Let A and B be two independent events of an e | vedclass

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(A) Given $P(A) = 0.3$ and $P(A \cup B) = 0.8$.Since $A$ and $B$ are independent,$P(A \cap B) = P(A)P(B)$.Using the addition theorem: $P(A \cup B) = P

Vedclass · Accepted Answer

$\frac{32}{35}$

Vedclass · Answer

(A) Given $P(A) = 0.3$ and $P(A \cup B) = 0.8$.Since $A$ and $B$ are independent,$P(A \cap B) = P(A)P(B)$.Using the addition theorem: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$0.8 = 0.3 + P(B) - 0.3P(B)$.$0.5 = 0.7P(B) \Rightarrow P(B) = \frac{0.5}{0.7} = \frac{5}{7}$.The conditional statement $A 	o B$ is equivalent to $
eg A \lor B$,so $P(A 	o B) = P(
eg A \cup B) = 1 - P(A \cap 
eg B)$.Since $A$ and $B$ are independent,$A$ and $
eg B$ are also independent.$P(A \cap 
eg B) = P(A)P(
eg B) = 0.3 	imes (1 - \frac{5}{7}) = 0.3 	imes \frac{2}{7} = \frac{3}{10} 	imes \frac{2}{7} = \frac{6}{70} = \frac{3}{35}$.Therefore,$P(A 	o B) = 1 - \frac{3}{35} = \frac{32}{35}$.

Let $A$ and $B$ be two independent events of an experiment. If $P(A) = 0.3$ and $P(A \cup B) = 0.8$,then find $P(A \to B)$,where $P(X)$ denotes the probability that statement $X$ is true.

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