Question 1197676: 4. There are 3 green cards, 4 red cards and 5 brown cards in a bag. What is the probability
a.) of getting a red card or a brown card after a green one (w/o replacement)?
b.) that a brown card is drawn first but not replaced, and a red card follows at the second draw?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
3 green + 4 red + 5 brown = 12 total
3/12 = 1/4 represents the probability of getting a green card.
Since the card is not put back, there are 12-1 = 11 cards left.
4 red + 5 brown = 9 cards we want out of 11 left over.
9/11 represents the probability of getting either red or brown on the second selection
(1/4)*(9/11) = 9/44 is the probability of getting green first, then red or brown second, where no replacement is made.
Answer: 9/44
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Part (b)
5 brown out of 12 total
5/12 = probability of getting a brown card
12-1 = 11 cards left
There are 4 red out of 11 left over
4/11 = probability of getting a red card after the first card is not replaced
(5/12)*(4/11) = (5*4)/(12*11)
(5/12)*(4/11) = (5*4)/(4*3*11)
(5/12)*(4/11) = 5/(3*11)
(5/12)*(4/11) = 5/33
This is the probability of getting brown first, then red next, assuming no replacements are made.
Answer: 5/33
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