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

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

Answers (1)

Answer: \frac{2}{15}

Hint: Use, P\left ( \bar{A}\cap \bar{B} \right )=P\left ( \bar{A} \right )P\left ( \bar{B} \right )

Given: A can solve the problem = \frac{2}{3}

            B can solve the problem= \frac{3}{5}

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 )

              A can solve the problem = \frac{2}{3}           

              i.e., P\left ( A \right )=\frac{2}{3}

            B can solve the problem= \frac{2}{3}

             i.e.,  P\left ( B \right )=\frac{3}{5}

              \begin{aligned} &\mathrm{P}(\mathrm{A})+\mathrm{P}(\overline{\mathrm{A}})=1 \\ &\mathrm{P}(\overline{\mathrm{A}})=1-\frac{2}{3} \\ &=\frac{1}{3} \end{aligned}

Similarly,P\left ( \bar{B} \right )=1-P\left ( B \right )

                             =1-\frac{3}{5}

                               =\frac{2}{5}

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

                       =\frac{1}{3}\times \frac{2}{5}

                        =\frac{2}{15}        

Posted by

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