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BITSAT 2023 Shift 1 Paper

Section: Mathematics

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Question : 130 of 130

Marks: +1, -0

If

P(B)=\frac{3}{5}, P(A \mid B)=\frac{1}{2}

P(A \cup B)=\frac{4}{5}

P(A \cup B)'+P(A' \cup B)=

$\frac{1}{5}$
$\frac{4}{5}$
$\frac{1}{2}$
1

Solution: 👈: Video Solution

\text{ } P(B)=\frac{3}{5}, P(A \mid B)=\frac{1}{2} \text{ and }

P(A \cup B)=\frac{4}{5}

P(A \cap B)=P(A \mid 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} \cdot P(A')=1-P(A)=\frac{1}{2}

We know,

P(A \cap B)+P(A' \cap B)=P(B)

[as

A \cap B

and

A' \cap B

are mutually exclusive events]

\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}

Now,

P(A' \cup B)=P(A')+P(B)-P(A' \cap B)

=\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}

\therefore P((A \cup B)')+P(A' \cup B)=\frac{1}{5}+\frac{4}{5}=1

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