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Question

$$A$$ and $$B$$ are two events such that $$P(A) = 0.54, P(B) = 0.69$$ and P(A $$\displaystyle \cap $$B) $$= 0.35$$
Find (i) P(A$$\displaystyle \cup $$B) 
(ii) P(A' $$\displaystyle \cap $$B')
(iii)  P(A $$\displaystyle \cap $$B')
(iv)  P(B $$\displaystyle \cap $$A')


Solution

It is given that $$P(A) = 0.54, P(B) = 0.69, P(A \displaystyle \cap  B) = 0.35$$
(i) We know that $$P$$ $$\displaystyle \left ( A \cup B \right )=P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )$$
$$\displaystyle \therefore P\left ( A\cup B \right )=0.54+0.69-0.35=0.88$$
(ii) $$\displaystyle A'\cap B'=\left ( A\cup B \right )',$$ [by De Morgan's law ]
$$\displaystyle \therefore P \left ( A'\cap B' \right )=P\left ( A\cup B \right )'=1-P\left ( A\cup B \right )=1-0.88=0.12$$
(iii) $$\displaystyle P\left ( A\cap B \right )'=P\left ( A \right )-P\left ( A\cap B \right )=0.54-0.35=0.19$$
(iv)We know that,  $$\displaystyle P\left ( B\cap A' \right )=P\left ( B \right )-P\left ( A\cap B \right )$$
$$\displaystyle \therefore P\left ( B\cap A' \right )= 0.69-0.35=0.34$$

Mathematics
NCERT
Standard XI

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