A sample of 1000 memory chips were selected from a production run and subjected
ID: 3228349 • Letter: A
Question
A sample of 1000 memory chips were selected from a production run and subjected to three tests. 40 of the chips have stuck-at faults (A), 50 have bridging faults (B) and 30 have open faults (C). of the 40 chips that failed test A, 33 also failed test B and 13 failed test c. only 8 chips failed both test B and C Assume these failures are representative of the whole production run. Determine the following for a chip selected at random from the full production run: a. Estimate the probability that this chip will fail either test A or C (or both). b. Estimate the probability that this chip will have some type of failure. c. Estimate the probability that this chip would pass all three tests. d. Suppose this chip fails test A. Estimate the probability that it will also fail test B, before you actually do the test. e. Suppose this chip fails test A and B. Estimate the probability that it will also fail test C, before you actually do the test. Suppose you knew that Dell had used two suppliers for power supplies in its GX300 computers, of which it had built 22,000, and that 15,000 had come from supplier An Assume that the power supply used in any particular system was randomly selected from the inventory. There are 5 Gx300's in the registrar's office. What is the probability that they all have the same power supply? (Read and think carefully!)Explanation / Answer
Answer to question# 1)
n = 1000
A = 40
B = 50
C = 30
A and B = 33
A and C = 13
B and C = 8
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Answer to part a)
As per the addition rule of probability we get
P(A or C) = P(A) + P(B) - P(A and B)
We need to find P(A or C) + P(both)
Thus we get:
P(A or C) + P(both) = P(A) + P(C)
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From the data provided above we get
P(A) = 40/1000 = 0.040
P(C) = 30/1000 = 0.030
P(A or C) + P(both A and C) = 0.040 + 0.030 = 0.070
Answer to part a) is 0.070
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Answer to part b)
P(chip will have some failure) = P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C)
P(A) = 0.04
P(B) = 50/1000 = 0.050
P(C) = 0.030
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P(A and B) = 33/1000 = 0.033
P(A and C) = 13/1000 = 0.0130
P(B and C) = 8/1000 = 0.008
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P( chip will have some failure) = 0.04 + 0.05 + 0.03 - 0.033 - 0.013 - 0.008
P( chip will have some failure) = 0.066
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Answer to part c)
Since there are only two possibilities , either a chip fails the test for which the probability is 0.066, or the chip clears all the tests. The sum of the probability of these two possibilities is 1.
P(chip passes all three tests) + P(chip will have some failure) = 1
P(chip passes all three tests) = 1 - P(chip will have some failure)
P(chip passes all three tests) = 1 - 0.066
P(chip passes all three tests) = 0.934
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Answer to part d)
P(B | A) = P(B and A) / P(A) .............[This is the formula of conditional probability]
P(B and A) = 0.033
P(A) = 0.040
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Thus P(B|A) = 0.033 / 0.040
P(B | A) = 0.825
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