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  • Need solution for RD sharma maths class 12 chapter 30 probablity exercise Very short answer question, question 13

Need solution for RD sharma maths class 12 chapter 30 probablity exercise Very short answer question, question 13

Answers (1)

Answer : \frac{1}{4}

Hint:

\begin{aligned} &P\left(\frac{B}{A \cup B^{\prime}}\right)=\frac{P\left(B \cap\left(A \cup B^{\prime}\right)\right)}{P\left(A \cup B^{\prime}\right)} \\ \end{aligned}

Given:

\begin{aligned} &P\left(A^{\prime}\right)=0.3 \\ &P(B)=0.4 \end{aligned} and

\begin{aligned} P\left(A \cap B^{\prime}\right)=0.5 \end{aligned} Then

Explanation:

We have

\begin{aligned} &P(\bar{A})=0.3, P(B)=0.4\end{aligned}  and \begin{aligned}& P(A \cap \bar{B})=0.5 \end{aligned}

According to bay’s theorem

\begin{aligned} P\left(\frac{B}{A \cap \bar{B}}\right)=\frac{P(B \cap(\bar{A} \cap \bar{B}))}{P(\bar{A} \cap \bar{B})}\\ =\frac{P(B \cap(\overline{A \cup B}))}{P(\bar{A} \cap \bar{B})} \end{aligned}

                           \begin{aligned} =\frac{P(\bar{B} \cap(\overline{A \cup B}))}{P(\bar{A} \cap \bar{B})}\\ =\frac{P(\bar{B} \cup(A \cup B))}{P(\bar{A} \cap \bar{B})} \end{aligned}

Now

\begin{aligned} \bar{B} \cup B=U=\phi \end{aligned}  Where U=universal set

So

\begin{aligned} P(\bar{B} \cup(A \cup B))=\phi \end{aligned}

Therefore \begin{aligned} P\left(\frac{B}{\bar {A} \cap \bar{B}}\right)=0 \end{aligned}

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