1. ## A tough one

There are three urns.
1. The first contains 2 white and 4 black cards
2. The second contains 8 white and four black cards
3. The third contains 1 white and 3 black card.

If one card is selected from each urn, find the probability that exactly 2 white cards are drawn"

2. What Plato is getting at is suppose the two white cards come from urns 1 and 2, repectively. Therefore, the third card, black, must come from urn 3.
So, (2/6)(8/12)(3/4)=1/6

Now, do the same for the other two cases and add them up. Supoose they come from urns 1 and 3 and urns 2 and 3

3. case 1 (urn 1 and 2)
(2/6)(8/12)(3/4) = 1/6

Case 2 (Urn 2 and 3)
(4/6)(8/12)(1/4) = 1/9

Case 3 (Urn 1 and 3)
(2/6)(4/12)(1/4) = 1/36

1/6 + 1/9 + 1/36 = 0.305

4. Originally Posted by Infiniti
case 1 (urn 1 and 2)
(2/6)(8/12)(3/4) = 1/6

Case 2 (Urn 2 and 3)
(4/6)(8/12)(1/4) = 1/9

Case 3 (Urn 1 and 3)
(2/6)(4/12)(1/4) = 1/36

1/6 + 1/9 + 1/36 = 0.305
Since you have the fractions, why not add them and put the answer in terms of a fraction?
$\displaystyle \frac{1}{6} + \frac{1}{9} + \frac{1}{36}$

$\displaystyle = \frac{6}{36} + \frac{4}{36} + \frac{1}{36}$

$\displaystyle = \frac{11}{36}$

The value of doing this is that this is the exact answer, not an approximation.

-Dan

5. Originally Posted by Infiniti
case 1 (urn 1 and 2)
(2/6)(8/12)(3/4) = 1/6

Case 2 (Urn 2 and 3)
(4/6)(8/12)(1/4) = 1/9

Case 3 (Urn 1 and 3)
(2/6)(4/12)(1/4) = 1/36