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PARADOXES IN MATHEMATICS

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WHAT IS A PARADOX? A paradox is an argument that produces an inconsistency within logic. Most logical paradoxes are known to be invalid arguments however there are exceptions.

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A few TYPES OF PARADOXES SELF-REFERENCE CONTRADICTION VICIOUS CIRCULARITY ALSO KNOWN AS INFINITE REGRESS

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SOME FAMILIAR EXAMPLES

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SELF-REFERENCE PARADOX THE BARBER PARADOX IS THE ANSWER TO THIS QUESTION NO?

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CONTRADICTION PARADOX THIS STATEMENT IS FALSE NOTHING IS IMPOSSIBLE

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VICIOUS CIRCULARITY PARADOX “THE FOLLOWING SENTENCE IS TRUE” “THE PREVIOUS SENTENCE IS FALSE” WHAT HAPPENS WHEN PINOCCHIO SAYS “MY NOSE WILL GROW NOW?”

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Why use paradoxes in math? They are good in promoting critical thinking. Some paradoxes have revealed errors in definitions assumed to be rigorous, and have caused axioms of mathematics and logic to be re-examined. One example is Russell’s paradox.

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TRHEE FAMOUS PARADOXES IN PROBABILITY THE THREE DOGS PARADOX BERTRAND’S BOX PARADOX THE TWO DOGS PARADOX

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THE THREE DOGS PARADOX There are three dogs, equally likely to be male or female. If one of them is male, what is the probability that all of them are male? are male?

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PROBABILITY OF EACH DOG to be male D1= 1/2 (male or female) D2= 1/2 (male or female) D3= 1/2 (male or female) But we are told that one of the dogs is male, say D1, so we are left with D2 and D3 to be males. so now we only have D1 and D2 unknown. D2= 1/2 D3=1/2

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From here, can we say that since the probability D2 and D3 to be both males is (1/2)(1/2)= 1/4 therefore the answer is 1/4?

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ANOTHER APPROACH Let S be a sample space for the three dogs, then. S={MMM, MMF, MFM,FMM, FFM, FMF, MFF} Note: {FFF} possibility is ruled out since we know that at least one of the dogs is male. Looking at the sample space we created. What is the probability of having all three dogs be male?

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Conclusion for the three dogs paradox 1. The probability for 3 dogs to be male is (1/8) 2. The probability for 2 dogs to be male is (1/4) 3. The probability for 3 dogs to be male given that one is male is (1/7). Hence the paradox.

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BERTRAND’S BOX PARADOX There are 3 jewelry boxes. Each box has 2 drawers with one gemstone in each drawer as follows. 1. B1- Diamond, Diamond 2. B2- Diamond, Emerald 3. B3- Emerald, Emerald One box is chosen at random and one of its drawers opened. If a diamond is found what is the probability that the other drawer of this box has the other diamond?

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ATTEMPT TO THE SOLUTION Let B=Box, D=Diamond, E= Emerald B1- D,D B2- D,E B3- E,E Since we are looking for the P(D,D) and we know we found one diamond, the B-(E,E) option is ruled out. We have left B1 and B2 therefore it seems like we have a 50% chance on each of the boxes because we could easily have the (D,E) or the (D,D), right?

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A DIFFERENT ATTEMPT Let’s assign individual probabilities to each jewelry box. Using probability notation to get a diamond on each box. IIf we have the (D,D) box the P(D,D)= 1 IIf we have the (D,E) box the P(D,E)=1/2 if we have the (E,E) box the P(D,E) =0 If we want the probability of the 2 diamonds on the same box, we must use the diamond path by Bayes theorem. P(D,D)= [P(D,D)*P(B1)] / [P(D,D)*P(B1)+P(D,E)*P(B2)+P(E,E)*P(B3)]

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Picture approach Box Diamond Found Box EMERALD

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Probability finding a second diamond Found P(0) P(1/3) P(0) P(1/3)

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LAST BUT NOT LEAST

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The two dogs paradox There are two dogs. One of them is male and was born on Sunday. What is the probability that the other dog is male? Assume that male and female dogs are equally likely to be born.

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We have two dogs, and we know that one is a male. We know they are equally likely to be born male and female. We have one more dog, so this dog can be male or female which give us the answer that the probability that this dog is male is.5 = 1/2. The problem also mentions that the dog was born on Sunday. Should that change anything? Let’s see.

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A MORE DETAILED LOOK AT THE PROBLEM We have two genders, Male and Female. We have Seven days of the week, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday. If we create a sample space for the two dogs we would have something like: S={SMSM, SMSF, SFSM,SMMF,SFMM,….,SFSF} In total our sample space would be 14^(2)=196 paired by gender/day. But this configuration includes all possibilities, so we need to take some out.

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Lets take out the samples that do not feature a male born on Sunday. FFemale/Female=49 NNon Sunday male/Non Sunday male=36 FFemale/Non Sunday male =42 NNon Sunday male/Female =42 49+36+42+42= 169 This gives us 169 outcomes that we can rule out since these don’t have a male born on Sunday. We have 196-169= 27 samples with a male born on Sunday. These could be our possibilities.

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Now we can count the samples with one male dog born on Sunday. SSunday male/Non Sunday male =6 NNon Sunday male/Sunday male=6 SSunday male/Female =7 FFemale/Sunday male=7 SSunday male/Sunday male=1 The probability that the second dog is male is produced by counting the days in which the two dogs where male.

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The tree samples in which the two dogs were male given that at least one was born on Sunday are: SSunday male/Non Sunday male =6 NNon Sunday male/Sunday male=6 SSunday male/Sunday male=1 We have 6+6+1=13 Since our possibilities were 27 we get the final result (13/27)< (1/2) Hence it matters that a dog is born on Sunday, Hence the paradox.

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