Let A and B be independent events with P(B) = | vedclass

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(A) Given that $A$ and $B$ are independent events,$P(A \cap B) = P(A)P(B)$.We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the val

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$4x^2 - 7x + 3 = 0$

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(A) Given that $A$ and $B$ are independent events,$P(A \cap B) = P(A)P(B)$.We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the values,$\frac{11}{20} = P(A) + \frac{2}{5} - P(A) 	imes \frac{2}{5}$.$\frac{11}{20} - \frac{8}{20} = P(A)(1 - \frac{2}{5})$.$\frac{3}{20} = P(A) 	imes \frac{3}{5}$.$P(A) = \frac{3}{20} 	imes \frac{5}{3} = \frac{1}{4}$.Since $A$ and $B$ are independent,$A'$ and $B$ are also independent.Therefore,$P(A' \mid B) = P(A') = 1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.Now,check which equation has $x = \frac{3}{4}$ as a root.For option $A$: $4(\frac{3}{4})^2 - 7(\frac{3}{4}) + 3 = 4(\frac{9}{16}) - \frac{21}{4} + 3 = \frac{9}{4} - \frac{21}{4} + \frac{12}{4} = 0$.Thus,$x = \frac{3}{4}$ is a root of $4x^2 - 7x + 3 = 0$.

Let $A$ and $B$ be independent events with $P(B) = \frac{2}{5}$ and $P(A \cup B) = \frac{11}{20}$. Then $P(A' \mid B)$ is a root of which equation?

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