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pishuonlain [190]
3 years ago
12

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.
Mathematics
1 answer:
Neporo4naja [7]3 years ago
8 0

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)

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