If P(B)=35,P(A∣B)=12 and P(A∪B)=45, then, P(A∪B)′+PA′∪B=

P(B)=35,P(A∣B)=12 and P(A∪B)=45P(A∩B)=P(A∣B)P(B)=12⋅35=310P(A∪B)=P(A)+P(B)−P(A∩B)P(A)=45−310=12PA′=1−P(A)=1−12=12We know P(A∩B)+PA′∩B=P(B) [as A∩B and A′∩B are mutually exclusive events⇒ 310+PA′∩B=35⇒ PA′∩B=35−310=310Now ,PA′∪B=PA′+P(B)−PA′∩B=12+35−310=5+6−310=45P(A∪B)′=1−P(A∪B)=1−45=15∴P(A∪B)′+PA′∪B=15+45=1

If P(B)=35,P(A∣B)=12 and P(A∪B)=45, then, P(A∪B)′+PA′∪B=

P(B)=35,P(A∣B)=12 and P(A∪B)=45P(A∩B)=P(A∣B)P(B)=12⋅35=310P(A∪B)=P(A)+P(B)−P(A∩B)P(A)=45−310=12PA′=1−P(A)=1−12=12We know P(A∩B)+PA′∩B=P(B) [as A∩B and A′∩B are mutually exclusive events⇒ 310+PA′∩B=35⇒ PA′∩B=35−310=310Now ,PA′∪B=PA′+P(B)−PA′∩B=12+35−310=5+6−310=45P(A∪B)′=1−P(A∪B)=1−45=15∴P(A∪B)′+PA′∪B=15+45=1

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If $P (B) = \frac{3}{5}, P (A ∣ B) = \frac{1}{2}$ and $P (A \cup B) = \frac{4}{5},$ then, $P ((A \cup B)^{'}) + P (A^{'} \cup B) =$

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$\frac{1}{5}$

$\frac{4}{5}$

$\frac{1}{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} P (B) = \frac{3}{5}, P (A ∣ B) = \frac{1}{2} and P (A \cup B) = \frac{4}{5} \\ P (A \cap B) = P (A ∣ B) P (B) = \frac{1}{2} \cdot \frac{3}{5} = \frac{3}{10} \\ P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ P (A) = \frac{4}{5} - \frac{3}{10} = \frac{1}{2} \\ P (A^{'}) = 1 - P (A) = 1 - \frac{1}{2} = \frac{1}{2} \end{array}$

We know $P (A \cap B) + P (A^{'} \cap B) = P (B)$

$[as A \cap B and A^{'} \cap B$ are mutually exclusive events

$\begin{array}{l} \Rightarrow \frac{3}{10} + P (A^{'} \cap B) = \frac{3}{5} \\ \Rightarrow P (A^{'} \cap B) = \frac{3}{5} - \frac{3}{10} = \frac{3}{10} \end{array}$

Now $, P (A^{'} \cup B) = P (A^{'}) + P (B) - P (A^{'} \cap B)$

$\begin{array}{l} = \frac{1}{2} + \frac{3}{5} - \frac{3}{10} = \frac{5 + 6 - 3}{10} = \frac{4}{5} \\ P ((A \cup B)^{'}) = 1 - P (A \cup B) = 1 - \frac{4}{5} = \frac{1}{5} \\ ∴ P ((A \cup B)^{'}) + P (A^{'} \cup B) = \frac{1}{5} + \frac{4}{5} = 1 \end{array}$