Probability of applying for an assembly job
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PROBLEM 4
Out of 11 people applying for an assembly job, 3 cannot do the work. Suppose two persons will be hired.
(a) How many distinct pairs are possible?
(b) In how many of the pairs will 0 or 1 people not be able to do the work?
(c) If two persons are chosen in a random manner, what is the probability that neither will be able to do the job?
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Solution Summary
PROBLEM 4
Out of 11 people applying for an assembly job, 3 cannot do the work. Suppose two persons will be hired.
(a) How many distinct pairs are possible?
(b) In how many of the pairs will 0 or 1 people not be able to do the work?
(c) If two persons are chosen in a random manner, what is the probability that neither will be able to do the job?
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Solution.
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<br>(a) Since there are 11 people, there are C(11,2)=11*10/2!=55 distinct pairs are possible.
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<br>(b) We count them separately.
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<br> First, we ...
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- BSc , Wuhan Univ. China
- MA, Shandong Univ.
Recent Feedback
- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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