If A- B and C are three events such that PB -3-4- PA - B -C-1-3 and PA- - B -C-1-3- then PB - C is equal to

We have P(B∩C')=P[(A ∪A') ∩(B∩C')] =P(A ∩ B ∩C')+P(A' ∩ B ∩C') =1/3+1/3=2/3 Now, P(B∩ C)= P(B) P(B ∩C') =3/4 2/3=1/12 ...

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Byju's Answer

Standard XIII

Mathematics

Complement of a Set

If A, B and C...

Question

If A, B and C are three events such that P(B) = $\frac{3}{4}, P (A \cap B \cap C^{'}) = \frac{1}{3} a n d P (A^{'} \cap B \cap C^{'}) = \frac{1}{3},$ then $P (B \cap C)$ is equal to

$\frac{1}{12}$

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$\frac{1}{6}$

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$\frac{1}{15}$

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$\frac{1}{9}$

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Solution

The correct option is A
$\frac{1}{12}$

We have
$P (B \cap C^{'}) = P [(A \cup A^{'}) \cap (B \cap C^{'})] = P (A \cap B \cap C^{'}) + P (A^{'} \cap B \cap C^{'}) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} N o w, P (B \cap C) = P (B) - P (B \cap C^{'}) = \frac{3}{4} - \frac{2}{3} = \frac{1}{12}$

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Similar questions

If A, B and C are three events such that P(B) = $\frac{3}{4}, P (A \cap B \cap C^{'}) = \frac{1}{3} a n d P (A^{'} \cap B \cap C^{'}) = \frac{1}{3},$ then $P (B \cap C)$ is equal to

Q. If A, B and C are three events such that

P (B) = \frac{3}{4}, P (A \cap B \cap C^{'}) = \frac{1}{3}

and

P (A^{'} \cap B \cap C^{'}) = \frac{1}{3}

, then

P (B \cap C)

is equal to

Q. If

A, B

and

C

are three events such that

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

;

P (A \cap B \cap C^{'}) = \frac{1}{3}

and

P (A^{'} \cap B \cap C^{'}) = \frac{1}{336}

then

P (B \cap C)

is equal to ...

Q. Assertion :If A, B, C are three events such that

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

P (B) = \frac{1}{6}

P (C) = \frac{2}{3}

then events A, B, C are mutually exclusive. Reason: If

P (A \cup B \cup C) = P (A) + P (B) + P (C)

then A, B, C are mutually exclusive events.

Q. If

A

and

B

are two events such that

P (A \cup B) = \frac{3}{4} . P (A \cap B) = \frac{1}{4}

and

P (¯ ¯¯ ¯ A) = \frac{2}{3}

, then what is

P (B)

equal to?

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If A, B and C are three events such that P(B) =34,P(A∩B∩C′)=13 and P(A′∩B∩C′)=13, then P(B∩C) is equal to

The correct option is A 112 We have P(B∩C′)=P[(A∪A′)∩(B∩C′)]=P(A∩B∩C′)+P(A′∩B∩C′)=13+13=23Now, P(B∩C)=P(B)−P(B∩C′)=34−23=112

If A, B and C are three events such that P(B) = $\frac{3}{4}, P (A \cap B \cap C^{'}) = \frac{1}{3} a n d P (A^{'} \cap B \cap C^{'}) = \frac{1}{3},$ then $P (B \cap C)$ is equal to

The correct option is A
$\frac{1}{12}$

We have
$P (B \cap C^{'}) = P [(A \cup A^{'}) \cap (B \cap C^{'})] = P (A \cap B \cap C^{'}) + P (A^{'} \cap B \cap C^{'}) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} N o w, P (B \cap C) = P (B) - P (B \cap C^{'}) = \frac{3}{4} - \frac{2}{3} = \frac{1}{12}$