Question 1190339: A jar contains 25 marbles, 10 of which are black and 15 are white. 2 marbles are drawn WITHOUT REPLACEMENT. Find out
a. The probability that both marbles drawn are white
b. The probability that at least one marble is black
c. The probability that exactly one marble is white
Answer by math_tutor2020(3817)   (Show Source):
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Part A
15 white, 25 total
m = P(1st is white) = 15/25 = 3/5
n = P(2nd is white, given 1st is white) = 14/24 = 7/12
P(2 white in a row) = m*n = (3/5)*(7/12) = 21/60 = 7/20
I subtracted 1 from the numerator and denominator when going from 15/25 to 14/24. This is directly due to the fact we aren't replacing the first marble.
Answer: 7/20
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Part B
P(2 white) + P(at least one black) = 1
P(at least one black) = 1 - P(2 white)
P(at least one black) = 1 - 7/20
P(at least one black) = 20/20 - 7/20
P(at least one black) = 13/20
This works because you either get 2 white in a row, or you get at least one black marble (pick one scenario only). The two events are complementary.
Answer: 13/20
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Part C
m = P(1st is white) = 15/25 = 3/5
n = P(2nd is black, given 1st is white) = 10/24 = 5/12
m*n = P(1st is white, 2nd is black)
m*n = (3/5)*(5/12)
m*n = 15/60
m*n = 1/4
q = P(1st is black) = 10/25 = 2/5
r = P(2nd is white, given 1st is black) = 15/24 = 5/8
q*r = P(1st is black, 2nd is white)
q*r = (2/5)*(5/8)
q*r = 10/40
q*r = 1/4
P(exactly 1 white) = m*n + q*r
P(exactly 1 white) = 1/4 + 1/4
P(exactly 1 white) = 2/4
P(exactly 1 white) = 1/2
It's a coincidence that we end up with exactly 1/2.
If the initial black and white counts were other values (say 16 black and 15 white) then the answer wouldn't be 1/2.
Answer: 1/2
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