Question

for a,b&c the chances of being selected as manager of a firm are in the ratio 4:1:2 respectively . the respective probabilities for them to introduce a radical change in marketing strategy are 0.3,0.8,& 0.5 .if the change does takes place , find the probability that it is due to the appointment of B or C

Answer 1

QUESTION:-

•For A,B&Cthe chances of being selected as manager of a firm are in the ratio 4:1:2 respectively . the respective probabilities for them to introduce a radical change in marketing strategy are 0.3,0.8,& 0.5 .if the change does takes place , find the probability that it is due to the appointment of B or C.

SOLUTION:-

Let $\sf E_1$:A is appointed as manager

Let $\sf E_2$:B is appointed as manager

Let $\sf E_3$:C is appointed as manager

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Then $\sf E_1,E_2,E_3\: are \:mutually \:exclusive\:and \:exhaustive.$

Moreover

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$\sf P(E_1)= \dfrac{4}{4+1+2} = \dfrac{4}{7}$

$\sf P(E_2)= \dfrac{1}{4 + 1 + 2} = \dfrac{1}{7}$

$\sf And \: P(E_3) = \dfrac{2}{4+1+2} = \dfrac{2}{7}$

Let E: There is a radical change in marketing strategy

then

$\sf P(E/E_1)=0.3= \dfrac{3}{10}$

$\sf P(E/E_2)=0.8= \dfrac{8}{10}$

$\sf And \: P(E/E_3)=0.5=\dfrac{5}{10}$

Using Baye's theorum the required probability is

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$:: \implies\sf P((E_2 \cap E_3) /E)$

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$::\implies \sf P(E_2/E)+P(E_3/E)$

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$\sf:: \implies \dfrac{P(E/E_2)P(E_2)}{P(E/E_1)P(E_1) + P(E/E_2)P(E_2)+P(E/E_3)P(E_3)} \\ \sf \: \: \: \: \: \: \: \: \: + \dfrac{P(E/E_3)P(E_3)}{P(E/E_1)P(E_1) + P(E/E_2)P(E_2)+P(E/E_3)P(E_3)}$

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$:: \implies \sf \dfrac{\dfrac{8}{10} \times \dfrac{1}{7} + \dfrac{5}{10} \times \dfrac{2}{7} }{ \dfrac{3}{10} \times \dfrac{4}{7} + \dfrac{8}{10} \times \dfrac{1}{7} + \dfrac{5}{10} \times \dfrac{2}{7} }$

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$:: \implies \sf \dfrac{8+10}{12+8+10}$

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$\bf:: \implies \dfrac{18}{30} = \boxed{ \bf\dfrac{3}{5}} \longrightarrow Required \:Answer$