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I can understand the table, but I cannot figure out how to solve it when they have removed the $M$.

I know that $L = \text{false}$ is $30\%$ since $L = \text{true}$ is $70\%$, but how to find out the $V = \text{false}$? Is there some formula for this or can it be done without?

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The answer should be: d) 0.925

Let F be false and T be true. Using the total law of probability, we can show that $P(V=F|L=F)$ is:

$$P(V=F|L=F,M=T)P(M=T)+P(V=F|L=F,M=F)P(M=F) \space ...(1)$$ $$=(1-0.3)(0.1)+(1-0.05)(0.9)$$ $$=\frac 7{100}+ \frac {171}{200}= \frac {185}{200}=0.925$$

In $(1)$ I used the fact that $$P(V=F|L=F,M=T)=1-P(V=T|L=F,M=T)$$ and $$P(V=F|L=F,M=F)=1-P(V=T|L=F,M=F)$$ Please let me know if I can clarify anything.

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  • $\begingroup$ Thank you very much, I clearly see how you got the result, and I can now also see how it is related to the formula. I just have one question. Why do we only calculate L = F, while we use both M = false and M = true, is it because the execise itself says L = F? $\endgroup$ – user504783 May 27 '18 at 12:15
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    $\begingroup$ No worries. So this stems from the Law of Total Probability. $P(X)=\sum_i P(X|A_i)$ with $\sum_i P(A_i)=1$. If, like in our example, we already had a conditional probability to start with, then we keep it constant throughout. e.g. lets say we had a few, then it holds that $P(X|A,B,C,D)=\sum_i P(X|A,B,C,D,E_i)P(E_i)$. Essentially what is happening here is that, if we already have information about A,B,C,D, we can use another variable say E that marginally sums to the total probability. Have a look here: en.wikipedia.org/wiki/Law_of_total_probability $\endgroup$ – Tony Hellmuth May 27 '18 at 12:27

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