SOLUTION: A production line manufactures 5-gal(18.93-liter) gasoline cans to a volume tolerance of 5%. The probability of any one can being out of tolerance is 0.03. If 4 cans are selected a

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Question 328408: A production line manufactures 5-gal(18.93-liter) gasoline cans to a volume tolerance of 5%. The probability of any one can being out of tolerance is 0.03. If 4 cans are selected at random:
a)What is the probability they are all out of tolerance?
b)What is the probability of exactly two being out?
c)What is the probability that all are in tolerance?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
a) P(4 bad)=%280.03%29%280.03%29%280.03%29%280.03%29=8.1x10%5E%28-7%29
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b) P(4 good)=%280.97%29%5E4=0.8853
Look at all of the possible combinations of exactly 2 bad (G-good, B-bad)
GGBB
GBGB
GBBG
BGGB
BGBG
BBGG
There are 6 of them.
Their individual probabilities are P=%280.97%29%5E2%280.03%29%5E2=0.00084681
Since there are 6 of them,
P(exactly 2 bad)=6%280.00084681%29=0.00508
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c) P(4 good)=%280.97%29%5E4=0.8853