$ \frac{xy}{xy+(1-x)(1-y)} $$.

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No, that's probability that both speak truth. – sdcvvc Nov 28 '11 at 16:54

Say the fact is True or False (T/F), and independent statements A, B are binary.

We are given P(A=1|T)=x and P(B=1|T)=y. By convexity P(A=0|T)=1-x and P(B=0|T)=1-y.

We can surmise P(A=0|F)=x and P(B=0|F)=y. Also by convexity P(A=1|F)=1-x and P(B=1|F)=1-y.

We want P(T|A=1,B=1).

By Bayes rule P(T|A=1,B=1) = P(A=1,B=1|T)P(T)/P(A=1,B=1).

By independence P(T|A=1,B=1) = P(A=1|T)P(B=1|T)P(T)/(P(A=1)P(B=1)).

Plugging in, the numerator is xyP(T).

The denominator requires P(A=1) = P(A=1|T)P(T) + P(A=1|F)P(F) = xP(T) + (1-x)P(F), same for P(B=1).

So I don’t think we are identified without knowing the marginal probability of the truth P(T).

Of course, for a given x and y you can provide P(T|A=1,B=1) as a function of P(T) between 0 and 1.

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