SOLUTION: Hi I have 3 questions that are stumping me and I'm not sure how to proceed. I see that there's at least 5 questions we can ask free of charge per day. If there's anyway you can hel
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-> SOLUTION: Hi I have 3 questions that are stumping me and I'm not sure how to proceed. I see that there's at least 5 questions we can ask free of charge per day. If there's anyway you can hel
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Question 656554: Hi I have 3 questions that are stumping me and I'm not sure how to proceed. I see that there's at least 5 questions we can ask free of charge per day. If there's anyway you can help me with these I'd really appreciate it... I've been spending hours working on this take home quiz and after skipping these questions, they are the last to finish. Thank you!!! :)
3) Use a simulation of 20 families to estimate the probability of having at least 3 girls in a
family of 5 children
4) a) Determine the number of ways that 8 girls and 2 boys can be arranged in a sequence.
b) Determine the probability of getting 8 girls and 2 boys among 10 births.
5) Water samples from 6 public swimming areas are combined for one test regarding
bacteria content. Further testing is done if the combined test fails. Past results have shown
that there is a 2% chance of an unacceptably high bacteria level. Determine the
probability that the combined sample will fail. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! 3) Use a simulation of 20 families to estimate the probability of having at least 3 girls in a family of 5 children
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Randomly generate a set of 5 digits where the selection choices are 0 and 1.
Let 0 represent girl birth; let 1 represent boy birth.
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Mske the selection 20 times. Count the number of times you get at least
3 0's. That number divided by 20 is the simulation probability.
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The theoretical probability of 3 girls in a fam of 5 is 5C3(1/2)^5 = 10/32
or 6.25/20
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4) a) Determine the number of ways that 8 girls and 2 boys can be arranged in a sequence.
Answer 10C2 = (10*9)/(1*2) = 45 ways
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b) Determine the probability of getting 8 girls and 2 boys among 10 births.
Ans: 45/2^10 = 45/1024
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5) Water samples from 6 public swimming areas are combined for one test regarding bacteria content. Further testing is done if the combined test fails. Past results have shown that there is a 2% chance of an unacceptably high bacteria level. Determine the probability that the combined sample will fail.
P(fail) = 1 - P(all samples are "good") = 1 - (0.98)^6 = 0.1142
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Cheers,
Stan H.