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Chapter 3: Q134SE (page 210)

Flawed Pentium computer chip.In October 1994, a flawwas discovered in the Pentium microchip installed in personalcomputers. The chip produced an incorrect result when dividing two numbers. Intel, the manufacturer of the Pentium chip, initially announced that such an error wouldoccur once in 9 billion divisions, or “once in every 27,000 years” for a typical user; consequently, it did not immediately offer to replace the chip.

Depending on the procedure, statistical software packages (e.g., Minitab) may perform an extremely large number of divisions to produce the required output. For heavy users of the software, 1 billion divisions over a short time frame is not unusual. Will the flawed chip be a problem for a heavy Minitab user? [Note:Two months after the flaw was discovered, Intel agreed to replace all Pentium chips free of charge.]

Short Answer

Expert verified

The probability is 0.111 and it is fairly common.

Step by step solution

01

Important formula

The formula for probability isP=favourableoutcomestotaloutcomes

02

Find that will the flawed chip be a problem for a heavy Minitab user.

Intel Corporationcommitted to replacing al Pentium chips free of charge two months after the error was detected. Then

P(U)=19=0.111

Therefore, the probability is 0.111 and it is fairly common.

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Most popular questions from this chapter

Study of why EMS workers leave the job.Refer to the Journal of Allied Health(Fall 2011) study of why emergencymedical service (EMS) workers leave the profession,Exercise 3.45 (p. 182). Recall that in a sample of 244former EMS workers, 127 were fully compensated whileon the job, 45 were partially compensated, and 72 hadnon-compensated volunteer positions. Also, the numbersof EMS workers who left because of retirement were 7 forfully compensated workers, 11 for partially compensatedworkers, and 10 for no compensated volunteers.

a.Given that the former EMS worker was fully compensatedwhile on the job, estimate the probability that theworker left the EMS profession due to retirement.

b.Given that the former EMS worker had a non-compensatedvolunteer position, estimate the probabilitythat the worker left the EMS profession due toretirement.

c.Are the events {a former EMS worker was fully compensatedon the job} and {a former EMS worker left thejob due to retirement} independent? Explain.

Blood diamonds.According to Global Research News(March 4, 2014), one-fourth of all rough diamonds producedin the world are blood diamonds. (Any diamond that is mined in a war zone—often by children—to finance a warlord’s activity, an insurgency, or an invading army’s effort is considered a blood diamond.) Also, 90% of the world’s rough diamonds are processed in Surat, India, and, of these diamonds one-third are blood diamonds.

a.Find the probability that a rough diamond is not a blood diamond.

b.Find the probability that a rough diamond is processed in Surat and is a blood diamond.

Do social robots walk or roll? Refer to the International Conference on Social Robotics (Vol. 6414, 2010) study of the trend in the design of social robots, Exercise 3.10 (p. 168). Recall that in a random sample of 106 social robots, 63 were built with legs only, 20 with wheels only, 8 with both legs and wheels, and 15 with neither legs nor wheels. Use the complements rule to find the probability that a randomly selected social robot is designed with either legs or wheels.

Male nannies. In a survey conducted by the International Nanny Association (INA) and reported on the INA Web site (www.nanny.org), 4,176 nannies were placed in a job in a given year. Only 24 of the nannies placed were men. Find the probability that a randomly selected nanny placed during the last year is a male nanny (a “mannie”).

The three-dice gambling problem. According toSignificance(December 2015), the 16th-century mathematician Jerome Cardan was addicted to a gambling game involving tossing three fair dice. One outcome of interest— which Cardan called a “Fratilli”—is when any subset of the three dice sums to 3. For example, the outcome {1, 1, 1} results in 3 when you sum all three dice. Another possible outcome that results in a “Fratilli” is {1, 2, 5}, since the first two dice sum to 3. Likewise, {2, 3, 6} is a “Fratilli,” since the second die is a 3. Cardan was an excellent mathematician but calculated the probability of a “Fratilli” incorrectly as 115/216 = .532.

a. Show that the denominator of Cardan’s calculation, 216, is correct. [Hint: Knowing that there are 6 possible outcomes for each die, show that the total number of possible outcomes from tossing three fair dice is 216.]

b. One way to obtain a “Fratilli” is with the outcome {1,1, 1}. How many possible ways can this outcome be obtained?

c. Another way to obtain a “Fratilli” is with an outcome that includes at least one die with a 3. First, find the number of outcomes that do not result in a 3 on any of the dice. [Hint: If none of the dice can result in a 3, then there are only 5 possible outcomes for each die.] Now subtract this result from 216 to find the number of outcomes that include at least one 3.

d. A third way to obtain a “Fratilli” is with the outcome {1, 2, 1}, where the order of the individual die outcomes does not matter. How many possible ways can this outcome be obtained?

e. A fourth way to obtain a “Fratilli” is with the outcome {1, 2, 2}, where the order of the individual die outcomes does not matter. How many possible ways can this outcome be obtained?

f. A fifth way to obtain a “Fratilli” is with the outcome {1, 2, 4}, where the order of the individual die outcomes does not matter. How many possible ways can this outcome be obtained? [Hint:There are 3 choices for the first die, 2 for the second, and only 1 for the third.]

g. A sixth way to obtain a “Fratilli” is with the outcome {1, 2, 5}, where the order of the individual die outcomes does not matter. How many possible ways can this outcome be obtained? [See Hintfor part f.]

h. A final way to obtain a “Fratilli” is with the outcome {1, 2, 6}, where the order of the individual die outcomes does not matter. How many possible ways can this outcome be obtained? [See Hintfor part f.]

i. Sum the results for parts b–h to obtain the total number of possible “Fratilli” outcomes.

j. Compute the probability of obtaining a “Fratilli” outcome. Compare your answer with Cardan’s.
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