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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability Exercise Multiple Choice Questions Question 37 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 30 Probability  Exercise Multiple Choice Questions  Question 37 Maths Textbook Solution.

Answers (1)

Answer:(d)

Hint: You must know the rules of finding probability.

Given:P(B)=\frac{3}{5}, P(A \cup B)=\frac{4}{5}, P\left(\frac{A}{B}\right)=\frac{1}{2}

Solution:

\begin{aligned} &P\left(\frac{A}{B}\right)=\frac{1}{2} \\ &\frac{P(A \cap B)}{P(B)}=\frac{1}{2} \end{aligned}

\begin{aligned} &P(A \cap B)=\frac{1}{2} \times P(B) \\ &P(A \cap B)=\frac{1}{2} \times \frac{3}{5}=\frac{3}{10} \end{aligned}

And

\begin{aligned} &P(B \cap \bar{A})=P(B)-P(A \cap B) \\ &=\frac{3}{5}-\frac{3}{10}=\frac{3}{10} \end{aligned}

Also,

\begin{aligned} &P(A \cup B)=P(A)+P(B)-P(A \cap B) \\ &\frac{4}{5}=P(A)+\frac{3}{5}-\frac{3}{10} \end{aligned}

\begin{aligned} &P(A)=\frac{4}{5}-\frac{3}{5}+\frac{3}{10} \\ &P(A)=\frac{8-6+3}{10} \\ &P(A)=\frac{5}{10} \\ &P(A)=\frac{1}{2} \end{aligned}

So,

\begin{aligned} &P(\bar{A})=1-P(A)=1-\frac{1}{2} \\ &P(\bar{A})=\frac{1}{2} \end{aligned}

Now,

\begin{aligned} &P\left(\frac{B}{\bar{A}}\right)=\frac{P(B \cap \bar{A})}{P(\bar{A})} \\ &=\frac{\left(\frac{3}{10}\right)}{\left(\frac{1}{2}\right)}=\frac{3 \times 2}{10}=\frac{6}{10} \\ &P\left(\frac{\bar{B}}{A}\right)=\frac{3}{5} \end{aligned}

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