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URN I contains 3 White,2 Black 2 Green Balls.
URN II contains 2 White,3 Black 4 Green Balls.
URN III contains 5 White,2 Black 2 Green Balls.
An urn is chosen at random and two balls are drawn,they happen to be black and green.
What is the probability that the come from urns
1) URN I
2) URN II
3) URN III ????
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2 Answers

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Best answer

Let P(E) = Probability of choosing 1Black & 1 Green balls from Urn. 

P(Ei) = Probability of selecting one of three urns .

Hence P(E1) = P(E2) =P(E3) = 1/3

P(E) =  Select Urn1 & choose 1Black & 1 Green balls + Select Urn1 & choose 1Black & 1 Green balls +Select Urn1 & choose 1Black & 1 Green balls

 = P(E1)* P(E / E) +  P(E2)* P(E / E2 ) + P(E3)* P(E / E3 )    

   =  (1/3)* 2C1*2C1  / (7C) + (1/3)* 3C1*4C1  / (9C) +(1/3)* 2C1*2C1  / (9C) = 4/63 + 1/9 + 1/27 = 40/189

Now we have to find probability that balls chose were from Urn 'i'

i.e P(Ei / E) =  ?

P(Ei / E)  = P(Ei ∩ E)/P(E)  = P(Ei)* P(E / Ei )  / P(E)

Now

Ans 1) for URN1 , put i =1, we have

P(E1 / E)  = P(E1 ∩ E)/P(E)  = P(E1)* P(E / E1)  / P(E) = (4/63) / (40/189) =    3/10 (Substituting values from above calculations)

Ans 2) for URN2 , put i =2, we have

P(E2 / E)  = P(E2 ∩ E)/P(E)  = P(E2)* P(E / E2)  / P(E) = (1/9) / (40/189) =    21/40 (Substituting values from above calculations)

Ans 3) for URN3 , put i =3, we have

P(E3 / E)  = P(E3 ∩ E)/P(E)  = P(E3)* P(E / E3)  / P(E) = (1/27) / (40/189) =    7/40 (Substituting values from above calculations)

 
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Nicely answered.Thanks Shashank !! :) :)
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Let the Event that ball happen to be black and green=E

Let the Event that ball taken from $\text{URN I}=U_I$

Let the Event that ball taken from $\text{URN II}=U_{II}$

Let the Event that ball taken from $\text{URN III}=U_{III}$


$$P(\frac{U_I}{E})=\frac{P(\frac{E}{U_I})\times P(U_I)}{P(\frac{E}{U_I})\times P({U_{I}})+P(\frac{E}{U_{II}})\times P({U_{II}})+P(\frac{E}{U_{III}})\times P({U_{III}})}$$

$$=\frac{\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{7}{2}}\times \frac{1}{3}}{\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{7}{2}}\times \frac{1}{3}+\frac{\binom{3}{1}\times \binom{4}{1} }{\binom{9}{2}}\times \frac{1}{3}+\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{9}{2}}\times \frac{1}{3}}=\frac{3}{10}$$


 



$$P(\frac{U_{II}}{E})=\frac{P(\frac{E}{U_I})\times P(U_{II})}{P(\frac{E}{U_I})\times P({U_{I}})+P(\frac{E}{U_{II}})\times P({U_{II}})+P(\frac{E}{U_{III}})\times P({U_{III}})}$$

$$=\frac{\frac{\binom{3}{1}\times \binom{4}{1} }{\binom{9}{2}}\times \frac{1}{3}}{\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{7}{2}}\times \frac{1}{3}+\frac{\binom{3}{1}\times \binom{4}{1} }{\binom{9}{2}}\times \frac{1}{3}+\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{9}{2}}\times \frac{1}{3}}=\frac{21}{40}$$


$$P(\frac{U_{II}}{E})=\frac{P(\frac{E}{U_II})\times P(U_{II})}{P(\frac{E}{U_I})\times P({U_{I}})+P(\frac{E}{U_{II}})\times P({U_{II}})+P(\frac{E}{U_{III}})\times P({U_{III}})}$$

$$=\frac{\frac{\binom{3}{1}\times \binom{4}{1} }{\binom{9}{2}}\times \frac{1}{3}}{\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{7}{2}}\times \frac{1}{3}+\frac{\binom{3}{1}\times \binom{4}{1} }{\binom{9}{2}}\times \frac{1}{3}+\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{9}{2}}\times \frac{1}{3}}=\frac{21}{40}$$


$$P(\frac{U_{III}}{E})=\frac{P(\frac{E}{U_III})\times P(U_{III})}{P(\frac{E}{U_I})\times P({U_{I}})+P(\frac{E}{U_{II}})\times P({U_{II}})+P(\frac{E}{U_{III}})\times P({U_{III}})}$$

$$=\frac{\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{9}{2}}\times \frac{1}{3}}{\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{7}{2}}\times \frac{1}{3}+\frac{\binom{3}{1}\times \binom{4}{1} }{\binom{9}{2}}\times \frac{1}{3}+\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{9}{2}}\times \frac{1}{3}}=\frac{9}{40}$$

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