Let A and B be two events such that P(A cap B | vedclass

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(A) Given, $P(A \cap B) = \frac{1}{6}$, $P(A \cup B) = \frac{31}{45}$, and $P(\bar{B}) = \frac{7}{10}$.Since $P(B) = 1 - P(\bar{B})$, we have $P(B) =

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$A$ and $B$ are independent

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(A) Given, $P(A \cap B) = \frac{1}{6}$, $P(A \cup B) = \frac{31}{45}$, and $P(\bar{B}) = \frac{7}{10}$.Since $P(B) = 1 - P(\bar{B})$, we have $P(B) = 1 - \frac{7}{10} = \frac{3}{10}$.Using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, we substitute the values:$\frac{31}{45} = P(A) + \frac{3}{10} - \frac{1}{6}$.$P(A) = \frac{31}{45} + \frac{1}{6} - \frac{3}{10} = \frac{62 + 15 - 27}{90} = \frac{50}{90} = \frac{5}{9}$.Now, check for independence: $P(A) 	imes P(B) = \frac{5}{9} 	imes \frac{3}{10} = \frac{15}{90} = \frac{1}{6}$.Since $P(A \cap B) = P(A) 	imes P(B)$, the events $A$ and $B$ are independent.

Let $A$ and $B$ be two events such that $P(A \cap B) = \frac{1}{6}$, $P(A \cup B) = \frac{31}{45}$, and $P(\bar{B}) = \frac{7}{10}$, then:

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