If P(B¯)={P(A∪B)}2, thenSTATEMENT – 1: P(A∪B) is at least 3−12STATEMENT – 2: For P(A∪B) to be minimum event A cannot happen given event B has not happend

P(A¯B¯)=1−P(A∪B)P(B¯)=1−xx2≤1⇒ x2+x−1≥0⇒ x≥5−12 or x≤−1−52∴ x is positivex≥5−12

If P(B¯)={P(A∪B)}2, thenSTATEMENT – 1: P(A∪B) is at least 3−12STATEMENT – 2: For P(A∪B) to be minimum event A cannot happen given event B has not happend

P(A¯B¯)=1−P(A∪B)P(B¯)=1−xx2≤1⇒ x2+x−1≥0⇒ x≥5−12 or x≤−1−52∴ x is positivex≥5−12

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If $P (\bar{B}) = {P (A \cup B)}^{2},$ then
STATEMENT – 1: $P (A \cup B)$ is at least $\frac{\sqrt{3} - 1}{2}$
STATEMENT – 2: For $P (A \cup B)$ to be minimum event A cannot happen given event B has not happend

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Both the statements are true and Statement2 is correct explanation of statement1

Both the statements are true and Statement2 is not the correct explanation of statement1

Statement 1 is true and Statement 2 is false

statemen-1 is false and Statement 2 is true

answer is D.

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Detailed Solution

$P (\frac{\bar{A}}{\bar{B}}) = \frac{1 - P (A \cup B)}{P (\bar{B})} = \frac{1 - x}{x^{2}} \leq 1$

$\Rightarrow x^{2} + x - 1 \geq 0$

$\Rightarrow x \geq \frac{\sqrt{5} - 1}{2} o r x \leq \frac{- 1 - \sqrt{5}}{2}$

$∴ x is positive$

$x \geq \frac{\sqrt{5} - 1}{2}$

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If P(B¯)={P(A∪B)}2, thenSTATEMENT – 1: P(A∪B) is at least 3−12STATEMENT – 2: For P(A∪B) to be minimum event A cannot happen given event B has not happend

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If $P (\bar{B}) = {P (A \cup B)}^{2},$ then
STATEMENT – 1: $P (A \cup B)$ is at least $\frac{\sqrt{3} - 1}{2}$
STATEMENT – 2: For $P (A \cup B)$ to be minimum event A cannot happen given event B has not happend