If A and B are two events such that P(A) = 0.4, P(B) = 0.3, and P(A union B) = 0.6, find P(B given A complement).

A. 1/3 — First, find P(A intersection B) using the addition rule. Then, use the conditional probability formula P(B|A^c) = P(B intersection A^c) / P(A^c).

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$If A and B are two events such that P (A) = 0.4, P (B) = 0.3, and P (A \cup B) = 0.6, find P (B ∣ A^{c}) .$

A. $1/3$
B. $1/2$
C. $2/3$
D. $3/4$

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Correct answer: A — $1/3$

Solution

P (A \cap B) = P (A) + P (B) - P (A \cup B) = 0.4 + 0.3 - 0.6 = 0.1 . P (B \cap A^{c}) = P (B) - P (A \cap B) = 0.3 - 0.1 = 0.2 . P (B ∣ A^{c}) = 0.2/ (1 - 0.4) = 0.2/0.6 = 1/3 .

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If A and B are two events such that P(A)=0.4, P(B)=0.3, and P(A∪B)=0.6, find P(B∣Ac).

Solution

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More Probability practice questions

$If A and B are two events such that P (A) = 0.4, P (B) = 0.3, and P (A \cup B) = 0.6, find P (B ∣ A^{c}) .$