Respuesta :
Answer:
Multiple answers
Step-by-step explanation:
The original urns have:
- Urn 1 = 2 red + 4 white = 6 chips
- 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 : [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:
- 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: [tex]\frac{2}{5}[/tex] = 40% ( 2 whites of 5 chips)
- In the second case the probability of taking a white one is of: [tex]\frac{1}{5}[/tex] = 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 :
[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)