- #1

laz0r

- 17

- 0

## 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?