Question 852604: A jar contains green wrapped candies and yellow wrapped candies. A person selects two wrapped candies without replacement. If the probability of selecting a green candy and a yellow candy is 3/8, and the probability of selecting green candy on the first draw is 11/16, find the probability of selecting the yellow candy on the second draw, given that the first candy selected was a green one. Give your answer as a decimal rounded to the hundredths place.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A person selects two wrapped candies without replacement. If the probability of selecting a green candy and a yellow candy is 3/8, and the probability of selecting green candy on the first draw is 11/16, find the probability of selecting the yellow candy on the second draw, given that the first candy selected was a green one. Give your answer as a decimal rounded to the hundredths place.
P(yellow |green) = P(green AND yellow)/P(green)
= (3/8)/(11/16)
= 6/11
= 0.55
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Cheers,
Stan H.
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Cheers,
Stan H.
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