Question 1192045: A box contains 20 balls that are exactly alike except that 10 are red, 6 are blue, and 4 are black. (a) If a single ball is drawn from a box, what is the probability tat it is red? (b) If a single ball is drawn from the box, what is the probability that it is blue or black? (c) If instead two balls are drawn from the box at the same time, what would be the probability that both are black? both are non green?
Answer by math_tutor2020(3817)   (Show Source):
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Answers:
a) 1/2
b) 1/2
c) 3/95 of getting two black; 100% of getting two non-green
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Explanation:
Part (a)
There are 10 red out of 20 total.
Therefore, 10/20 = 1/2 is the probability of selecting red. It happens half the time.
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Part (b)
We get the same answer as part (a) because there are 6 blue + 4 black = 10 that are either blue or black.
This leads to 10/20 = 1/2 like before.
Half of the balls are red, the other half are either blue or black.
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Part (c)
I'll assume non-replacement is happening here. Meaning that whatever is selected is not put back.
4 black, 20 total
4/20 = 1/5 = probability of selecting one black
3/19 = probability of selecting another black
Note the numerator and denominator subtracted by 1.
(4/20)*(3/19) = 12/380 = 3/95 is the probability of selecting 2 black in a row, assuming non replacement.
The probability of getting two non-green is 100% because there are no green balls in the box. The only colors we have are red, blue, and black.
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