The probability that exactly one of the independent events $A$ and $B$ occurs is equal to

P (A) + P (B) - 2 P (A \cap B)

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

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P (¯ A) + P (¯ B) - 2 P (¯ A \cap ¯ B)

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None of the above

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Solution

The correct options are
B $P (¯ A) + P (¯ B) - 2 P (¯ A \cap ¯ B)$
D $P (A) + P (B) - 2 P (A \cap B)$
$P (A \cap ¯ ¯¯ ¯ B) + P (B \cap ¯ ¯¯ ¯ A) = P (A) - P (A \cap B) + P (B) - P (A \cap B) = P (A) + P (B) - 2 P (A \cap B)$
Also
$P (A \cap ¯ ¯¯ ¯ B) + P (B \cap ¯ ¯¯ ¯ A) = 1 - P (¯ ¯¯ ¯ A) + 1 - P (¯ ¯¯ ¯ B) - 2 {1 - P (¯ ¯¯ ¯ A \cup ¯ ¯¯ ¯ B)} = 1 - P (¯ ¯¯ ¯ A) + 1 - P (¯ ¯¯ ¯ B) - 2 {1 - P (¯ ¯¯ ¯ A) - P (¯ ¯¯ ¯ B) + P (¯ ¯¯ ¯ A \cap ¯ ¯¯ ¯ B)} = P (¯ ¯¯ ¯ A) + P (¯ ¯¯ ¯ B) - P (¯ ¯¯ ¯ A \cap ¯ ¯¯ ¯ B)$

Suggest Corrections

Similar questions

P (¯ A \cup B) = P (¯ A) + P (B) - P (¯ A \cap B)

P (A \cap ¯ B) = P (¯ B) - P (A \cap B)

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

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

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

P (¯ A / B) = \frac{3}{4}

P (¯ B / A) = \frac{1}{2}

P (B / A) = P (B)

A

B

P (A) = 0.3

P (B) = 0.6

P (¯ A \cap ¯ B)

P (A - B) = P (A) - P (A ⋂ B) = P (A ⋃ B) - P (B) = P (A ⋂ ¯ B) = 1 - P (¯ A ⋂ B)

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