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Probability I 9.07 2/10/2004 Probability and gambling • De Mere: “Which is more likely, rolling at least one 6 in 4 rolls of a single die, or rolling at least one double 6 in 24 rolls of a pair of dice?” • De Mere reasoned they should be the same: – Chance of one 6 in one roll = 1/6 – Average number in 4 rolls = 4·(1/6) = 2/3 – Chance of one double 6 in one roll = 1/36 – Average number in 24 rolls = 24·(1/36) = 2/3 • Why, then, did it seem like he lost more often with the second gamble? • He asked his friend Pascal, and Pascal & Fermat worked out the theory of probability. Class details • Reminder: HW 1 due on Friday. • HW2 is now on the web. It’s due Friday of next week. • Readings in Probability now on the web. • Reminder: Office hours today, 3-4 pm Basic definitions • Random experiment = observing the outcome of a chance event. • Elementary outcome = any possible result of the random experiment = Oi • Sample space = the set of all elementary outcomes. 1Example sample spaces Sample space for a pair of dice • • • Tossing a single coin: –{H, T} Tossing two coins: – {HH, TH, HT, TT} One roll of a single die: Each pair is an elementary outcome. Fair coin or die Properties of probabilities • For a fair coin or die, the elementary •P(Oi) ≥ 0 outcomes have equal probability – Negative probabilities are meaningless – P(H) = P(T) = 0.5 • The total probability of the sample space – P(1 spot) = P(2) = P(3) = P(4) = P(5) = P(6) = must equal 1. – If you roll the die, one of the elementary • Of course, the coin or die might not be fair outcomes must occur. P = .15 .10 .25 .15 .15 .20 2 1/63How do we decide what these probabilities are?• 1. Probability = event’s relative frequency in the population.– Look at every member of the population, and record the relative frequency of each event.– Often you simply can’t do this.• 2. Estimate probability based on the relative frequency in a (large) sample.– Not perfect, but feasible.• 3. Classical probability theory: probability based on an assumption that the game is fair.– E.G. heads and tails equally likely.– Similarly, might otherwise have a theoretical model for the probabilities.Events• An event is a set of elementary outcomes.• The probability of an event is the sum of the probabilities of the elementary outcomes.• E.G. tossing a pair of dice:Event A: Dice sum to 3 Event B: Dice sum to 64Event C: White die = 1 Event D: Black die = 1Combining events•E AND F: both event E and event F occur•E OR F: either event E occurs, or event F does, or both•NOT E: event E does not occurC OR D: W=1 OR B=15The addition rule• P(W=1) = 6/36• P(B=1) = 6/36•P(W=1 or B=1) ≠ P(W=1) + P(B=1)• P(E or F) = P(E) + P(F) – P(E and F)Subtract the region ofoverlap, so you don’tcount it twice.A or B: Dice sum to 3, or sum to 6Mutually exclusive events• Events E and F are mutually exclusive if the two events could not have both occurred.– P(E and F) = 0.– The events have no elementary outcomes in common. (There’s no overlap in our sample space diagram.)• If E and F are mutually exclusive,– P(E or F) = P(E) + P(F)• The elementary outcomes are mutually exclusive.– P(any Oi) = P(O1)+P(O2)+···P(ON) = 1Another example: P(sum=7) = ?P(sum=7) = 6/36.6Another example: P(B=5 or 4) = ?P(B=5 or 4) = 6/36 + 6/36 = 12/36.P(sum=7 or (B=5 or 4)) = ?• P(sum=7 and (B=5 or 4)) = P({2, 5}, {3, 4}) = 2/36• P(sum=7 or (B=5 or 4)) = 6/36 + 12/36 – 2/36 = 16/36P(sum=7 or (B=5 or 4))P(A or B) = P(A) + P(B) – P(A and B) = 6/36 + 12/36 – 2/36 = 16/36.De Mere revisited• Wanted to know what is the probability of getting at least one 6 in 4 tosses of a die.•P(1st=6 or 2nd=6 or 3rd=6 or 4th=6)•P(1st=6) = P(2nd=6) = P(3rd=6)=P(4th=6) = 1/6• Are these events mutually exclusive?– No, you could get a 6 on both the 1st& 2ndtosses, for example.– So De Mere was incorrect. P ≠ 4·(1/6)7The addition formula, continued• P(E or F) = P(E) + P(F) – P(E and F)• You will probably rarely use this formula except for simple cases! It gets complicated quickly if you want to compute P(E or F or G or …)• Example: P(at least one 6 in 3 tosses of a die)=?2 tosses of a die1sttoss (W)2ndtoss (B)3 tosses of a die3rd=1 3rd=2 3rd=33rd=4 3rd=5 3rd=6P(one 6 in 3 rolls) = P(1st=6) + P(2nd=6) + P(3rd=6) +…3 tosses of a die3rd=1 3rd=2 3rd=33rd=4 3rd=5 3rd=6P(one 6 in 3 rolls) = ... – P(6 in 1st& 2nd) – …83 tosses of a die3rd=1 3rd=2 3rd=33rd=4 3rd=5 3rd=6P(one 6 in 3 rolls) = ... – P(6 in 1st& 3rd) – …3 tosses of a die3rd=1 3rd=2 3rd=33rd=4 3rd=5 3rd=6P(one 6 in 3 rolls) = ... – P(6 in 2nd & 3rd) – …3 tosses of a die• P(at least one 6 in 3 tosses) = P(6 in 1st) + P(6 in 2nd) + P(6 in 3rd) –P(6 in 1st& 2nd) – P(6 in 1st& 3rd) –P(6 in 2nd& 3rd) + P(6 in 1st& 2nd& 3rd)= 1/6+1/6+1/6-1/36-1/36-1/36+1/216= 91/216• Phew... It only gets worse from here. De Mere probably doesn’t want to calculate P(at least one 6 in 4 tosses) this way. Luckily there are other ways to go about this.3 tosses of a die – Venn diagramA = 6 in 1stB = 6 in 2ndC = 6 in 3rdP(A + B + C) = P(A) + P(B) + P(C) …“or”3 tosses of a die – Venn diagram 3 tosses of a die – Venn diagram A and B=AB AB P(A + B + C) = P(A + B + C) = P(A) + P(B) + P(C) P(A) + P(B) + P(C) –P(AB)… – P(AB) – P(AC)… C 3 tosses of a die – Venn diagram 3 tosses of a die – Venn diagram P(A + B + C) = P(A + B + C) = P(A) + P(B) + P(C) P(A) + P(B) + P(C) – P(AB) – P(AC) – P(AB) – P(AC) – P(BC)… – P(BC) + P(ABC) 93 tosses of a die – Venn diagram P(A + B + C) = P(A) + P(B) + P(C) – P(AB) – P(AC) – P(BC) + P(ABC) Conditional probability • The probability that event A will occur, given that event C has already occurred. • P(A|C) • P(dice sum to 3) = P({1,2},{2,1}) = 2/36. • Suppose we have already tossed the black die, and got a 2. Given that this has already occurred, what is the probability that the dice will sum to 3? Venn diagrams, II • Just as with sample space diagrams, lack of overlap means two events are mutually exclusive. • Consider the event “A, but not A and B)” = A-AB. • Are the events …


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MIT 9 07 - Probability I

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