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1 probability. Specify Sample Space 1-1: Toss a coin two times and note the sequence of heads and tails. 1-2: Toss a coin three times and note the number.

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Presentation on theme: "1 probability. Specify Sample Space 1-1: Toss a coin two times and note the sequence of heads and tails. 1-2: Toss a coin three times and note the number."— Presentation transcript:

1 1 probability

2 Specify Sample Space 1-1: Toss a coin two times and note the sequence of heads and tails. 1-2: Toss a coin three times and note the number of heads. 1-3: Pick two real numbers at random between zero and one. 1-4: Pick a real number X at random between zero and one, then pick a number Y at random between zero and X.

3 HW1-1 (to be posted) Three systems are depicted, each consisting of 3 unreliable components. The series system works if and only if (abbreviated as iff) all components work; the parallel system works iff at least one of the components works; and the 2- out-of-3 system works iff at least 2 out of 3 components work. Find the event that each system is functioning.

4 Prove the theorems 1-5: P(  ) = 0 1-6: A  B => P(A)  P(B) 1-7: P(A)  1 1-8: P(A c ) = 1 – P(A) 1-9: P(A  B) = P(A) + P(B) – P(A  B)

5 1-10: Assign the probability Probability: The random experiment is to throw a fair die. How can we define sample space S, and probability law P to an arbitrary event E (that belongs to 2 S )?

6 1-11: Find the probability

7 1-12: Conditional Prob. An urn contains two black balls, numbered 1 and 2, and two white balls, numbered 3 and 4. S = {(1,b), (2,b), (3,w), (4,w)} A: black balls are selected, B: even-numbered balls are selected C: number of selected ball is greater than 2 Assuming that the four outcomes are equally likely, find P[A|B] and P[A|C]

8 1-13: Bayes’ rule (1/2) Young actors are more drug-addictive among all actors? Sample space is young and old actors Drug-addicted young actors old actors Y O D+D-

9 1-13 Bayes’ rule (2/2) P[Y|D+] = P[Y  D+] / P[D+] P[Y] = 30/90 P[Y  D+] = P[D+|Y] * P[Y] = 10/30 * 30/90 P[D+] = P[D+|Y] * P[Y] + P[D+|O] * P[O] = 10/30 * 30/ /60 * 60/90 = 20/90 P[Y] = 30/90 P[Y|D+] = 10/90 / 20/90 = 1/2

10 HW 1-2 (to be posted) Suppose a drug test is 99% true positive and 99% true negative results. Suppose that 0.5% of people are users of the drug. If a guy tests positive, what is the probability he is a real drug user?

11 HW 1-3 (to be posted) Go back to young actor drug problem. Which values are to be in the blanks if we want to conclude that young actors are not more drug- addictive? Drug-addicted young actors ?? ?? 90 old actors Y O D+D-

12 A ACAC BCBC B


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