We have three urns labeled 1, 2 and 3. The urns contain, respectively, three white and three black balls, four white and two black balls, and one white and two black balls. An experiment consists of selecting an urn at random then drawing a ball from it. Define the joint probability distribution over U and C, where U is the chosen urn with values 1, 2 and 3; and C is the color of the ball, with values black and white.

So far I have calculated as follows based on the fact that first we select random urn and then we select ball.

 P(U = 1 , color = white) = 1/6   
 P(U = 1 , color = black) = 1/6   
 P(U = 2 , color = white) = 1/6  
 P(U = 2 , color = black) = 1/6  
 P(U = 3 , color = white) = 1/3  
 P(U = 3 , color = black) = 1/3  

Is this correct?. I quite not sure on my approach. Need advice on this.



The probabilities can be calculated by multiplying the individual probabilities for events in the sequences given by the experiment. For example, the probability of choosing the second urn is 1/3. Given that the second urn has been chosen, the probability of choosing a white ball is 2/3. Hence, $P(U=2,\text{color=white})=1/3*2/3=2/9$. So it seems like you have the right idea but the calculations are not quite correct.

You might also want to check the definition of conditional probability which may help in understanding this:

$P(A|B) = \frac{P(A,B)}{P(B)} \Leftrightarrow P(A,B)=P(B)P(A|B)$

for events A and B.

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  • $\begingroup$ typo it should be P(U=2,color=white)=1/3∗2/3=2/9 . Thanks for the approach $\endgroup$ – webNash Mar 4 '16 at 8:29

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