The probabilities of three events A- B- and C are P A -0-6- P B -0-4- P C -0-5 - If P A - B -0-8- P A - C -0-3- P A - B - C -0-2and P A - B - C - 0-85- then find the range of P B - C A- 0-25B- 1-25C- 0-35D- 0-75

The correct option is C 0.35We have, P(A ∩ B) = P(A) + P(B) P(A ∪ B)=0.6 + 0.4 0.8 = 0.2 Also, P(A ∪ B ∪ C) = P(A) + P(B) + P(C) P(A ∩ C) – P(A ∩ C) – P( ...

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Standard X

Mathematics

Probability

The probabili...

Question

The probabilities of three events A, B, and C are P(A) = 0.6, P(B) = 0.4, P(C) = 0.5. If $P (A \cup B) = 0.8, P (A \cap C) = 0.3, P (A \cap B \cap C) = 0.2$ , and $P (A \cup B \cup C) \geq 0.85$ , then find the range of $P (B \cap C)$

0.25

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1.25

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0.35

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0.75

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Solution

The correct option is C 0.35
We have,
$P (A \cap B) = P (A) + P (B) - P (A \cup B) = 0.6 + 0.4 - 0.8 = 0.2$
Also,
$P (A \cup B \cup C) = P (A) + P (B) + P (C) - P (A \cap C) - P (A \cap C) - P (B \cap C) + P (A \cap B \cap C)$
$\Rightarrow P (B \cap C) = 1.2 - P (A \cup B \cup C) . . . . . . . . . . (1)$
Now,
$0.85 \leq P (A \cup B \cup C) \leq 1$
$F r o m E q . (1), 0.2 \leq P (B \cap C) \leq 0.35$

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The probabilities of three events A, B, and C are P(A) = 0.6, P(B) = 0.4, P(C) = 0.5. If P(A∪B)=0.8,P(A∩C)=0.3,P(A∩B∩C)=0.2, and P(A∪B∪C)≥0.85, then find the range of P(B∩C)

The correct option is C 0.35We have, P(A∩B)=P(A)+P(B)−P(A∪B)=0.6+0.4−0.8=0.2 Also, P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩C)–P(A∩C)–P(B∩C)+P(A∩B∩C) ⇒P(B∩C)=1.2−P(A∪B∪C)..........(1) Now, 0.85≤P(A∪B∪C)≤1 From Eq.(1),0.2≤P(B∩C)≤0.35

The probabilities of three events A, B, and C are P(A) = 0.6, P(B) = 0.4, P(C) = 0.5. If $P (A \cup B) = 0.8, P (A \cap C) = 0.3, P (A \cap B \cap C) = 0.2$ , and $P (A \cup B \cup C) \geq 0.85$ , then find the range of $P (B \cap C)$