Given right triangle A w/a hypotenuse length of 3y + x and a leg of y-x, and right triangle B, w/ a hypotense length of y+5 and
1 answer:
You might be interested in
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 : (2 red from 6 chips(2/6=1/2))
For a white one is: (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:
= 40% ( 2 whites of 5 chips)
- In the second case the probability of taking a white one is of:
= 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 :
x
=
= 26.66% (
the probability of taking a white chip from the urn 1,
the probability of taking a white chip from urn two)
For the second case we multiply:
x
=
= .06% (
the probability of taking a red chip from the urn 1,
the probability of taking a white chip from the urn two)