Answer:
Multiple answers
Step-by-step explanation:
The original urns have:
We take one chip from the first urn, so we have:
The probability of take a red one is : [tex]\frac{1}{3}[/tex] (2 red from 6 chips(2/6=1/2))
For a white one is: [tex]\frac{2}{3}[/tex](4 white from 6 chips(4/6=(2/3))
Then we put this chip into the second urn:
We have two possible cases:
If we select a chip from the urn two:
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 :
[tex]\frac{4}{6}[/tex] x [tex]\frac{2}{5}[/tex] = [tex]\frac{4}{15}[/tex] = 26.66% ( [tex]\frac{4}{6}[/tex] the probability of taking a white chip from the urn 1, [tex]\frac{2}{5}[/tex] the probability of taking a white chip from urn two)
For the second case we multiply:
[tex]\frac{1}{3}[/tex] x [tex]\frac{1}{5}[/tex] = [tex]\frac{1}{30}[/tex] = .06% ( [tex]\frac{1}{3}[/tex] the probability of taking a red chip from the urn 1, [tex]\frac{1}{5}[/tex] the probability of taking a white chip from the urn two)