conditional probability on Truth and lies Mike tells truth with probability 1/3 and lies with probability 2/3. Independently, David tells truth with probability 3/4 and lies with probability 1/4. Both watch a soccer match. David tells you that Spain won, Mike tells you that Spain lost. What probability will you assign to Spain's win?

Suppose there are 10 people instead of 2. Now, with each of them telling truth or lie, the probability of win will decrease? Or in the same way if i calculate P(loss) it will be 1-P(win)?

how should i approach for this question?
Mike tells truth with probability 1/3 and lies with probability 2/3. Independently, David tells truth with probability 3/4 and lies with probability 1/4. Both watch a soccer match. David tells you that Spain won, Mike tells you that Spain lost. 


Edited: Full credit is due to joriki for this amended answer.

Correction: We're looking for Spain winning given what they said.  

Letting $D_S, M_S$ be the events "David said Spain won" and "Mike said Spain won".  These will be a lie or truth depending on what the truth is.

$$\begin{align}\mathsf P(S \mid D_S, \neg M_S) &amp; = \frac{\mathsf P(S, D_S, \neg M_S)}{\mathsf P(S, D_S, \neg M_S)+\mathsf P(\neg S, D_S, \neg M_S)}
\\ &amp; = \frac{\mathsf P(S)\mathsf P(D_S\mid S)\mathsf P(\neg M_S\mid S)}{\mathsf P(S)\mathsf P(D_S\mid S)\mathsf P(\neg M_S\mid S)+\mathsf P(\neg S)\mathsf P(D_S\mid \neg S)\mathsf P(\neg M_S\mid \neg S)} \end{align}$$

We have &nbsp; $\mathsf P(\neg M_S\mid S)=2/3 \\ \mathsf P(\neg M_S\mid\neg S)=1/3 \\ \mathsf P(D_S\mid S)=3/4 \\ \mathsf P(D_S\mid\neg S)=1/4$ .

What we don't have is: $\mathsf P(S)$, the prior probability that Spain won.

$$\begin{align}\mathsf P(S \mid D_S, \neg M_S) &amp; = \frac{3\mathsf P(S)}{3\mathsf P(S)+2\mathsf P(\neg S)} \end{align}$$
Mike tells truth with probability 1/3 and lies with probability 2/3. Independently, David tells truth with probability 3/4 and lies with probability 1/4. Both watch a soccer match. David tells you that Spain won, Mike tells you that Spain lost. 


Edited: Full credit is due to joriki for this amended answer.

Correction: We're looking for Spain winning given what they said.  

Letting $D_S, M_S$ be the events "David said Spain won" and "Mike said Spain won".  These will be a lie or truth depending on what the truth is.

$$\begin{align}\mathsf P(S \mid D_S, \neg M_S) &amp; = \frac{\mathsf P(S, D_S, \neg M_S)}{\mathsf P(S, D_S, \neg M_S)+\mathsf P(\neg S, D_S, \neg M_S)}
\\ &amp; = \frac{\mathsf P(S)\mathsf P(D_S\mid S)\mathsf P(\neg M_S\mid S)}{\mathsf P(S)\mathsf P(D_S\mid S)\mathsf P(\neg M_S\mid S)+\mathsf P(\neg S)\mathsf P(D_S\mid \neg S)\mathsf P(\neg M_S\mid \neg S)} \end{align}$$

We have &nbsp; $\mathsf P(\neg M_S\mid S)=2/3 \\ \mathsf P(\neg M_S\mid\neg S)=1/3 \\ \mathsf P(D_S\mid S)=3/4 \\ \mathsf P(D_S\mid\neg S)=1/4$ .

What we don't have is: $\mathsf P(S)$, the prior probability that Spain won.

$$\begin{align}\mathsf P(S \mid D_S, \neg M_S) &amp; = \frac{3\mathsf P(S)}{3\mathsf P(S)+2\mathsf P(\neg S)} \end{align}$$