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Standard XII

Mathematics

Conditional Probability

Let E 1 and E...

Question

Let $E_{1}$ and $E_{2}$ be two independent events such that $P (E_{1}) = P_{1}$ and $P (E_{2}) = P_{2}$ . Describe in words of the events whose probabilities are

$1 - (1 - P_{1}) (1 - P_{2})$

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Solution

$P (E_{1}) = P_{1}$ and $P (E_{2}) = P_{2}$

$1 - (1 - P_{1}) (1 - P_{2}) = 1 - P (E_{1})^{'} P (E_{2})^{'} = 1 - P (E_{1}^{'} \cap E_{2}^{'}) = 1 - [1 - P (E_{1} \cup E_{2})] = P (E_{1} \cup E_{2})$
So, either $E_{1}$ or $E_{2}$ or both $E_{1}$ and $E_{2}$ occurs.

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Similar questions

Let $E_{1}$ and $E_{2}$ be two independent events such that $P (E_{1}) = P_{1}$ and $P (E_{2}) = P_{2}$ . Describe in words of the events whose probabilities are

(i) $P_{1} P_{2}$

(ii) $(1 - P_{1}) P_{2}$

(iii) $1 - (1 - P_{1}) (1 - P_{2})$

(iv) $P_{1} + P_{2} - 2 P_{1} P_{2}$

Q. Let A and B be two independent events such that P(A) = p₁ and P(B) = p₂. Describe in words the events whose probabilities are:

(i) p₁p₂ (ii) (1

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p₁)p₂ (iii) 1

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p₁)(1

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p₂) (iv) p₁ + p₂

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2p₁p₂ [NCERT EXEMPLAR]

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}) =

Q. If

E_{1}

and

E_{2}

are two events such that

P (E_{1}) = \frac{1}{4}, P (\frac{E_{2}}{E_{1}}) = \frac{1}{2}

and

P (\frac{E_{1}}{E_{2}}) = \frac{1}{4},

then which of the following is/are correct?

E_{1}

and

E_{2}

are two events such that

P (E_{1}) = \frac{1}{4}

P (\frac{E_{2}}{E_{1}}) = \frac{1}{2}

and

P (\frac{E_{1}}{E_{2}}) = \frac{1}{4}

, then

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Let E1 and E2 be two independent events such that P(E1)=P1 and P(E2)=P2. Describe in words of the events whose probabilities are 1−(1−P1)(1−P2)

P(E1)=P1 and P(E2)=P2 1−(1−P1)(1−P2)=1−P(E1)′P(E2)′=1−P(E′1∩E′2)=1−[1−P(E1∪E2)]=P(E1∪E2) So, either E1 or E2 or both E1 and E2 occurs.

Let $E_{1}$ and $E_{2}$ be two independent events such that $P (E_{1}) = P_{1}$ and $P (E_{2}) = P_{2}$ . Describe in words of the events whose probabilities are

$1 - (1 - P_{1}) (1 - P_{2})$

$P (E_{1}) = P_{1}$ and $P (E_{2}) = P_{2}$

$1 - (1 - P_{1}) (1 - P_{2}) = 1 - P (E_{1})^{'} P (E_{2})^{'} = 1 - P (E_{1}^{'} \cap E_{2}^{'}) = 1 - [1 - P (E_{1} \cup E_{2})] = P (E_{1} \cup E_{2})$
So, either $E_{1}$ or $E_{2}$ or both $E_{1}$ and $E_{2}$ occurs.