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Classical Definition of Probability

Let Ec denote...

Question

Let $E^{c}$ denote the complement of an event $E$ . Let $E_{1}, E_{2}$ and $E_{3}$ be any pairwise independent events with $P (E_{1}) > 0$ and $P (E_{1} \cap E_{2} \cap E_{3}) = 0$ . Then $P (\frac{E_{2}^{c} \cap E_{3}^{c}}{E_{1}})$ is equal to:

A

P (E_{3}^{c}) - P (E_{2}^{c})

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B

P (E_{3}) - P (E_{2}^{c})

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C

P (E_{3}^{c}) - P (E_{2})

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D

P (E_{2}^{c}) + P (E_{3})

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Solution

The correct option is C $P (E_{3}^{c}) - P (E_{2})$
$P (\frac{E_{2}^{c} \cap E_{3}^{c}}{E_{1}}) = \frac{P (E_{2}^{c} \cap E_{3}^{c} \cap E_{1})}{P (E_{1})}$

$= \frac{P (E_{1}) - P (E_{1} \cap E_{2}) - P (E_{1} \cap E_{3}) + P (E_{1} \cap E_{2} \cap E_{3})}{P (E_{1})}$

$= \frac{P (E_{1}) - P (E_{1}) . P (E_{2}) - P (E_{1}) . P (E_{3}) + 0}{P (E_{1})} = 1 - P (E_{2}) - P (E_{3})$

$= P (E_{3}^{c}) - P (E_{2})$

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Similar questions

P (E_{1}) = 0.2, P (E_{2}) = 0.4

P (E_{3}) = 0.6

E_{1}, E_{2}

E_{3}

are independent events, then the probability that at least one of these events

E_{1}, E_{2}

E_{3}

Q. For any two independent events

E_{1}

E_{2}, P {(E_{1} \cup E_{2}) \cap (_{1}) \cap (_{2})}

E = E_{1} E_{2} E_{3} E_{4} E_{5}

P (E_{1}) = \frac{95}{100}

P (E_{2} / E_{1}) = \frac{94}{99}

P (E_{1} / E_{1} E_{2}) = \frac{93}{98}

P (E_{4} / E_{1} E_{2} E_{3}) = \frac{92}{97}

P (E_{5} / E_{1} E_{2} E_{3} E_{4}) = \frac{91}{96}

P (E)

.

E_{1} a n d E_{2}

are events in a probability space satisfying the following constraints

P (E_{1}) = P (E_{2}); P (E_{1} \cup E_{2}) = 1; E_{1} a n d E_{2}

are independent then

P (E_{1}) =

E_{1}, E_{2}, E_{3}

be three mutually exclusive events such that

P (E_{i}) = p_{i}

i = 1, 2, 3

P (E_{1} \cap E_{2}^{'}) + P (E_{2} \cap E_{3}^{'}) + P (E_{3} \cap E_{1}^{'})

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