SOLUTION: A bag contains 12 jellybeans 4 black 3 yellow 3 green 2 red if u draw 1 jellybean what is the probability of drawing:
a red then a green(replaced)
a red then a green (not replace
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-> SOLUTION: A bag contains 12 jellybeans 4 black 3 yellow 3 green 2 red if u draw 1 jellybean what is the probability of drawing:
a red then a green(replaced)
a red then a green (not replace
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Question 183410: A bag contains 12 jellybeans 4 black 3 yellow 3 green 2 red if u draw 1 jellybean what is the probability of drawing:
a red then a green(replaced)
a red then a green (not replaced) Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A bag contains 12 jellybeans 4 black 3 yellow 3 green 2 red if u draw 1 jellybean what is the probability of drawing:
a red then a green(replaced)
The events are independent because the 1st marble is replaced before the
2nd marble is drawn.
If A and B are independent events, P(A and B) = P(A)*P(B)
P(red and green) = P(red)*P(green) = 2/12 * 3/12 = 6/12^2 = 6/144= 1/24
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The events are dependent when the 1st is not replaced before the 2nd is drawn.
P(A and B) = P(A)P(B|A)
a red then a green (not replaced)
P(red and green) = P(red)*P(green|red) = (2/12)*(3/11) = 1/22
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Cheers,
Stan H.
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