The lecture on 2/8/2011 mainly focused on independence. Two events A and B are independent if P(A|B)=P(A) and are dependent otherwise. If two events are independent then they cannot be mutually exclusive. Reliability: Parallel= 1-(1-P(A))*(1-P(A2))*...(1-P(An)) Series= P(A)*P(A1)*P(An)

Janel Wasilewski February 8, 2011

HW 0: Lecture Summary

Example:

We’re popping confetti poppers for Daniel’s birthday. He will only have a happy birthday if all 21 of them pop. All poppers pop independently. Pi = probability ith popper pops Ai = event that ith popper pops The events are in series, so to find the probability of a happy birthday: Pr(A1 ∩ A2 ∩ A3 … A21) = Pr(A1)*Pr(A2)*Pr(A3) … Pr(A21) If Pr(Ai) = .95, Pr(happy b-day) = (.95)21 = .34056 Now Daniel will have a happy birthday if any of the 21 poppers pop. The events are in parallel, so the probability of a happy birthday: Pr(A1 ∪ A2 ∪ A3 … A21) = 1 – Pr(A1 ∪ A2 ∪ A3 … A21)’ Pr(happy b-day) = 1- (1 - .95)21

Definition: Events Ai, …, An are mutually independent if for every k (2 through n) and every subset of indices i1, i2, …, ik, P(Ai1 ∩ Ai2 … Aik) = P(Ai1)*P(Ai2) … P(Aik).

Example:

We want to find P(system lifetime exceeds t0). There are two different ways to do this: directly or using the compliment. We may use either, but when there are more numbers, it is more efficient to use the compliment. We have two components in parallel, each with probability .9. We group them together to find the probability of either. Direct: (.9)(.9) – (.9)(.9) = .99 Compliment: 1 – (1 - .9)2 = .99

Example:

In this problem, A1 and A2 are in parallel. A3 and A4 are in series. The group of A1 and A2 is in parallel with the group of A3 and A4. Pr(A1 ∪ A2) = .9 + .9 – (.9)(.9) = .99 Pr(A3 ∩ A4) = (.9)2 = .81 Pr(system works) = .99 + .81 – (.99)(.81) = .9981

-When in parallel, the final probability will be better than the probability of the best component. -In series, the final probability will be worse than the probability of the worst component.

Example:

Components A1 and A2 are in parallel. A3 and A4 are in series. A5 and A6 are in series. The series A3/A4 is in parallel with the series A5/A6. The parallel combination of A1/A2 is in series with the combination of A3/A4/A5/A6 and the sole component A7. Pr(A1 ∪ A2) = .9 + .9 – (.9)(.9) = .99 Pr(A3 ∩ A4) = (.9)2 = .81

Pr(A5 ∩ A6) = (.9)2 = .81 Pr(A3/A4 ∪ A5A6) = .81 +.81 – (.81)(.81) = .9639 Pr(A1/A2 ∩ A3/A4/A5/A6 ∩ A7) = (.99)(.9639)(.9) = .8588

-Mutually exclusive and exhaustive events can be thought of as a jigsaw puzzle.

Conditional probability: P(Aj | B) = [P(Aj ∩ B)] / [P(B)]

Example (rare disease):

In this example, we made a tree diagram, showing P(has disease) = .001 and P(no disease) = .999. From has disease, P(+ test) = .99 and P(- test) = .01. From no disease, P(+ test) = .02 and P(- test) = .98. We calculated the probability of having the disease and getting a positive test result to be .00099. Then we calculated the probability of not having the rare disease and getting a positive test result to be .01998. Finally, we calculated the probability of having disease given that a positive test result was achieved: P(A1 | B) = .00099 / .02097 = .047

Example:

We briefly took a look at the Project Pluto problem. We discussed the probability that the mission succeeds/Beer Cooler fails and the probability all 3 systems fail. We also discussed the conditional probabilities that Gronk did not fail, Beer Cooler failed, and Gronk and Frab both failed, given more than one system failed. To determine those conditional probabilities, we would need to make a tree diagram, having the first branch be the unconditional probability, and the next branch being the conditional probabilities.

ST371 Page 1

ST371 Page 2

ST371 Page 3

Notes for Class on February 8, 2011

Independent Events 2 events A and B are independent if P(A|B) = P(A). The 2 events, A and B, are dependent otherwise. If events A and B are independent then, P(A and B) = P(A) x P(B) If this rule is proven to be true, then all events are considered mutually independent. Systems Systems may contain subsystems that are parallel or in series of each other. Each cell in each system is considered independent of each other. The probability of the entire system working properly can be calculated from the probabilities of each cell. For Example ______ 1 _______ _______| |________ _________| |_______ 2 _______| |_______________ | | |____ 3 ___________________ 4 ____| The probability of all cells (1,2,3,4) = 0.9. Parallel The subsystem that cells 1 and 2 are in, is considered a parallel system. To calculate the probability that the system in parallel works you would use the following equation. 1 – ( 1 – (probability that component works))^(the number of components) So for the example above the probability that component 1 or 2 works is 0.9, so your equation would be 1 – ( 1 – ( 0.9 ) ) ^ (2) = 0.99 Series The subsystem that cells 3 and 4 are in, is considered a series system. To calculate the probability that the system in series works you would use the following equation. (probability of component) x (probability of next component) x ……… So for the example above the probability that component 3 or 4 works is 0.9, so your equation would be ( 0.9 ) x ( 0.9 ) = 0.81 Probability that entire system works The entire system is considered parallel so the equation used would be 1 – ( 1 – 0.99 ) x ( 1 – 0.81 ) = 0.9981 Law of Total Probability If there are a number of events, A1 A2 A3,…, that are mutually exclusive and exhaustive events, then for any other event B, it must occur with exactly one of the A events. The partitioning of B is illustrated as a “jigsaw puzzle” with event B occurring with all events A in the illustration. Bayes’ Theorem If there are a number of events, A1, A2, Ak,…., that are mutually exclusive and exhaustive events, with prior probabilities, then for any other event B for which P(B)>0, then the posterior probability of A1 given that event B has occurred is

P(Aj|B) = P(Aj and B) = P(B|Aj) P(Aj) j= 1,…..,k P(B) k Σ P(B|Ai)xP(Ai) i=1 Figure 2.12 on page 73 gives a good tree diagram example of Bayes’ Theorem.

Phillip Lin

ST371-002

Homework 0

In class recently we have covered a range of topics that, the most important

of which include conditional probability with the Bayes’ Theorem, reliability in

direction of current due to the arrangement circuits with parallel and series, and

systems and subsystems coupled with the Law of Total Probability.

The Bayes’ Theorem deals with conditional probability when there of a

hypothesis in terms of the prior probability. The Bayes’ Theorem states that the

probability of an event A given an event B depends not only on the relationship

between A and B but also the marginal probability in each event.

We covered circuits as well involving parallel and series. It shows how

different elements are arranged in a circuit depending on whether it is a parallel or

series arrangement, and measures the reliability of either set up. Elements in a

series has the current run through every element in that series, whereas with a

parallel arrangement the current can choose which path to take.

The last vital and all-inclusive thing I am listing is when we learned the Law

of Total Probability, which covers the probability including all samples separately.

The Law of Total Probability states that if two events called A and B are mutually

exclusive then the probability of event A union event B equals the probability of A

plus the probability of B.