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Chapter 3: Q109SE (page 204)

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

An experiment results in one of three mutually exclusive events, A, B, or C. It is known that P (A)= .30, P(B)= .55 , and P(C)= .15. Find each of the following probabilities:

a. P(A∪B)

b.P(A∩C)

c. P (A/B)

d. P(B∪C)

e. Are B and C independent events? Explain.

Three fair coins are tossed and either heads(H) or tails(T) are observed for each coin.

  1. List the sample points for the experiment.
  2. Assign probabilities to the sample points.
  3. Determine the probability of observing each of the following events:

A= {Three heads are observed}

B= {Exactly two heads are observed}

C= {At least two heads are observed}

Likelihood of a tax return audit. At the beginning of each year, the Internal Revenue Service (IRS) releases information on the likelihood of a tax return being audited. In 2013, the IRS audited 1,242,479 individual tax returns from 145,236,429 filed returns; also, the IRS audited 25,905 returns from the total of 1,924,887 corporation returns filed (IRS 2014 Data Book).

a. Suppose an individual tax return is randomly selected. What is the probability that the IRS audited the return?

b. Refer to part a. Determine the probability that the IRS did not audit an individual return.

c. Suppose a corporation tax return is randomly selected. What is the probability that the IRS audited the return?

d. Refer to part c. Determine the probability that the IRS did not audit a corporation's return.

Exit poll candidates and voters.In an exit poll, 45% of voters said that the main issue affecting their choice of candidates was the economy, 35% said national security, and the remaining 20% were not sure. Suppose we select one of the voters who participated in the exit poll at random and ask for the main issue affecting his or her choice of candidates.

  1. List the sample points for this experiment.
  2. Assign reasonable probabilities to the sample points.
  3. What is the probability that the main issue affecting randomly selected voters’ choice was either the economy or national security?

Consider the experiment depicted by the Venn diagram, with the sample space S containing five sample points. The sample points are assigned the following probabilities:

P (E1) = .20, P (E2) = .30, P (E3)= .30, P (E4) = .10, P (E5) = .10.

a. Calculate P (A), P (B), and P (A∩B).

b. Suppose we know that event A has occurred, so that the reduced sample space consists of the three sample points in A—namely, E1, E2, and E3. Use the formula for conditional probability to adjust the probabilities of these three sample points for the knowledge that A has occurred [i.e., P (Ei/A)]. Verify that the conditional probabilities are in the same proportion to one another as the original sample point probabilities.

c. Calculate the conditional probabilityP (E1/A)in two ways: (1) Add the adjusted (conditional) probabilities of the sample points in the intersection A∩B, as these represent the event that B occurs given that A has occurred; (2) use the formula for conditional probability:

P (B/A) =P (A∩B)P (A)

Verify that the two methods yield the same result.

d. Are events A and B independent? Why or why not?
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