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bahirNaik
asked
in Probability
Jan 1, 2016

1,880 views
3 votes

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 ????

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 ????

6 votes

Best answer

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

P(E_{i}) = Probability of selecting one of three urns .

Hence P(E_{1}) = 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(E_{1})* P(E / E_{1 }) + P(E_{2})* P(E / E_{2}_{ }) + P(E_{3})* P(E / E_{3}_{ })

= (1/3)* ^{2}C_{1}*^{2}C_{1 }/ (^{7}C_{2 }) + (1/3)* ^{3}C_{1}*^{4}C_{1 }/ (^{9}C_{2 }) +(1/3)* ^{2}C_{1}*^{2}C_{1 }/ (^{9}C_{2 }) = 4/63 + 1/9 + 1/27 = 40/189

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

i.e P(E_{i} / E) = ?

P(E_{i} / E) = P(E_{i} ∩ E)/P(E) = P(E_{i})* P(E / E_{i}_{ }) / P(E)

Now

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

P(E_{1} / E) = P(E1 ∩ E)/P(E) = P(E_{1})* 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(E_{2} / E) = P(E_{2} ∩ E)/P(E) = P(E_{2})* P(E / E_{2}) / P(E) = (1/9) / (40/189) = 21/40 (Substituting values from above calculations)

Ans 3) for URN3 , put i =3, we have

P(E_{3} / E) = P(E_{3} ∩ E)/P(E) = P(E_{3})* P(E / E_{3}) / P(E) = (1/27) / (40/189) = 7/40 (Substituting values from above calculations)

0 votes

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 $\text{URN III}=U_{III}$

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

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

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

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

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

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

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

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