SOLUTION: Two balls are drawn from an urn containing 2 white, 3 red and 4 black balls one by one without replacement. What is the probability that at least one ball is red. Answer A) 1/

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Question 328004: Two balls are drawn from an urn containing 2 white, 3 red and 4 black balls one by one without replacement. What is the probability that at least one ball is red.
Answer
A) 1/9.
B) 7/12.
C) 1/132.
D) 2/9
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
White : 2+
Red :3
Black :4
Total=9
P(at least 1 red)=P(W,R) or P(B,R) or P(R,R) or P(R,W) or P(R,B)
P(at least 1 red)=P%28W%2CR%29%2BP%28B%2CR%29%2BP%28R%2CR%29%2BP%28R%2CW%29%2BP%28R%2CB%29
.
.
Look at the two ball draws,
P%28W%29=2%2F9
P%28R%29=3%2F8
P%28W%2CR%29=6%2F72
.
.
P%28R%29=3%2F9
P%28R%2CW%29=2%2F8
P%28R%2CW%29=6%2F72
.
.
P%28B%29=4%2F9
P%28R%29=3%2F8
P%28B%2CR%29=12%2F72
.
.
P%28R%29=3%2F9
P%28B%29=4%2F8
P%28R%2CB%29=12%2F72
.
.
P%28R%29=3%2F9
P%28R%29=2%2F8
P%28R%2CR%29=6%2F72
.
.
.
P(at least 1 red)=6%2F72%2B6%2F72%2B12%2F72%2B12%2F72%2B6%2F72
P(at least 1 red)=42%2F72=highlight%287%2F12%29
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.
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We could also have looked at the probability of no red balls.
P(no reds)=P%28W%2CW%29%2BP%28B%2CB%29%2BP%28W%2CB%29%2BP%28B%2CW%29
P(no reds)=
P(no reds)=2%2F72%2B12%2F72%2B8%2F72%2B8%2F72
P(no reds)=30%2F72=5%2F12
Since
P(no reds)+P(at least 1 red)=1
P(at least 1 red)=1-P(no reds)
P(at least 1 red)=1-5%2F12
P(at least 1 red)=highlight%287%2F12%29