SOLUTION: Urn 1 contains 5 red marbles and 3 green marbles ,and Urn 2 contains 2 Red marbles and 4 green marbles,A Marble is drawn from Urn 1 and another marble is drawn from Urn 2.
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Question 1143195: Urn 1 contains 5 red marbles and 3 green marbles ,and Urn 2 contains 2 Red marbles and 4 green marbles,A Marble is drawn from Urn 1 and another marble is drawn from Urn 2.
1. Determine the probability that both are red.
2.Determine the probability that red marble is drawn from Urn 1 and Green marble is drawn from Urn 2.
3. Determine the probability that green marble is drawn from Urn 1 and red marble is drawn fron Urn 2.
4.Determine the probability that the ball drawn is ine green and one Red marble.
Found 2 solutions by greenestamps, Boreal:
Answer by greenestamps(13206) (Show Source): You can put this solution on YOUR website!
In answering these questions, you might as well get some practice by finding the probabilities of all possible outcomes and verifying that the sum of the probabilities is 1. Then you will have all you need to answer the specific questions.
In most cases, the final answers to the questions should be fractions in simplified form. However, note that when doing these calculations it is easiest NOT to reduce fractions to lowest terms....
(a) P(R,R) = (5/8)(2/6) = 10/48
(b) P(R,G) = (5/8)(4/6) = 20/48
(c) P(G,R) = (3/8)(2/6) = 6/48
(d) P(G,G) = (3/8)(4/6) = 12/48
The sum is 48/48, so the calculations should be correct.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
From the first it is 5/8 and from the second it is 2/6 or 1/3
They are independent, so the joint probability is the product, or 5/24
This is 5/8*2/3 or 5/12
3. This is 3/8*1/3=1/8
4. GR has a probability of 1/8 from above
add to that RG, which is 5/12 from above, and that sum is 13/24.
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