Question 1136595: Bowl A contains 4 red chips and 6 green chips. Bowl B contains 7 red chips and 3 green chips. A chip is selected at random from Bowl A and is placed in Bowl B. Subsequently, a chip is selected at random from Bowl B. What is the probability it is a red chip? Set up both a Sample Space and a probability measure.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
P(R,R) = (4/10)*(8/11) = 32/110
P(G,R) = (6/10)*(7/11) = 42/110
P(R from bowl B) = 32/110+42/110 = 74/110
(Simplify all those probabilities if required....)
Calculate the probabilities for the other outcomes in the sample space to verify that the sum of all the probabilities is 1:
P(R,G) = (4/10)*(3/11) = 12/110
P(G,G) = (6/10)*(4/11) = 24/110
Yes, we're okay: 32+42+12+24 = 110
ANSWER: The probability that the chip drawn from bowl B is red is 74/110 = 37/55.
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