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Union of Sets

If A and ...

Question

If $A$ and $B$ are arbitrary events, then

P (A \cap B) \geq P (A) + P (B)

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P (A \cup B) \leq P (A) + P (B)

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P (A \cap B) = P (A) + P (B)

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None of these

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Solution

The correct option is B $P (A \cup B) \leq P (A) + P (B)$
If $A$ and $B$ are arbitrary events, we know that
$P (A \cup B) = P (A) + P (B) - P (A \cap B)$
SInce $P (A \cup B)$ can be written as $P (A) + P (B)$ $-$ some positive quantity, we can say that
$P (A \cup B) \leq P (A) + P (B)$ ..

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Similar questions

If A and B are mutually exclusive events then
(a) PA≤PB
(b) PA≥PB
(c) PA<PB
(d) None of these

A

(A)

(B)

P (A / B) > P (A)

P (B / A) > P (B)

P (A / B) = \frac{P (A \cap B)}{P (B)}

P (A / B) = P (B / A)

P (A) = P (B)

(A \cap B) \neq ϕ

P (A / B) = \frac{P (A \cap B)}{P (B)}

P (B / A) = \frac{P (A \cap B)}{P (A)}

(i) P (A) = \frac{1}{3}, P (B) = \frac{1}{4} and P (A \cup B) = \frac{5}{12}, then find P (A | B) and P (B | A) . (ii) P (A) = \frac{6}{11}, P (B) = \frac{5}{11} and P (A \cup B) = \frac{7}{11}, then find P (A \cap B), P (A | B) and P (B | A) . (iii) P (A) = \frac{7}{13}, P (B) = \frac{9}{13} and P (A \cap B) = \frac{4}{13}, then find P (A | B) . (iv) P (A) = \frac{1}{2}, P (B) = \frac{1}{3} and P (A \cap B) = \frac{1}{4}, then find P (A | B), P (B | A), P (A | B) and P (A | B) .

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