Conditional Probability Question

[laz0r]

Homework Statement

Suppose there are two urns that contain white and yellow balls. Urn 1 contains
10 white and 5 yellow balls, and Urn 2 contains 6 white and 12 yellow balls. You
are going to draw 3 balls without replacement from one of the urns. To decide
which urn to draw from, you will roll an ordinary six-sided (balanced) die. If the
die comes up with a 1, you will draw 3 balls from Urn 1, otherwise you will draw
3 balls from Urn 2.
(a) What is the probability you draw exactly two white balls?
(b) Given you draw exactly two white balls, what is the probability you rolled a
1?

Homework Equations

(10,2) for instance would read ten choose 2 in combinatorics

The Attempt at a Solution

Ok, so I'm not entirely sure how to do this, but this is my attempt..

Urn #1)
10 white balls
5 yellow

((10,2)(5,1))/(15,3) = P(choose exactly 2 white balls from urn #1) = P(#1) = 45/91

Urn #2)

6 white balls
12 yellow

((6,2)(12,1))/(18,3) = P(choose exactly 2 white balls from urn #2) = P(#2) = 15/68

-----

P(draw from urn 1) = 1/6 = P(#11)
P(draw from urn 2) = 5/6 = P(#22)

P(draw exactly 2 white balls) = P(#1)P(#11) + P(#2)P(#22) = (45/91)(1/6) + (15/68)(5/6) = 0.26624

PART B)

P(roll a 1 | draw 2 white balls) = P(roll a 1 and draw 2 white balls)/P(draw 2 white balls)

Am I on the right track here?

 

[BvU]

Wrong kind of question for this forum: PF is not in the business of stamp-approving homework. Wouldn't be good for our relations with the poor teachers of this world.

Show some self-assurance !

I have no clue what would be the right track, but I enjoyed following yours. Thanks for sharing this with us !