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

Question

12 days ago

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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:

Zina [9.1K] · Answer 1 · 2026-03-16T03:13:05+03:00

There are several possible outcomes. The initial composition of the urns is as follows: Urn 1 contains 2 red chips and 4 white chips, totaling 6 chips, whereas Urn 2 has 3 red and 1 white, amounting to 4 chips. When a chip is drawn from the first urn, the probabilities are as follows: for a red chip, it is probability is (2 red from 6 chips = 2/6 = 1/2); for a white chip, it is (4 white from 6 chips = 4/6 = 2/3). After the chip is transferred to the second urn, two scenarios can arise: if the chip drawn from the first urn is white, then Urn 2 will contain 3 red and 2 white chips, making a total of 5 chips, creating a 40% chance for drawing a white chip. Conversely, if a red chip is drawn first, Urn 2 will contain 4 red and 1 white chip, which results in a 20% chance of drawing a white chip. This scenario exemplifies a dependent event, as the outcome hinges on the type of chip drawn first from Urn 1. For the first scenario, the combined probability is (the probability of drawing a white chip from Urn 1) multiplied by (the probability of drawing a white chip from Urn 2), equaling 26.66%. For the second scenario, the probabilities yield a value of 6%.