Home › JEE Advanced › Probability & Statistics › Let $E_1,E_2,\ldots,E_n$ be mutually exclusive c…
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 $P(E)=\sum_{i=1}^{n} P(E_i)\,P(E/E_i)$.
Related questions
Tossing two distinguishable coins can end in:Rolling one die gives a sample space of size:Classical probability is the ratio of favourable outcomes to:The classical theory fails for experiments whose outcomes are:The classical theory assumes that all outcomes are:Kolmogorov was a mathematician from:Kolmogorov published that approach in the year:The axiomatic approach to probability was developed by: