Question 1171227: A bag has 4 red, 2 green, and 5 orange balls. If a ball is randomly selected from the bag, what is the theoretical probability that it is the color specified in parts (a) through (d).
a) green
b) not green
c) orange
d) not orange
I believe the probabilities to be 4/11, 2/11, and 5/11, but I'm not sure if that's right and I'm not sure if that's how to solve the question if it's theoretical probability. Thank you.
Found 2 solutions by Solver92311, math_tutor2020:
Answer by Solver92311(821)   (Show Source):
Answer by math_tutor2020(3817)  (Show Source):
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We have 4 red, 2 green and 5 orange.
This leads to a total of 4+2+5 = 6+5 = 11.
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Part (a)
There are 2 green out of 11 total. The probability of getting green is 2/11.
Answer: 2/11
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Part (b)
There are 4 red and 5 orange, so 4+5 = 9 are not green.
The probability of not green is 9/11.
Note how 2/11 from part (a) adds to 9/11 to get 11/11 = 1.
So you could do 1 - (2/11) = 9/11.
Answer: 9/11
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Part (c)
There are 5 orange out of 11 total. So we get 5/11 as the theoretical probability of randomly selecting an orange ball.
Answer: 5/11
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Part (d)
P(orange) = probability of orange
P(not orange) = probability of not orange
P(orange)+P(not orange) = 1
P(not orange) = 1 - P(orange)
P(not orange) = 1 - (5/11)
P(not orange) = (11/11) - (5/11)
P(not orange) = (11-5)/11
P(not orange) = 6/11
Or we could note there are 4 red and 2 green, so 4+2 = 6 non orange balls out of 11 total.
This probability is complementary to the result from part (c).
Answer: 6/11
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