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Probability $$P\left( A \right) =0.7,P\left( B \right) =0.4,P\left( A\cap B \right) =0.3$$, then $$P\left( A\cap { B }^{ \prime  } \right) $$ is equal to


A
0.1
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B
0.3
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C
0.2
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D
0.4
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Solution

The correct option is D $$0.4$$
Given, $$P\left( A \right) =0.7,P\left( B \right) =0.4$$

and $$P\left( A\cap B \right) =0.3$$

We know that,

$$P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right) $$

$$\Rightarrow P\left( A\cup B \right) =0.7+0.4-0.3=0.8$$

Now, $$P\left( A\cap { B }^{ \prime  } \right) =P\left( A\cup B \right) -P\left( B \right) $$ $$=0.8-0.4=0.4$$

Mathematics

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