Explain solution RD Sharma class 12 chapter Probability exercise 30.7 question 24 maths

Answers (1)

Answer:

 \frac{7}{10}

Hint:

 Use Baye’s theorem.

Given:

 The person A, B, C applies for job of manager in private company. Chance of their selection is in ratio 1:2:4. The probability that A, B, C can introduce change to improve profit of company are 0.8, 0.5, 0.3 respectively.

Solution:

Let E1 ,E2,E3  be event that the selection of A, B, C.

\begin{aligned} &P(E_1)=\text { Probability of selection of A }=\frac{1}{7}\\ &P(E_2)=\text { Probability of selection of B }=\frac{2}{7}\\ &P(E_3)=\text { Probability of selection of C }=\frac{4}{7}\\ &P\left ( \frac{A}{E_1} \right )=\text { Probability that A doesnot introduce }=0.2\\ &P\left ( \frac{A}{E_2} \right )\text { Probability that B doesnot introduce }=0.5\\ &P\left ( \frac{A}{E_3} \right )=\text { Probability that C doesnot introduce }=0.7\\ \end{aligned}

Using Baye’s theorem we get

\begin{aligned} &P\left (\frac{E_3}{A} \right )=\frac{P(E_3).P\left ( \frac{A}{E_3} \right )}{P(E_1)\times P\left ( \frac{A}{E_1} \right )+P(E_2)\times P\left ( \frac{A}{E_2} \right )+P(E_3)\times P\left ( \frac{A}{E_3} \right )}\\ &=\frac{{\frac{4}{7}\times 0.7}}{\frac{1}{7}\times 0.2+\frac{2}{7}\times 0.5+\frac{4}{7}\times 0.7}\\ &=\frac{28}{0.2+1+2.8}\\ &=\frac{28}{4}\\ &=\frac{7}{10} \end{aligned}

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