If two events $A$ and $B$ are such that $P\left( A^{c} \right) = 0.3,\ P(B) = 0.4\ and\ P\left( A \cap B^{c} \right) = 0.5,$ then $P\left\lbrack \frac{B}{(A \cup B^{c})} \right\rbrack$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
$P\left( \frac{B}{A \cup B^{C}} \right) = \frac{P(B \cap \left( A \cup B^{c} \right)}{P(A \cup B^{c})}$
$= \frac{P(A \cap B)}{P(A) + P\left( B^{c} \right) - P(A \cap B^{c})}$
$= \frac{P(A) - P(A \cap B^{c})}{P(A) + P\left( B^{c} \right) - P(A \cap B^{c})}$
$= \frac{0.7 - 0.5}{0.8} = \frac{1}{4}$

Q. If two events $A$ and $B$ are such that $P\left( A^{c} \right) = 0.3,\ P(B) = 0.4\ and\ P\left( A \cap B^{c} \right) = 0.5,$ then $P\left\lbrack \frac{B}{(A \cup B^{c})} \right\rbrack$ is equal to

Explanation

Reviewed by Dr. Abubacker

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