- Feb 27, 2003
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I have the following stat questions over probability due tomorrow, being a summer course I'm kind of tight on time to learn, I'm just posting these to make sure I have completed them correctly, especially numbers 6 and 7.
1. The Center for Child Care reports on parental status of 539 children. There are 333 married, 182 divorced, and 24 widowed. What is the probability a particular child chosen at random will have a parent who is divorced?
My solution:
182/539 = .339
2. Of 500 employees, 200 participate ina company's profit sharing plan (p), 400 have major-medical insurance coverage (m), and 200 employees participate in both programs. What is the probability that a randomly selected employee will participate in at least one of the two programs?
My solution:
((400+200)-200)/500 = 4/5
3. Sample of Dakota Worlwide employees is to be surveyed about a new pension plan. The employees are classified as follows:
Supervisors 120
Maintenance 50
Production 1460
Management 302
Secretarial 68
What is the probability that the first person chosen at random is NOT in management?
My solution:
sum of all = 2000
2000-302 = 1698
1698/2000 = .849
4. Goodyear claims that the Z-Ray tire will last 60,000 miles. However, they know that the probability that this will occur is .9. What is the probability that all four tires you purchased will last at least 60,000 miles?
My solution:
.9 - independent so it doesn't matter how many tires there are, correct?
5. 80% of vinyl material received from Vendor A is of exceptional quality while only 50% of the vinyl received from vendor B is of exceptional quality. However, the manufacturing cpacity of Vendor A is limited, and for this reason only 40% of the vinyl material purchased by our firm comes from Vendor A. The other 60% comes from Vendor B. An incoming shipment of vinyl material is inspected, and found to be of exceptional quality. What is the probability that it came from Vendor A?
My solution:
p(a|e) = (p(a)*p(e|a))/(p(a)*p(e|a)+p(b)*p(e|b))
p(a|e) = (.4*.8) / (.4*.8+.6*.5) = .2253
6. The telephone switchboard of a company is usually busy 30 percent of the time. If a customer calls at randomly selected times, what is the probability that a busy signal willbe obtained on two of three sucessive calls?
My solution:
I'm confused on this one
I drew a tree diagram for the possible scenarios do I multiply the percentage chances of happening through the ones with 2x busy calls out and add them all up?
(.3 * .3 * .3) + (.3 * .3 * .7) + (.3 * .7 * .3) + (.7 * .3 * .3) = 0.216 ?
7. Ninety students will graduate this spring. Of the 90 students, 50 are planning to attend college. Two students are picked at random. What is the probability one of the two selected students plan to attend college?
My solution:
I think its similar to the above problem
<edit removed failed attempt at tree diagram>
((5 / 9) * (4 / 9)) + ((5 / 9) * (4 / 9)) + ((5 / 9) * (5 / 9)) = 0.802469136 ?
8. A contracotr in the building-wrecking has observed that his major competitor submits bids on about 70% of the jobs on which the contractor bids. The contractor has observed that he has been awarded jobs about 40% of the time when his major competitor also submitted a bid and about 60% of the time when the competitor did not bid on the job. What is the probability that the contracotr will not be awarded the contract?
My solution:
(.7 * .6) + (.3 * .4) = 0.54
1. The Center for Child Care reports on parental status of 539 children. There are 333 married, 182 divorced, and 24 widowed. What is the probability a particular child chosen at random will have a parent who is divorced?
My solution:
182/539 = .339
2. Of 500 employees, 200 participate ina company's profit sharing plan (p), 400 have major-medical insurance coverage (m), and 200 employees participate in both programs. What is the probability that a randomly selected employee will participate in at least one of the two programs?
My solution:
((400+200)-200)/500 = 4/5
3. Sample of Dakota Worlwide employees is to be surveyed about a new pension plan. The employees are classified as follows:
Supervisors 120
Maintenance 50
Production 1460
Management 302
Secretarial 68
What is the probability that the first person chosen at random is NOT in management?
My solution:
sum of all = 2000
2000-302 = 1698
1698/2000 = .849
4. Goodyear claims that the Z-Ray tire will last 60,000 miles. However, they know that the probability that this will occur is .9. What is the probability that all four tires you purchased will last at least 60,000 miles?
My solution:
.9 - independent so it doesn't matter how many tires there are, correct?
5. 80% of vinyl material received from Vendor A is of exceptional quality while only 50% of the vinyl received from vendor B is of exceptional quality. However, the manufacturing cpacity of Vendor A is limited, and for this reason only 40% of the vinyl material purchased by our firm comes from Vendor A. The other 60% comes from Vendor B. An incoming shipment of vinyl material is inspected, and found to be of exceptional quality. What is the probability that it came from Vendor A?
My solution:
p(a|e) = (p(a)*p(e|a))/(p(a)*p(e|a)+p(b)*p(e|b))
p(a|e) = (.4*.8) / (.4*.8+.6*.5) = .2253
6. The telephone switchboard of a company is usually busy 30 percent of the time. If a customer calls at randomly selected times, what is the probability that a busy signal willbe obtained on two of three sucessive calls?
My solution:
I'm confused on this one
I drew a tree diagram for the possible scenarios do I multiply the percentage chances of happening through the ones with 2x busy calls out and add them all up?
(.3 * .3 * .3) + (.3 * .3 * .7) + (.3 * .7 * .3) + (.7 * .3 * .3) = 0.216 ?
7. Ninety students will graduate this spring. Of the 90 students, 50 are planning to attend college. Two students are picked at random. What is the probability one of the two selected students plan to attend college?
My solution:
I think its similar to the above problem
<edit removed failed attempt at tree diagram>
((5 / 9) * (4 / 9)) + ((5 / 9) * (4 / 9)) + ((5 / 9) * (5 / 9)) = 0.802469136 ?
8. A contracotr in the building-wrecking has observed that his major competitor submits bids on about 70% of the jobs on which the contractor bids. The contractor has observed that he has been awarded jobs about 40% of the time when his major competitor also submitted a bid and about 60% of the time when the competitor did not bid on the job. What is the probability that the contracotr will not be awarded the contract?
My solution:
(.7 * .6) + (.3 * .4) = 0.54
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