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  • Provide solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 6

Provide solution for RD Sharma maths class 12 chapter Probability exercise 30.7 question 6

Answers (1)

Answer:

 \frac{2}{9}

Hint:

 Use baye’s theorem.

Given:

 Two groups are competing the computation for position of board of director of cooperation. The probability that the first and second group will win 0.6 and 0.4 respectively. Further if she first group win the probability of introducing new product is 0.7 and corresponding 0.3 if second group win, find probability of new product introduce was by second group.

Solution:

Let E = first group

F = second group

G = new product

We need to find probability that new product introduce was by second group.

i.e,

P\left (\frac{F}{G} \right )

So,

\begin{aligned} &P\left (\frac{F}{G} \right )=\frac{P(F).P\left ( \frac{G}{F} \right )}{P(E)\times P\left ( \frac{G}{E} \right )+P(F)\times P\left ( \frac{G}{F} \right )}\\ \end{aligned}

P(E) = probability of first group win = 0.6

P(F) = probability of second group win = 0.4

\begin{aligned} &P\left (\frac{G}{E} \right )= \text { probability of new product of first group } =0.7\\ &P\left (\frac{G}{F} \right )= \text { probability of new product of second group } =0.3\\ \end{aligned}

Putting values in Baye’s theorem we get

\begin{aligned} &P\left (\frac{E_2}{A} \right )=\frac{0.4\times 0.3}{0.6\times 0.7+0.4\times 0.3 }\\ &=\frac{0.12}{0.42+0.12}\\ &=\frac{0.12}{0.54}\\ &=\frac{2}{9} \end{aligned}

Posted by

Gurleen Kaur

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