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  • A and B are events such that P(A) = 0.4, P(B) = 0.3 and P(A ∪ B) = 0.5. Then P (B′ ∩ A) equals A. 2|3 B. 1|2 C. 3|10 D. 1|5

A and B are events such that P(A) = 0.4, P(B) = 0.3 and P(A ∪ B) = 0.5. Then P (B′ ∩ A) equals
A. 2|3
B. 1|2
C. 3|10
D. 1|5

Answers (1)

Given-

\\\mathrm{P}(\mathrm{A})=0.4, \mathrm{P}(\mathrm{B})=0.3$ and \\$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.5$ \\$P(A \cup B)=P(A)+P(B)-P(A \cap B)[$ Additive Law of Probability $]$ \\$\Rightarrow 0.5=0.4+0.3-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$ \\$\Rightarrow P(A \cap B)=0.7-0.5$ \\$\Rightarrow P(A \cap B)=0.2$ \\$\therefore \mathrm{P}\left(\mathrm{B}^{\prime} \cap \\\mathrm{A}\right)=P(A)-P(A \cap B)$ \\$=0.4-0.2$ \\$=0.2$ \\$=\frac{1}{5}

Hence, the correct option is D

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infoexpert22

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