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  • If P(B) = 3|5, P(A|B) = 1|2 and P(A ∪ B) = 4|5, then P(A ∪ B)′ + P(A′ ∪ B) = A. 1|5 B. 4|5 C. 1|2 D. 1

    If P(B) = 3|5, P(A|B) = 1|2 and P(A ∪ B) = 4|5, then P(A ∪ B)′ + P(A′ ∪ B) =
A. 1|5
B. 4|5
C. 1|2
D. 1

Answers (1)

Given-

\begin{aligned} &P(B)=\frac{3}{5}, P(A \mid B)=\frac{1}{2} \text { and } P(A \cup B)=\frac{4}{5}\\ &\text { As we know, }\\ &\mathrm{P}(\mathrm{A} \mid \mathrm{B}) \times \mathrm{P}(\mathrm{B})=\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \text { [Property of Conditional Probability] }\\ &\Rightarrow \frac{1}{2} \times \frac{3}{5}=P(A \cap B)\\ &\Rightarrow P(A \cap B)=\frac{3}{10}\\ &\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \text { [Additive Law of Probability] }\\ &\therefore \frac{4}{5}=P(A)+\frac{3}{5}-\frac{3}{10}\\ &\Rightarrow P(A)=\frac{4}{5}-\frac{3}{5}+\frac{3}{10}\\ &\Rightarrow P(A)=\frac{1}{5}+\frac{3}{10}\\ &\Rightarrow P(A)=\frac{2+3}{10} \end{aligned}

\\ \Rightarrow P(A)=\frac{5}{10}=\frac{1}{2} \\ \therefore P(A \cup B)^{\prime}=P\left[A^{\prime} \cap B^{\prime}\right] \\ =1-P(A \cup B) \\ =1-\frac{4}{5} \\ =\frac{1}{5} \\ \text { and } P\left(A^{\prime} \cup B\right)=1-P\left(A^{\prime} \cap B\right) \\ =1-[P(A)-P(A \cap B)]

\\ =1-\left(\frac{1}{2}-\frac{3}{10}\right) \\ =1-\left(\frac{5-3}{10}\right) \\ =1-\frac{2}{10} \\ =\frac{10-2}{10} \\ =\frac{4}{5} \\ \Rightarrow P(A \cup B)^{\prime}+P\left(A^{\prime} \cup B\right)=\frac{1}{5}+\frac{4}{5} \\ =1

Hence, the correct option is D

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