SOLUTION: I have a couple questions here that I do not understand how to do. Can someone please give me some guidance on how to do them? 1) A box contains blue chips and red chips. A pers

Algebra ->  Probability-and-statistics -> SOLUTION: I have a couple questions here that I do not understand how to do. Can someone please give me some guidance on how to do them? 1) A box contains blue chips and red chips. A pers      Log On


   



Question 134555: I have a couple questions here that I do not understand how to do. Can someone please give me some guidance on how to do them?
1) A box contains blue chips and red chips. A person selects two chips without replacement. If the probability of selecting a blue chip and a red chip is 32/128, and the probability of selecting a blue chip on the first draw is 5/16, find the probability of selecting the red chip on the second draw, given the first chip selected was a blue chip?
a) 1/5
b) 2/5
c)3/5
d) 4/5
2) A recent poll by the American automobile club found that 30% of those surveyed are worried about aggressive drivers on the road. If three people are selected at random, what is the probability that all three will be worried about agressive drivers on the road?
a) .900
b) 0.027
c) 0.081
d) 0.300

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
1) A box contains blue chips and red chips. A person selects two chips without replacement. If the probability of selecting a blue chip and a red chip is 32/128, and the probability of selecting a blue chip on the first draw is 5/16, find the probability of selecting the red chip on the second draw, given the first chip selected was a blue chip?
P(red on 2nd|blue on 1st) = [P(red and blue)/P(blue) = (32/128)/(5/16)= 4/5
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a) 1/5
b) 2/5
c)3/5
d) 4/5
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2) A recent poll by the American automobile club found that 30% of those surveyed are worried about aggressive drivers on the road. If three people are selected at random, what is the probability that all three will be worried about agressive drivers on the road?
P(W and W and W) = [P(W)]^3 = (0.3)^3 = 0.027
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a) .900
b) 0.027
c) 0.081
d) 0.300
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Cheers,
Stan H.