If P(A cup B) = frac{2}{3},P(A cap B) = frac{ | vedclass

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(A) Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the given values: $\frac{2}{3} = \frac{1}{3} + P

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$A$ and $B$ are independent events.

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(A) Using the addition theorem of probability: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.Substituting the given values: $\frac{2}{3} = \frac{1}{3} + P(B) - \frac{1}{6}$.$\frac{2}{3} = \frac{1}{6} + P(B) \implies P(B) = \frac{2}{3} - \frac{1}{6} = \frac{4-1}{6} = \frac{3}{6} = \frac{1}{2}$.Now,check for independence: $P(A) 	imes P(B) = \frac{1}{3} 	imes \frac{1}{2} = \frac{1}{6}$.Since $P(A \cap B) = P(A) 	imes P(B) = \frac{1}{6}$,the events $A$ and $B$ are independent.

$P(A)$	$P(B)$	$P(A \cap B)$	$P(A \cup B)$
$0.5$	$0.35$	$.........$	$0.7$

If $P(A \cup B) = \frac{2}{3}$,$P(A \cap B) = \frac{1}{6}$,and $P(A) = \frac{1}{3}$,then which of the following is true?

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Fill in the blanks in the following table:

$P(A)$ $P(B)$ $P(A \cap B)$ $P(A \cup B)$

$0.5$ $0.35$ $.........$ $0.7$

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