If P(B)=(3)/(5), P(A B)=(1)/(2) and P(A B)=(4)/(5) , then P(A B)^ +P (A^ B)=

Correct answer: D. 1 P(B)=(3)/(5), P(A B)=(1)/(2) and P(A B)=(4)/(5) P(A B)=P(A B) P(B)=(1)/(2) · (3)/(5)=(3)/(10) P(A B)=P(A)+P(B)-P(A B) P(A)=(4)/(5)-(3)/(10)=(1)/(2). P(A^ )=1-P(A)=(1)/(2) We know, P(A B)+P(A^ B)=P(B) [as A B and A^ B are mutually exclusive events] (3)/(10)+P(A^ B)=(3)/(5) P(A^ B)=(3)/(5)-(3)/(10)=(3)/(10) Now, P(A^ B)=P (A^ )+P(B)-P(A^ B) =(1)/(2)+(3)/(5)-(3)/(10)=(5+6-3)/(10)=(4)/(5) P((A B)^ )=1-P(A B)=1-(4)/(5)=(1)/(5) P((A B)^ )+P(A^ B)=(1)/(5)+(4)/(5)=1

BITSAT 2023Paper 1

Exam: BITSAT
Year: 2023
Paper: Paper 1
Subject: Mathematics
Chapter: Probability

BITSAT 2023 · Paper 1 · Mathematics · Probability

If P(B)=(3)/(5), P(A B)=(1)/(2) and P(A B)=(4)/(5) , then P(A B)^...

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

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

, then

\mathrm{P}(\mathrm{A} \cup \mathrm{B})^{\prime}+\mathrm{P}

\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)=

Options

A $\frac{1}{5}$
B $\frac{4}{5}$
C $\frac{1}{2}$
D1

Correct answer

D. 1

Step-by-step solution

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

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}

\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})

\mathrm{P}(\mathrm{A})=\frac{4}{5}-\frac{3}{10}=\frac{1}{2}. \quad \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})=\frac{1}{2}

We know,

\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\mathrm{P}(\mathrm{B})

[as

A \cap B

and

A^{\prime} \cap B

are mutually exclusive events]

\Rightarrow \quad \frac{3}{10}+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\frac{3}{5}

\Rightarrow \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\frac{3}{5}-\frac{3}{10}=\frac{3}{10}

Now,

\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)=\mathrm{P} \left(\mathrm{A}^{\prime}\right)+\mathrm{P}(\mathrm{B})-\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)<br />=\frac{1}{2}+\frac{3}{5}-\frac{3}{10}=\frac{5+6-3}{10}=\frac{4}{5}<br /> \mathrm{P}\left((\mathrm{A} \cup \mathrm{B})^{\prime}\right)=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})=1-\frac{4}{5}=\frac{1}{5}<br />\therefore \mathrm{P}\left((\mathrm{A} \cup \mathrm{B})^{\prime}\right)+\mathrm{P}\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)=\frac{1}{5}+\frac{4}{5}=1

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