Question

Suppose you have two urns with poker chips in them. Urn I contains two red chips and four white chips. Urn II contains three red chips and one white chip. You randomly select one chip from urn I and put it into urn II. Then you randomly select a chip from urn II.
(a) What is the probability that the chip you select from urn II is white?
b) Is selecting a white chip from urn I and selecting a white chip from urn II independent? Justify your answer numerically.

Answer 1

Answer:

Multiple answers

Step-by-step explanation:

The original urns have:

1. Urn 1 = 2 red + 4 white = 6 chips
2. Urn 2 = 3 red + 1 white = 4 chips

We take one chip from the first urn, so we have:

The probability of take a red one is : $\frac{1}{3}$ (2 red from 6 chips(2/6=1/2))

For a white one is: $\frac{2}{3}$(4 white from 6 chips(4/6=(2/3))

Then we put this chip into the second urn:

We have two possible cases:

• First if the chip we got from the first urn was white. The urn 2 now has 3 red + 2 whites = 5 chips
• Second if the chip we got from the first urn was red. The urn two now has 4 red + 1 white = 5 chips

If we select a chip from the urn two:

• In the first case the probability of taking a white one is of:  $\frac{2}{5}$ = 40%  ( 2 whites of 5 chips)
• In the second case the probability of taking a white one is of:  $\frac{1}{5}$ = 20%  ( 1 whites of 5 chips)

This problem is a dependent event because the final result depends of the first chip we got from the urn 1.

For the fist case we multiply :

$\frac{4}{6}$ x $\frac{2}{5}$ = $\frac{4}{15}$ = 26.66%   ( $\frac{4}{6}$ the probability of taking a white chip from the urn 1, $\frac{2}{5}$  the probability of taking a white chip from urn two)

For the second case we multiply:

$\frac{1}{3}$ x $\frac{1}{5}$ = $\frac{1}{30}$ = .06%   ( $\frac{1}{3}$ the probability of taking a red chip from the urn 1, $\frac{1}{5}$   the probability of taking a white chip from the urn two)