How to condition this problem involving UrnsThere are two urns one with 6 blue objects an 4 yellow objects, and the other with 3 blue objects and 4 yellow objects. We choose an object at random.(1) If an object we select is yellow, what is the probability it came from the first urn?(2) Assume we put the yellow object from (1) back in the urn it was drawn from and then draw an object from the same urn. This object is also yellow. What is the probability we chose the first urn?I have already determined that P ( { We drew from the first urn } | { The object was yellow } ) = 1 2 . P ( { We drew from the first urn } | { The object was yellow } )We compute this using Bayes' theorem. P ( { The object was yellow } | { We drew from the first urn } ) P ( { We drew from the first urn } ) P { The object was yellow } Implicit in the question are the following probabilities: P ( { The object was yellow } ) = 8 17 P ( { The object was yellow } | { We drew from the first urn } ) = 4 10 P ( { We drew from the first urn } ) = 10 17 Combining these results together, we have that P ( { We drew from the first urn } | { The object was yellow } ) = 4 10 10 17 8 17 = 1 2 Part (b) looks like it is asking to compute P ( { We drew the second object the first urn } | { the set of outcomes such that the first and second draws are from the same urn and both objects are yellow } )

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How to condition this problem involving UrnsThere are two urns one with 6 blue objects an 4 yellow objects, and the other with 3 blue objects and 4 yellow objects. We choose an object at random.(1) If an object we select is yellow, what is the probability it came from the first urn?(2) Assume we put the yellow object from (1) back in the urn it was drawn from and then draw an object from the same urn. This object is also yellow. What is the probability we chose the first urn?I have already determined that      P    (  {  We drew from the first urn  }      |    {  The object was yellow  }  )  =      1    2    .      P    (  {  We drew from the first urn  }      |    {  The object was yellow  }  )We compute this using Bayes&#039; theorem.                    P            (      {      The object was yellow      }              |            {      We drew from the first urn      }      )              P            (      {      We drew from the first urn      }      )                      P            {      The object was yellow      }      Implicit in the question are the following probabilities:      P    (  {  The object was yellow  }  )  =      8    17        P    (  {  The object was yellow  }      |    {  We drew from the first urn  }  )  =      4    10        P    (  {  We drew from the first urn  }  )  =      10    17  Combining these results together, we have that      P    (  {  We drew from the first urn  }      |    {  The object was yellow  }  )  =                    4        10                    10        17                    8      17        =      1    2  Part (b) looks like it is asking to compute      P    (  {  We drew the second object the first urn  }      |    {  the set of outcomes such that the first and second draws are from the same urn and both objects are yellow  }  )

Cavalascamq · Accepted Answer

Step 1Here is my understanding for part (2), let A= &quot;from 1st urn | ball is yellow&quot;  P  (  A  )  =  P  (  A  ∩  part(1) from 1st urn  )  +  P  (  A  ∩  part(1) from 2nd urn  )  P  (  A  ∩  part(1) from 1st urn  )  =  P  (  A         |       part(1) from 1st urn  )  ⋅  P  (  part(1) from 1st urn  )Step 2  P  (  A  ∩  part(1) from 2nd urn  )  =  0, because it says &quot;we put the yellow object from (1) back in the urn it was drawn from and then draw an object from the same urn&quot;, so if the first time in part(1) you draw from the 2nd urn, then for part(2), you still draw from 2nd urn (from the same urn), so event A cannot happen, so this part is zero.

4enevi · Accepted Answer

Step 1Remember, each urn has equal a priori probability of       1    2   of being chosen  =            (      U      r      n            1            ∩            y      e      l      l      o      w      )              (      U      r      n      1            ∩            y      e      l      l      o      w      )      +      (      u      r      n      2            ∩            y      e      l      l      o      w      )            =                              1          2                          4          10                                                  1            2                          (                  4          10                +                  4          7                )              =            7      17      Step 2For the second part, we are returning the yellow to the same urn from which drawn,Thus P(drawn second time from urn   1  )  =            7      17                  4      10        =            14      85      If the first time, yellow was drawn from urn 2,the probability of drawing second time from Urn 1 is obviously 0 zero, so it won&#039;t enter the calculations for second part

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