Let $E_1,E_2,\ldots,E_n$ be mutually exclusive cases that cover the sample space. According to the theorem of total probability, $P(E)$ is equal to
A$\sum_{i=1}^{n} P(E_i)+P(E/E_i)$
B$\sum_{i=1}^{n} P(E_i)\,P(E/E_i)$
C$\prod_{i=1}^{n} P(E_i)\,P(E/E_i)$
D$\sum_{i=1}^{n} \dfrac{P(E/E_i)}{P(E_i)}$
Answer & Solution
Correct answer: B. $\sum_{i=1}^{n} P(E_i)\,P(E/E_i)$
The theorem of total probability decomposes event $E$ across mutually exclusive and exhaustive cases $E_i$. Thus $P(E)=\sum_{i=1}^{n} P(E_i)\,P(E/E_i)$.
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