## Independent Experiments and The law of Total Probability:

** Independent Experiments:** Two random experiments are said to be independent, if for every pair of events A and B, where A is associated with the first and B with the second experiment, the probability of simultaneous occurrence of the events A and B when the two experiments are performed together, is equal to the product of the probabilities P(A) and P(B), calculated separately on the basis of two experiments.

In other words, if the occurrence of an event in one experiment does affect in a way the occurrence of any event in the other experiment, then the experiments are independent.

** Note:** If A and B are events associated with experiments that are not independent, then

P(AB) = P(A) P(B/A) = P(B) P(A/B) |

** The law of Total Probability-** Let S be the sample space and E

_{1}, E

_{2}, E

_{3}, ……., E

_{n}be n mutually exclusive and totally exhaustive events associated with a random experiment. If A is any event that occurs with E

_{1}or E

_{2}or ….. or E

_{n}, then-

P(A) = P(E_{1}) . P(A/E_{1}) + P(E_{2}) . P(A/E_{2}) + …….. + P(E_{n}) . P(A/E_{n}) |

EProof:_{1}, E_{2}, E_{3}, ……., E_{n} are n mutually exclusive and totally exhaustive events.∴ S = E _{1} ∪ E_{2} ∪ E_{3} ∪ ……. ∪ E_{n} and E_{i} ∩ E_{j} = Φ for i ≠ jWe have A = (A ∩ E _{1}) ∪ (A ∩ E_{2}) ∪ (A ∩ E_{3}) ∪ ……. ∪ (A ∩ E_{n})∴ P(A) = P(A ∩ E _{1}) + P(A ∩ E_{2}) + P(A ∩ E_{3}) + ……. + P(A ∩ E_{n}) [by addition theorem]Also, using the multiplication theorem of probability, we have- P(A ∩ E _{i}) = P(E_{i}) . P(A/E_{i}), for i = 1, 2, 3, …..Hence, P(A) = P(E _{1}) . P(A/E_{1}) + P(E_{2}) . P(A/E_{2}) + …….. + P(E_{n}) . P(A/E_{n}) |

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