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There are two urns on a table. Urn number 1 contains four white marbles and two red marbles each. Urn number 2 contains one white marbles and five red marble. Without looking, a marble is selected at random from urn number 1 and placed in urn number 2. Finally, a marble is selected at random from urn number 2. What is the probability that the marble selected from urn number 2 is red? Here is what I did: 6/7*5/7=0.6122

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    https://en.wikipedia.org/wiki/Law_of_total_probability – 2017-02-08

1 Answers 1

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Okay, so there's a 1/3$ chance the marble transferred from urn 1 is red.

If it's red, then urn 2 has one white marble and six red marbles, so there is a $6/7$ probability of drawing red.

If the marble transferred is white (which happens $2/3$ of the time) then urn 2 has two white marbles and five red marbles, so the probability of drawing red is $5/7.$

Thus the total probability of drawing red is $$ \frac{1}{3}\frac{6}{7} + \frac{2}{3}\frac{5}{7} = \frac{16}{21}$$

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    That seems right. I appreciate your help. I posted another question on lie detector test and I will provide my answers shortly. If you can please look at it that will be great. – 2017-02-08