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(A) Since $A$ and $B$ are mutually exclusive events,$A \cap B = \phi$,which implies $P(A \cap B) = 0$. Given $P(A) = \frac{1}{4}$ and $P(B) = \frac{3}

Vedclass · Accepted Answer

$\frac{7}{19}$

Vedclass · Answer

(A) Since $A$ and $B$ are mutually exclusive events,$A \cap B = \phi$,which implies $P(A \cap B) = 0$. Given $P(A) = \frac{1}{4}$ and $P(B) = \frac{3}{7}$. We need to find $P(A / A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)}$. Since $A \subset (A \cup B)$,we have $A \cap (A \cup B) = A$,so $P(A \cap (A \cup B)) = P(A) = \frac{1}{4}$. Also,$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{4} + \frac{3}{7} - 0 = \frac{7 + 12}{28} = \frac{19}{28}$. Therefore,$P(A / A \cup B) = \frac{1/4}{19/28} = \frac{1}{4} 	imes \frac{28}{19} = \frac{7}{19}$.

If $A$ and $B$ are mutually exclusive events with $P(A)=\frac{1}{4}$ and $P(B)=\frac{3}{7}$,then what is the value of $P(A / A \cup B)$?

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