Probability Whenever we carry out an experiment, we cannot predict - - PowerPoint PPT Presentation

ST 370 Probability and Statistics for Engineers

Probability

Whenever we carry out an experiment, we cannot predict exactly what the outcome will be. If we could, there would be no need to carry it out! We can describe all the possible outcomes, and we use probabilities to describe which outcomes are more or less likely to occur. Sometimes those probabilities are determined mathematically, but

• ften we can only estimate them from observations.

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ST 370 Probability and Statistics for Engineers

Example: Acceptance sampling A business receives a shipment of 200 electronic components, and one is chosen at random, and tested for compliance with its requirements. The two possible outcomes are “item is compliant” or “item is non-compliant”. Suppose the shipment contains 5 non-compliant items; the definition

• f “chosen at random” is that each item in the shipment has the

same chance of being chosen, namely 1 in 200, or 0.005. So the probability that the chosen item is non-compliant is 5 in 200,

• r 0.025.

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ST 370 Probability and Statistics for Engineers

Example: Camera flash Suppose that the specifications for a cell phone camera flash require it to recharge in no longer than 0.6 seconds. A phone is chosen at random from a production line and tested for compliance; let p be the probability of being in compliance. We have no simple way to calculate p, but we could estimate it by testing a random sample of phones. Note These are objective probabilities: two engineers seeing the same data would agree on the probability; the probability that the next President will be a Republican is subjective, not objective.

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ST 370 Probability and Statistics for Engineers

Probabilities from counting Recall the shipment: 200 items, 5 non-compliant. Suppose two items are chosen at random: The first is equally likely to be any of the 200; The second is equally likely to be any of the remaining 199. What is the probability that exactly one of them is non-compliant?

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ST 370 Probability and Statistics for Engineers

Either: The first is non-compliant (probability = 5/200) and the second is compliant (probability = 195/199); or The first is compliant (probability = 195/200) and the second is non-compliant (probability = 5/199). So the probability is 5 200 × 195 199 + 195 200 × 5 199 = 2 × 5 × 195 200 × 199 = 0.048995. This is an example of the hypergeometric distribution.

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ST 370 Probability and Statistics for Engineers

Conditional probability

Recall the cell phone camera flash units. Suppose the flash intensity is also required to be at least 800 watt-seconds: A = recharge time ≤ 0.6 seconds; B = intensity ≥ 800 watt-seconds. Suppose that 90% of flash units have acceptable recharge time, and 99% of those have acceptable intensity. Then the percentage of acceptable units is 0.9 × 0.99 × 100% = 89.1%.

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ST 370 Probability and Statistics for Engineers

In terms of probabilities for a randomly chosen flash unit, the probability of an acceptable recharge time is 0.9; we write P(A) = 0.9. Then given an acceptable recharge time, the probability of an acceptable intensity is 0.99; we write P(B given A) = 0.99. This is the conditional probability of B given A, written P(B|A).

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ST 370 Probability and Statistics for Engineers

Complement of an event The event that A does not happen, the complement of A, is A′ = not A = recharge time > 0.6 seconds; The probability is P(A′) = 1 − P(A) = 0.1. Often, in solving problems, P(A′) is easier to calculate than P(A), and P(A) is found from P(A) = 1 − P(A′).

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ST 370 Probability and Statistics for Engineers

Multiplication rule The flash unit is compliant if both recharge time and intensity are acceptable; the probability is P(A and B) = 0.9 × 0.99 = P(A) × P(B|A). This is written P(A ∩ B): P(A ∩ B) = P(A) × P(B|A). Consequently P(B|A) = P(A ∩ B) P(A) , which is sometimes taken as the definition of conditional probability.

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ST 370 Probability and Statistics for Engineers

Probabilistic independence Suppose that the percentage of flash units with acceptable intensity is the same 99%, regardless of whether the recharge time is acceptable: P(B) = 0.99 = P(B|A). That is, the conditional probability P(B|A) is the same as the unconditional probability P(B). Then P(A ∩ B) = P(A) × P(B|A) = P(A) × P(B). We say that A and B are independent (probabilistically, or stochastically).

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ST 370 Probability and Statistics for Engineers

It is easy to show that, when A and B are probabilistically independent, then also: A′ and B are probabilistically independent; A and B′ are probabilistically independent; A′ and B′ are probabilistically independent.

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ST 370 Probability and Statistics for Engineers

Note that non-compliance of difference aspects of the same unit are

• ften not independent.

For instance, a bad connection to a capacitor might lead to both slow recharge and low intensity. Often P(not B|not A) = P(B′|A′) > P(B′) so that A′ and B′ are (probabilistically) dependent, not independent. We might sometimes assume, however, that compliance of a cell phone flash unit is (probabilistically) independent of compliance of the touch screen, for example if they are supplied from different sources.

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ST 370 Probability and Statistics for Engineers

Total probability

Example: semiconductor manufacturing Chips are occasionally exposed to high levels of contamination (H), leading to an increased risk of failure (F): P(F|H) = 0.100, P(F|H′) = 0.005, P(H) = 0.2. What is the probability of failure, P(F)?

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ST 370 Probability and Statistics for Engineers

Failure can occur after high contamination, or not after high contamination, but not both, so P(Failure) = P(Failure and High contamination) + P(Failure and not High contamination) That is, P(F) = P(F ∩ H) + P(F ∩ H′) = P(F|H)P(H) + P(F|H′)P(H′) = 0.1 × 0.2 + 0.005 × (1 − 0.2) = 0.024.

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ST 370 Probability and Statistics for Engineers

Bayes’ Rule

Suppose that a chip is observed to fail. What is the probability that it was exposed to high level of contamination? That is, what is P(H|F)? P(H|F) = P(H ∩ F) P(F) = P(F|H)P(H) P(F) = 0.1 × 0.2 0.024 = 0.833.

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ST 370 Probability and Statistics for Engineers

Using the total probability expression for P(F), we could write P(H|F) = P(F|H)P(H) P(F|H)P(H) + P(F|H′)P(H′) which is the usual way to write Bayes’ rule. In general, more than two terms may appear in the denominator. For example, contamination could be Low (C1), Medium (C2), or High (C3), each with its own risk of subsequent failure: P(Ci|F) = P(F|Ci)P(Ci) 3

j=1 P(F|Cj)P(Cj)

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ST 370 Probability and Statistics for Engineers

Prior and posterior probabilities P(Ci) is sometimes called the prior probability of Ci, because it is known before we find out whether the chip fails. Similarly, P(Ci|F) is called the posterior probability of Ci, because it can be calculated after we find out that the chip has failed. Bayes’ rule shows how the information that the chip failed changes the probabilities of the levels of contamination: P(Ci|F) = P(Ci) × P(F|Ci) 3

j=1 P(F|Cj)P(Cj)

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ST 370 Probability and Statistics for Engineers

Odds ratio The odds on the event H that a chip is exposed to the high level of contamination are P(H) P(H′) = P(H) 1 − P(H) = 0.2 0.8 = 1 4, or 1 : 4. The posterior odds ratio, given that the chip failed, is P(H|F) P(H′|F) = P(H) P(H′) × P(F|H) P(F|H′) = 1 4 × 0.100 0.005 = 1 4 × 20,

• r 5 : 1.

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ST 370 Probability and Statistics for Engineers

That is, posterior odds = prior odds × Bayes factor, where the “Bayes factor” is P(F|H) P(F|H′). The Bayes factor is also known as the likelihood ratio, the ratio of the probabilities of what was observed (chip failure) given the two possible histories (contaminated versus uncontaminated).

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ST 370 Probability and Statistics for Engineers

A diagnostic testing example A certain population has a 1.48% prevalence of bowel cancer. The fecal occult blood screen test has 67% sensitivity and 91% specificity. A person is randomly chosen from the population, and tested. Write C for the event that the person suffers from the cancer, and T for the event that the test is positive for the presence of cancer. Then P(C) = 0.0148 (prevalence) P(T|C) = 0.67 (sensitivity) P(T|C ′) = 1 − 0.91 = 0.09 (specificity) and hence P(T) = 0.67 × 0.0148 + 0.09 × (1 − 0.0148) = 0.0986.

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ST 370 Probability and Statistics for Engineers

Positive test result If the test is positive for the presence of cancer: P(C|T) = P(T|C)P(C) P(T) = 0.67 × 0.0148 0.0986 = 0.1006 P(C ′|T) = 1 − P(C|T) = 0.8994. That is, among people who test positive, only 10% have cancer, and the other 90% are cancer-free. This is a screening test; a positive result requires further investigation to make a definite diagnosis.

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ST 370 Probability and Statistics for Engineers

Negative test result If the test is negative for the presence of cancer: P(C|T ′) = P(T ′|C)P(C) P(T ′) = (1 − 0.67) × 0.0148 1 − 0.0986 = 0.0054 P(C ′|T ′) = 1 − P(C|T) = 0.9946. That is, among people who test negative, 0.5% have cancer, and the

• ther 99.5% are cancer-free.

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