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  • Provide Solution For R. D. Sharma Maths Class 12 Chapter 30 Probaility Exercise 30.4 Question 9 Maths Textbook Solution.

Provide Solution For  R. D. Sharma Maths Class 12 Chapter 30 Probaility  Exercise 30.4 Question 9 Maths Textbook Solution.

Answers (1)

Answer:\begin{aligned} &P(B)=\frac{1}{2} \text { then } P(A)=\frac{1}{3} \\ \end{aligned}

             \begin{aligned} &P(B)=\frac{1}{3} \text { then } P(A)=\frac{1}{2} \end{aligned}

Hint: Form events as a equation to solve

Given, A and B are independent events

              P\left ( A\cap B \right )=\frac{1}{6}

               P\left ( \bar{A}\cap \bar{B} \right )=\frac{1}{3}

Solution:

As we know, Two events are said to be independent if the product of the events are equal to their intersection. i.e.P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )

Now, 

            \begin{aligned} &P(A \cap B)=P(A) P(B) \\ &\frac{1}{6}=P(A) P(B) \\ &P(A)=\frac{1}{6(P(B))} \end{aligned}                                            ...................(1)

Now, for P(\bar{A} \cap \bar{A})=P(\bar{A}) P(\bar{B}) \\

                 \begin{aligned} &\frac{1}{3}=[1-\mathrm{P}(\mathrm{A})][1-\mathrm{P}(\mathrm{B})] \\ &\frac{1}{3}=\left[1-\frac{1}{6 \mathrm{P}(\mathrm{B})}\right][1-\mathrm{P}(\mathrm{B})] \\ &\frac{1}{3}=\left[1-\frac{1}{6 \mathrm{x}}\right][1-\mathrm{x}] \end{aligned}

                Where, P\left ( B \right )=x

\begin{aligned} &\frac{1}{3}=\left[\frac{6 x-1}{6 x}\right][1-x] \\ &2 x=6 x-6 x^{2}-1+x \\ &2 x-6 x+6 x^{2}+1-x=0 \\ \end{aligned}

\begin{aligned} &6 x^{2}-5 x+1=0 \\ &6 x^{2}-3 x-2 x+1=0 \\ &3 x(2 x-1)-1(2 x-1)=0 \\ &(2 x-1)(3 x-1)=0 \\ \end{aligned}

\begin{aligned} &x=\frac{1}{2}, \frac{1}{3} \\ &P(B)=\frac{1}{2} \text { then } P(A)=\frac{1}{3} \\ &P(B)=\frac{1}{3} \text { then } P(A)=\frac{1}{2} \end{aligned}

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