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  • Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 23

Need solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 23

Answers (1)

Answer:

 \frac{3}{5}

Hint:

 Use Baye’s theorem.

Given:

 From A, B, C chance of being selected as manager of firm in ratio 4:1:2 respectively. The respectively probability for them a radical change in marketing strategy, 0.3, 0.8, 0.5.

Solution:

Let E1 ,E2,E3  denote that event that the change take place is selected B is selected C is selected.

\begin{aligned} &P(E_1)=\frac{4}{7}\\ &P(E_2)=\frac{1}{37}\\ &P(E_3)=\frac{2}{7}\\ &P\left ( \frac{A}{E_1} \right )=0.3\\ &P\left ( \frac{A}{E_2} \right )=0.8\\ &P\left ( \frac{A}{E_3} \right )=0.5\\ \end{aligned}

Using Baye’s theorem we get

\begin{aligned} &P\left (\frac{E_1}{A} \right )=\frac{P(E_1).P\left ( \frac{A}{E_1} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )}\\ &=\frac{{\frac{4}{7}\times 0.3 }}{\frac{4}{7}\times 0.3+\frac{1}{7}\times 0.8+\frac{2}{7}\times 0.5}\\ &=\frac{1.2}{1.2+0.8+1}\\ &=\frac{1.2}{3}\\ &=\frac{12}{30}\\ &=\frac{2}{5}\\ &\text { Required probability }=1-P\left (\frac{E_1}{A} \right )=1-\frac{2}{5}=\frac{3}{5} \end{aligned}

Posted by

Gurleen Kaur

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