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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 of heads . 1-3: Pick two real numbers at random between zero and one .
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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.
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.
Prove the theorems • 1-5: P() = 0 • 1-6: A B => P(A) P(B) • 1-7: P(A) 1 • 1-8: P(Ac) = 1 – P(A) • 1-9: P(AB) = P(A) + P(B) – P(AB)
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 2S)?
1-11: Find the probability • A lifetime of a computer chip is measured. We find that the probability that its lifetime exceeds t is given by • , >0 • Find the probability that the lifetime of a chip falls into (r, s]
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]
Drug-addicted D+ D- young actors Y 10 20 30 old actors 10 50 60 O 20 90 70 1-13: Bayes’ rule (1/2) • Young actors are more drug-addictive among all actors? • Sample space is young and old actors
1-13 Bayes’ rule (2/2) • P[Y|D+] = P[YD+] / P[D+] • P[Y] = 30/90 • P[YD+] = P[D+|Y] * P[Y] = 10/30 * 30/90 • P[D+] = P[D+|Y] * P[Y] + P[D+|O] * P[O] = 10/30 * 30/90 + 10/60 * 60/90 = 20/90 • P[Y] = 30/90 • P[Y|D+] = 10/90 / 20/90 = 1/2
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?
Drug-addicted D+ D- young actors Y 10 20 30 old actors ?? ?? 60 O ?? 90 ?? 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?
AC A B BC