If two events A and B are such that P(AC) = 0.3, P(B) = 0.4 and P(A ∩ Bc) = 0.5, then P[B/(A ∪ Bc)] is equal to

Question

If two events A and B are such that P(AC) = 0.3, P(B) = 0.4 and P(A ∩ Bc) = 0.5, then P[B/(A ∪ Bc)] is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

P[B/(A ∪ Bc)] = (P(B ∩ (A ∪ Bc))/P(A ∪ Bc)) = (P(A ∩ B)/P(A) + P(Bc) - P(A ∩ Bc)) = (P(A) - P(A ∩ Bc)/P(A) + P(Bc) - P(A ∩ Bc)) = (0.7 - 0.5 /0.8) = (1/4)

Q. If two events $A$ and $B$ are such that $P (A^{C}) = 0.3, P (B) = 0.4$ and $P (A \cap B^{c}) = 0.5$ , then $P [B / (A \cup B^{c})]$ is equal to

$P [B / (A \cup B^{c})] = \frac{P ( B \cap ( A \cup B ^{c} ))}{P ( A \cup B ^{c} )}$
$= \frac{P ( A \cap B )}{P ( A ) + P ( B ^{c} ) - P ( A \cap B ^{c} )}$
$= \frac{P ( A ) - P ( A \cap B ^{c} )}{P ( A ) + P ( B ^{c} ) - P ( A \cap B ^{c} )}$
$= \frac{0.7 - 0.5}{0.8} = \frac{1}{4}$

Q. If two events A and B are such that P(AC)=0.3,P(B)=0.4 and P(A∩Bc)=0.5, then P[B/(A∪Bc)] is equal to

P[B/(A∪Bc)]=P(A∪Bc)P(B∩(A∪Bc))​ =P(A)+P(Bc)−P(A∩Bc)P(A∩B)​ =P(A)+P(Bc)−P(A∩Bc)P(A)−P(A∩Bc)​ =0.80.7−0.5​=41​

Q. If two events $A$ and $B$ are such that $P (A^{C}) = 0.3, P (B) = 0.4$ and $P (A \cap B^{c}) = 0.5$ , then $P [B / (A \cup B^{c})]$ is equal to

$P [B / (A \cup B^{c})] = \frac{P ( B \cap ( A \cup B ^{c} ))}{P ( A \cup B ^{c} )}$
$= \frac{P ( A \cap B )}{P ( A ) + P ( B ^{c} ) - P ( A \cap B ^{c} )}$
$= \frac{P ( A ) - P ( A \cap B ^{c} )}{P ( A ) + P ( B ^{c} ) - P ( A \cap B ^{c} )}$
$= \frac{0.7 - 0.5}{0.8} = \frac{1}{4}$