Ch.2: Probability

Chapter 2 of the book "Probability for the Enthusiastic Beginner" by David Morin (2016) Definition of probability The rules of probability Examples Four classic problems Bayes' theorem Stirling's formula Problems

167 flashcards · Mathematics · by lakesare

Part of Textbooks › Probability for the Enthusiastic Beginner (David Morin) · 7 courses

  1. [Definition of probability] How does Morin define the probability ppp of an event?

    In a very large number of identical trials of a process, the event occurs in a fraction ppp of the trials, on average. Trials: flipping a coin, rolling a die, picking a ball from a box (with replacement).Example: a 5 on a 6-sided die has probability 1/61/61/6 , so in many rolls a 5 comes up in about 1/61/61/6 of them.

  2. [Definition of probability] A box contains 3 red balls and 7 blue balls. Why is the probability of picking a red ball 3/103/103/10 ?

    Each of the 10 balls is picked 1/101/101/10 of the time, on average, and 3 of the balls are red. This assumes that each ball is equally likely to be picked. Likewise a blue ball is picked 7/107/107/10 of the time.

  3. [Definition of probability] If all possible outcomes are equally likely, what is the probability ppp of an event?

    p=number of desired outcomestotal number of possible outcomes\boldsymbol{p = \dfrac{\text{number of desired outcomes}}{\text{total number of possible outcomes}}}p=total number of possible outcomesnumber of desired outcomes​ So computing a probability reduces to counting (Chapter 1).Example: rolling an even number on a die: desired outcomes 2, 4, 6, so p=3/6=1/2p = 3/6 = 1/2p=3/6=1/2 .

  4. [Definition of probability] p=number of desired outcomestotal number of possible outcomesp = \dfrac{\text{number of desired outcomes}}{\text{total number of possible outcomes}}p=total number of possible outcomesnumber of desired outcomes​ Under which assumption does this formula hold?

    All possible outcomes are equally likely. Morin: this usually isn't a restriction in his setups, e.g. permutations and subgroups (Chapter 1) have equally likely outcomes.Example: a coin, a fair die, a ball picked at random from a box.

  5. [Definition of probability] Do "probability", "chance" and "odds" mean the same thing?

    "Probability" and "chance" do; "odds" means something different. Example: "a 40% chance that the bus is late" means "a 40% probability that the bus is late".

  6. [Definition of probability] Definition of probability: in a very large number of identical trials, an event of probability ppp occurs in a fraction ppp of the trials, on average.Why can't the words "on average" be dropped, i.e. why can't the event occur in exactly a fraction ppp of the trials?

    The number of times an event occurs is an integer, and a fraction ppp of the trials usually isn't. Example: with one roll of a die, "exactly" would mean getting 1/61/61/6 of a 1; the number of 1s is 0 or 1.

  7. [Definition of probability] A die is rolled 6 times. Does each number come up exactly (1/6)⋅6=1(1/6) \cdot 6 = 1(1/6)⋅6=1 time?

    No, not necessarily: more likely some numbers appear more than once and others not at all. This is why the definition of probability speaks of a "very large number" of trials and of the fraction "on average".

  8. [Definition of probability] Why does the definition of probability require a very large number of trials?

    For a small number of trials, the fractions of the time that the events occur usually don't look like their probabilities. The larger the number of trials, the closer the fractions generally are to the probabilities (explained in Chapter 5).Example: roll a die a huge number of times, and each number appears in approximately 1/61/61/6 of the rolls.

  9. [Definition of probability] A fair coin is flipped 100 times. About how likely is it to get 49, 50 or 51 Heads?

    About 24%. So with 100 flips the fraction of Heads has a decent chance of deviating moderately from 1/21/21/2 .

  10. [Definition of probability] A fair coin is flipped 100,000 times. How likely is it that the fraction of Heads lies between 49% and 51%?

    99.999999975%, so the fraction of Heads almost surely stays close to 1/21/21/2 . With 10910^9109 flips, the fraction lies between 49.99% and 50.01% with the same probability 99.999999975%.

  11. [The rules of probability] The probability that two events AAA and BBB both occur is called their joint probability.

    Example: for two dice, the joint probability of a 2 on the left die and a 5 on the right die is 1/361/361/36 .

  12. [The rules of probability] When are two events independent?

    When the occurrence of one doesn't affect the probability that the other occurs. Example: rolling two dice, A={a 2 on the left die}A = \{\text{a 2 on the left die}\}A={a 2 on the left die} and B={a 5 on the right die}B = \{\text{a 5 on the right die}\}B={a 5 on the right die} : BBB has probability 1/61/61/6 whatever the left die shows.

  13. [The rules of probability] One card is drawn from a 52-card deck. Can A={the card is a king}A = \{\text{the card is a king}\}A={the card is a king} and B={the same card is a heart}B = \{\text{the same card is a heart}\}B={the same card is a heart} be two independent events?

    Yes: the card has two qualities (value and suit), and each gives an event. P(heart)=1/4P(\text{heart}) = 1/4P(heart)=1/4 whether or not the card is a king, and P(king)=1/13P(\text{king}) = 1/13P(king)=1/13 whether or not the card is a heart.

  14. [The rules of probability] What is the difference between an outcome and an event?

    An outcome is a result of an experiment; an event is a set of outcomes. A given outcome belongs to many events.Example: drawing a card has 52 outcomes; the event "drawing a heart" contains 13 of them. The king of hearts belongs to "a heart", "a king", "a red card", "a value higher than 8", ...

  15. [The rules of probability] The event that consists of all possible outcomes is called the sample space.

    It occurs with probability 1. The empty set is also an event, with probability 0. Example: for one roll of a die, the sample space is {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}{1,2,3,4,5,6} .

  16. [The rules of probability] "And" rule for independent events AAA , BBB : P(A and B)= ?\ P(A \text{ and } B) = \; ? P(A and B)=?

    P(A and B)=P(A)⋅P(B)\boldsymbol{P(A \text{ and } B) = P(A) \cdot P(B)}P(A and B)=P(A)⋅P(B) P(X)P(X)P(X) denotes the probability of the event XXX .Example: a 2 on the left die and a 5 on the right die: 16⋅16=136\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}61​⋅61​=361​ , matching 1 of the 36 ordered pairs.Example: a king and a heart: 113⋅14=152\frac{1}{13} \cdot \frac{1}{4} = \frac{1}{52}131​⋅41​=521​ , the king of hearts.

  17. [The rules of probability] For independent events AAA and BBB : P(A and B)=P(A)⋅P(B)P(A \text{ and } B) = P(A) \cdot P(B)P(A and B)=P(A)⋅P(B) .How does Morin justify this rule with NNN trials ( NNN very large)?

    On average P(A)⋅NP(A) \cdot NP(A)⋅N trials have AAA ; by independence, a fraction P(B)P(B)P(B) of those also have BBB . So P(A)⋅P(B)⋅NP(A) \cdot P(B) \cdot NP(A)⋅P(B)⋅N trials have both, which is a fraction P(A)⋅P(B)P(A) \cdot P(B)P(A)⋅P(B) of the NNN trials.Example: two dice: (1/6)⋅N(1/6) \cdot N(1/6)⋅N trials have a 2 first, and 1/61/61/6 of these have a 5 second.

  18. [The rules of probability] In a probability square for independent events AAA and BBB (area = probability, total area 1), why are the dividing lines straight across the whole square?

    BBB takes the same fraction of the AAA strip and of the "not AAA " strip, so BBB occurs with the same probability whether or not AAA occurs. P(A)=0.2P(A) = 0.2P(A)=0.2 , P(B)=0.4P(B) = 0.4P(B)=0.4 : the " AAA and BBB " region has area 0.2⋅0.4=0.080.2 \cdot 0.4 = 0.080.2⋅0.4=0.08 , a pictorial proof of P(A and B)=P(A)⋅P(B)P(A \text{ and } B) = P(A) \cdot P(B)P(A and B)=P(A)⋅P(B) .

  19. [The rules of probability] When are two events dependent?

    When the occurrence of one does affect the probability that the other occurs. Example: A={it rains at 6:00}A = \{\text{it rains at 6:00}\}A={it rains at 6:00} , B={you walk to the store at 6:00}B = \{\text{you walk to the store at 6:00}\}B={you walk to the store at 6:00} : people walk less in the rain.

  20. [The rules of probability] A box contains 2 red and 3 blue balls. Two balls are picked in succession, without replacement. What is the probability that the second ball is blue, if the first is red, and if the first is blue?

    3/43/43/4 if the first is red, 2/42/42/4 if the first is blue. After a red pick, 3 blue and 1 red remain; after a blue pick, 2 blue and 2 red remain. So "red first" and "blue second" are dependent.

  21. [The rules of probability] What does the conditional probability P(B∣A)P(B|A)P(B∣A) denote, and how is it read?

    The probability that BBB occurs, given that AAA occurs. Read: "the probability of BBB , given AAA ". Example: two picks without replacement from a box with 2 red and 3 blue balls: P(blue second ∣ red first)=3/4P(\text{blue second} \,|\, \text{red first}) = 3/4P(blue second∣red first)=3/4 .

  22. [The rules of probability] General "And" rule (for any events AAA , BBB ): P(A and B)= ?\ P(A \text{ and } B) = \; ? P(A and B)=?

    P(A and B)=P(A)⋅P(B∣A)\boldsymbol{P(A \text{ and } B) = P(A) \cdot P(B|A)}P(A and B)=P(A)⋅P(B∣A) P(B∣A)P(B|A)P(B∣A) is the probability of BBB given AAA . Also written P(A∩B)P(A \cap B)P(A∩B) .Example: box with 2 red and 3 blue balls, two picks without replacement: P(Red1 and Blue2)=25⋅34=310P(\text{Red}_1 \text{ and } \text{Blue}_2) = \frac{2}{5} \cdot \frac{3}{4} = \frac{3}{10}P(Red1​ and Blue2​)=52​⋅43​=103​ .

  23. [The rules of probability] P(A and B)=P(A)⋅P(B∣A)P(A \text{ and } B) = P(A) \cdot P(B|A)P(A and B)=P(A)⋅P(B∣A) , where P(B∣A)P(B|A)P(B∣A) is the probability of BBB given AAA .Which identity with counts over NNN trials is this rule? ( nAn_AnA​ = number of trials in which AAA occurs, etc.)

    nA and BN=nAN⋅nA and BnA\boldsymbol{\dfrac{n_{A \text{ and } B}}{N} = \dfrac{n_A}{N} \cdot \dfrac{n_{A \text{ and } B}}{n_A}}NnA and B​​=NnA​​⋅nA​nA and B​​ The three fractions are P(A and B)P(A \text{ and } B)P(A and B) , P(A)P(A)P(A) and P(B∣A)P(B|A)P(B∣A) : of the trials with AAA , the fraction P(B∣A)P(B|A)P(B∣A) also have BBB .

  24. [The rules of probability] P(A and B)=P(A)⋅P(B∣A)P(A \text{ and } B) = P(A) \cdot P(B|A)P(A and B)=P(A)⋅P(B∣A) , where P(B∣A)P(B|A)P(B∣A) is the probability of BBB given AAA .What does this rule say in a Venn diagram where area is probability?

    Area of the overlap = area of AAA times the fraction of AAA that the overlap takes up. The fraction of the AAA region taken by " AAA and BBB " is P(B∣A)P(B|A)P(B∣A) .

  25. [The rules of probability] P(A and B)P(A \text{ and } B)P(A and B) can be written with a conditional probability in which two ways?

    P(A and B)=P(A)⋅P(B∣A)=P(B)⋅P(A∣B)\boldsymbol{P(A \text{ and } B) = P(A) \cdot P(B|A) = P(B) \cdot P(A|B)}P(A and B)=P(A)⋅P(B∣A)=P(B)⋅P(A∣B) Because P(A and B)=P(B and A)P(A \text{ and } B) = P(B \text{ and } A)P(A and B)=P(B and A) . These two forms are the basis of Bayes' theorem.Example: a king and a heart from one card: 113⋅14=14⋅113=152\frac{1}{13} \cdot \frac{1}{4} = \frac{1}{4} \cdot \frac{1}{13} = \frac{1}{52}131​⋅41​=41​⋅131​=521​ .

  26. [The rules of probability] A box contains red balls 1, 2 and blue balls 3, 4, 5. Two are picked without replacement. How do you get P(red first and blue second)P(\text{red first and blue second})P(red first and blue second) by listing outcomes?

    There are 5⋅4=205 \cdot 4 = 205⋅4=20 equally likely ordered pairs, and 2⋅3=62 \cdot 3 = 62⋅3=6 of them are red then blue, so 6/20=3/106/20 = 3/106/20=3/10 . 2nd\1st123451−21314151212−32425231323−43534142434−54515253545−\begin{array}{c|cc|ccc} \text{2nd} \backslash \text{1st} & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & - & 21 & 31 & 41 & 51 \\ 2 & 12 & - & 32 & 42 & 52 \\ \hline 3 & 13 & 23 & - & 43 & 53 \\ 4 & 14 & 24 & 34 & - & 54 \\ 5 & 15 & 25 & 35 & 45 & - \end{array}2nd\1st12345​1−12131415​221−23242…

  27. [The rules of probability] In a probability square for two picks without replacement from 2 red and 3 blue balls, why is the horizontal line at a different height in the two vertical strips?

    The probabilities on the second pick depend on the first pick. R1R_1R1​ strip: red second with probability 1/41/41/4 . B1B_1B1​ strip: red second with probability 2/42/42/4 . The " R1R_1R1​ and B2B_2B2​ " region has area 25⋅34=310\frac{2}{5} \cdot \frac{3}{4} = \frac{3}{10}52​⋅43​=103​ .

  28. [The rules of probability] Is it necessary to use the "And" rules P(A and B)=P(A)⋅P(B∣A)P(A \text{ and } B) = P(A) \cdot P(B|A)P(A and B)=P(A)⋅P(B∣A) to compute probabilities?

    No: you can count all the outcomes and solve from scratch; the rules are a shortcut. The shortcut matters when the numbers are large and listing all outcomes is difficult.

  29. [The rules of probability] General "And" rule: P(A and B)=P(A)⋅P(B∣A)P(A \text{ and } B) = P(A) \cdot P(B|A)P(A and B)=P(A)⋅P(B∣A) .Why is the "And" rule for independent events a special case?

    For independent events P(B∣A)=P(B)P(B|A) = P(B)P(B∣A)=P(B) , so the rule becomes P(A and B)=P(A)⋅P(B)P(A \text{ and } B) = P(A) \cdot P(B)P(A and B)=P(A)⋅P(B) . Example: two dice: P(5 on right ∣ 2 on left)=P(5 on right)=1/6P(\text{5 on right} \,|\, \text{2 on left}) = P(\text{5 on right}) = 1/6P(5 on right∣2 on left)=P(5 on right)=1/6 .

  30. [The rules of probability] Two balls are picked without replacement from red balls 1, 2 and blue balls 3, 4, 5. Does P(Red1∣Blue2)P(\text{Red}_1 | \text{Blue}_2)P(Red1​∣Blue2​) (first red, given second blue) make sense, although the second pick happens later?

    Yes: P(Red1∣Blue2)=6/12=1/2P(\text{Red}_1 | \text{Blue}_2) = 6/12 = 1/2P(Red1​∣Blue2​)=6/12=1/2 , read off from the 20 equally likely ordered pairs. Of the 12 pairs with a blue second ball, 6 have a red first ball. Compare P(Red1∣Red2)=2/8=1/4P(\text{Red}_1 | \text{Red}_2) = 2/8 = 1/4P(Red1​∣Red2​)=2/8=1/4 .The second pick has no causal influence on the first; the knowledge of the second pick changes the probability of what the first pick was.

  31. [The rules of probability] Give a trivial yet extreme example of dependent events.

    AAA and "not AAA ". If AAA occurs, "not AAA " occurs with probability 0; if AAA doesn't occur, "not AAA " occurs with probability 1.

  32. [The rules of probability] Why does P(B∣A)=P(B)P(B|A) = P(B)P(B∣A)=P(B) imply P(A∣B)=P(A)P(A|B) = P(A)P(A∣B)=P(A) ?

    Equate P(A)⋅P(B∣A)=P(B)⋅P(A∣B)P(A) \cdot P(B|A) = P(B) \cdot P(A|B)P(A)⋅P(B∣A)=P(B)⋅P(A∣B) , substitute P(B∣A)=P(B)P(B|A) = P(B)P(B∣A)=P(B) and cancel P(B)P(B)P(B) . So if BBB is independent of AAA , then AAA is independent of BBB : independence has no direction.

  33. [The rules of probability] What are three equivalent conditions for events AAA and BBB to be independent?

    P(B∣A)=P(B)\boldsymbol{P(B|A) = P(B)}P(B∣A)=P(B) , P(A∣B)=P(A)\ \boldsymbol{P(A|B) = P(A)} P(A∣B)=P(A) , P(A and B)=P(A)⋅P(B)\ \boldsymbol{P(A \text{ and } B) = P(A) \cdot P(B)} P(A and B)=P(A)⋅P(B) The third is equivalent to the first because P(A and B)=P(A)⋅P(B∣A)P(A \text{ and } B) = P(A) \cdot P(B|A)P(A and B)=P(A)⋅P(B∣A) holds for all events.Example: one card, AAA = king, BBB = heart: P(king and heart)=152=113⋅14P(\text{king and heart}) = \frac{1}{52} = \frac{1}{13} \cdot \frac{1}{4}P(king…

  34. [The rules of probability] When are two events exclusive?

    When they can't both happen. Example: one die, A={a 2}A = \{\text{a 2}\}A={a 2} , B={a 5}B = \{\text{a 5}\}B={a 5} .Example: one card, A={a diamond}A = \{\text{a diamond}\}A={a diamond} , B={a heart}B = \{\text{a heart}\}B={a heart} .

  35. [The rules of probability] "Or" rule for exclusive events AAA , BBB : P(A or B)= ?\ P(A \text{ or } B) = \; ? P(A or B)=?

    P(A or B)=P(A)+P(B)\boldsymbol{P(A \text{ or } B) = P(A) + P(B)}P(A or B)=P(A)+P(B) The regions of AAA and BBB don't overlap, so the area of their union is the sum of their areas.Example: a 2 or a 5 on one die: 16+16=13\frac{1}{6} + \frac{1}{6} = \frac{1}{3}61​+61​=31​ .Example: a diamond or a heart: 14+14=12\frac{1}{4} + \frac{1}{4} = \frac{1}{2}41​+41​=21​ .

  36. [The rules of probability] "Not" rule: P(not A)= ?\ P(\text{not } A) = \; ? P(not A)=? And how does it follow from the "Or" rule for exclusive events?

    P(not A)=1−P(A)\boldsymbol{P(\text{not } A) = 1 - P(A)}P(not A)=1−P(A) AAA and "not AAA " are exclusive and every outcome belongs to one of them, so 1=P(A or (not A))=P(A)+P(not A)1 = P(A \text{ or } (\text{not } A)) = P(A) + P(\text{not } A)1=P(A or (not A))=P(A)+P(not A) .Example: not rolling a 6: 1−16=561 - \frac{1}{6} = \frac{5}{6}1−61​=65​ .

  37. [The rules of probability] When are two events nonexclusive?

    When it is possible for both to happen. Example: one die, A={an even number}A = \{\text{an even number}\}A={an even number} , B={a multiple of 3}B = \{\text{a multiple of 3}\}B={a multiple of 3} : a 6 gives both.Example: one card, A={a king}A = \{\text{a king}\}A={a king} , B={a heart}B = \{\text{a heart}\}B={a heart} : the king of hearts gives both.

  38. [The rules of probability] General "Or" rule (for any events AAA , BBB ): P(A or B)= ?\ P(A \text{ or } B) = \; ? P(A or B)=?

    P(A or B)=P(A)+P(B)−P(A and B)\boldsymbol{P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)}P(A or B)=P(A)+P(B)−P(A and B) "Or" is the inclusive or: AAA or BBB or both. Also written P(A∪B)P(A \cup B)P(A∪B) . Example: an even number or a multiple of 3 on one die: 12+13−16=23\frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{2}{3}21​+31​−61​=32​ (the numbers 2, 3, 4, 6).

  39. [The rules of probability] P(A or B)=P(A)+P(B)−P(A and B)P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)P(A or B)=P(A)+P(B)−P(A and B) Why is P(A and B)P(A \text{ and } B)P(A and B) subtracted?

    Otherwise the overlap of AAA and BBB is counted twice. An outcome in both AAA and BBB counts once for " AAA or BBB ", like any other outcome of the union. Example: two overlapping bird cages with 7 and 5 birds, 2 of them in the overlap: 7+5−2=107 + 5 - 2 = 107+5−2=10 birds.

  40. [The rules of probability] What is the probability that one card from a 52-card deck is a king or a heart?

    413\boldsymbol{\dfrac{4}{13}}134​ 113+14−152=1652\frac{1}{13} + \frac{1}{4} - \frac{1}{52} = \frac{16}{52}131​+41​−521​=5216​ : the 13 hearts plus the 3 other kings; the king of hearts is subtracted once so it isn't counted twice.

  41. [The rules of probability] Instead of P(A)+P(B)−P(A and B)P(A) + P(B) - P(A \text{ and } B)P(A)+P(B)−P(A and B) , how else can P(A or B)P(A \text{ or } B)P(A or B) be split, and why is it usually less convenient?

    P(only A)+P(only B)+P(A and B)P(\text{only } A) + P(\text{only } B) + P(A \text{ and } B)P(only A)+P(only B)+P(A and B) ; but the probabilities of only AAA and only BBB are often hard to compute. Example: bird cages with 7 and 5 birds, 2 in the overlap: 5+3+2=105 + 3 + 2 = 105+3+2=10 .

  42. [The rules of probability] General "Or" rule: P(A or B)=P(A)+P(B)−P(A and B)P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)P(A or B)=P(A)+P(B)−P(A and B) .Why is the "Or" rule for exclusive events a special case?

    For exclusive events P(A and B)=0P(A \text{ and } B) = 0P(A and B)=0 , so the rule becomes P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B)P(A or B)=P(A)+P(B) . Example: a 2 or a 5 on one die: 16+16−0=13\frac{1}{6} + \frac{1}{6} - 0 = \frac{1}{3}61​+61​−0=31​ .

  43. [The rules of probability] Which of the four combinations of (in)dependence and (non)exclusiveness are possible for two events?

    Exclusive and independent Not possible (unless one event has probability 0). Exclusive and dependent Possible. Example: A={a 2 on a die}A = \{\text{a 2 on a die}\}A={a 2 on a die} , B={a 5 on the same die}B = \{\text{a 5 on the same die}\}B={a 5 on the same die} . Nonexclusive and independent Possible. Example: A={a 2 on a die}A = \{\text{a 2 on a die}\}A={a 2 on a die} , B={a 5 on another die}B = \{\text{a 5 on another die}\}B={a 5 on another die} . Nonexclusive and dependent Possible. Example…

  44. [The rules of probability] Why can't two events AAA and BBB be both exclusive and independent (unless BBB has probability 0)?

    Exclusive gives P(B∣A)=0P(B|A) = 0P(B∣A)=0 ; independent gives P(B∣A)=P(B)P(B|A) = P(B)P(B∣A)=P(B) ; so P(B)=0P(B) = 0P(B)=0 . Equivalently: independent events have P(A and B)=P(A)⋅P(B)>0P(A \text{ and } B) = P(A) \cdot P(B) > 0P(A and B)=P(A)⋅P(B)>0 when both probabilities are nonzero, so both can happen.

  45. [The rules of probability] What follows about dependence from exclusiveness, and about exclusiveness from independence?

    Exclusive events are dependent; independent events are nonexclusive. The second statement is the contrapositive of the first. Exclusive events have the extreme value P(B∣A)=0P(B|A) = 0P(B∣A)=0 .Example: a 2 and a 5 on the same die are exclusive and dependent; a 2 and a 5 on two different dice are independent and nonexclusive.

  46. [The rules of probability] One die is rolled. A={a 2}A = \{\text{a 2}\}A={a 2} , B={an even number}B = \{\text{an even number}\}B={an even number} . Why are AAA and BBB dependent?

    P(B∣A)=1≠P(B)=1/2P(B|A) = 1 \ne P(B) = 1/2P(B∣A)=1=P(B)=1/2 (and P(A∣B)=1/3≠P(A)=1/6P(A|B) = 1/3 \ne P(A) = 1/6P(A∣B)=1/3=P(A)=1/6 ). They can both happen, so they are nonexclusive and dependent.

  47. [The rules of probability] UUU = "a person is under 6'3''", ZZZ = "the person's last name begins with Z" (independent). In the probability square, with areas a,b,c,da, b, c, da,b,c,d as shown, what are P(Z∣U)P(Z|U)P(Z∣U) and P(U∣Z)P(U|Z)P(U∣Z) ?

    P(Z∣U)=ba+b≈0\boldsymbol{P(Z|U) = \frac{b}{a+b} \approx 0}P(Z∣U)=a+bb​≈0 , P(U∣Z)=bb+c≈1\ \boldsymbol{P(U|Z) = \frac{b}{b+c} \approx 1} P(U∣Z)=b+cb​≈1 P(Z∣U)P(Z|U)P(Z∣U) : the fraction of the area below the horizontal line that lies right of the vertical line. P(U∣Z)P(U|Z)P(U∣Z) : the fraction of the area right of the vertical line that lies below the horizontal line.

  48. [The rules of probability] Is P(A∣B)=P(B∣A)P(A|B) = P(B|A)P(A∣B)=P(B∣A) in general?

    No. The order of AAA and BBB in a conditional probability matters. Confusing the two is the "prosecutor's fallacy".Example: P(human ∣ has a brain)P(\text{human} \,|\, \text{has a brain})P(human∣has a brain) is very small, but P(has a brain ∣ human)=1P(\text{has a brain} \,|\, \text{human}) = 1P(has a brain∣human)=1 .

  49. [The rules of probability] UUU and ZZZ are independent. Why is "so P(Z∣U)=P(U∣Z)P(Z|U) = P(U|Z)P(Z∣U)=P(U∣Z) " wrong, and what is correct?

    Correct: each conditional probability equals the unconditional one, P(Z∣U)=P(Z)P(Z|U) = P(Z)P(Z∣U)=P(Z) and P(U∣Z)=P(U)P(U|Z) = P(U)P(U∣Z)=P(U) , and P(Z)P(Z)P(Z) and P(U)P(U)P(U) can be very different. Example: UUU = "under 6'3''", ZZZ = "last name begins with Z": P(Z)≈0P(Z) \approx 0P(Z)≈0 and P(U)≈1P(U) \approx 1P(U)≈1 .

  50. [Examples] All possible outcomes of an experiment are equally likely. What is the probability ppp of an event?

    p=number of desired outcomestotal number of possible outcomes\boldsymbol{p = \dfrac{\text{number of desired outcomes}}{\text{total number of possible outcomes}}}p=total number of possible outcomesnumber of desired outcomes​ This formula is valid only when the outcomes are equally likely.Example: rolling an even number with one die: p=3/6=1/2p = 3/6 = 1/2p=3/6=1/2 .

  51. [Examples] What is the strategy Morin calls the "art of not" for finding P(A)P(A)P(A) ?

    Calculate P(not A)P(\text{not } A)P(not A) and use P(A)=1−P(not A)P(A) = 1 - P(\text{not } A)P(A)=1−P(not A) . Useful when "not AAA " is simpler than AAA .Example: P(at least one 6 in three rolls)=1−(5/6)3P(\text{at least one 6 in three rolls}) = 1 - (5/6)^3P(at least one 6 in three rolls)=1−(5/6)3 .

  52. [Examples] The event "not AAA " is called the complement of the event AAA .

    P(A)+P(not A)=1P(A) + P(\text{not } A) = 1P(A)+P(not A)=1 .

  53. [Examples] Why does the "art of not" ( P(A)=1−P(not A)P(A) = 1 - P(\text{not } A)P(A)=1−P(not A) ) suit questions of the form "what is the probability of at least one ...?"

    The only way to not get at least one is to get exactly zero, so one probability replaces many. "At least one" means exactly one, or two, or three, etc.: adding all those probabilities is messy or intractable.

  54. [Examples] Three dice are rolled. What is the probability of obtaining at least one 6?

    1−(5/6)3=91/216≈42%\boldsymbol{1 - (5/6)^3 = 91/216 \approx 42\%}1−(5/6)3=91/216≈42% Zero 6's needs a non-6 on each die: (5/6)3=125/216(5/6)^3 = 125/216(5/6)3=125/216 , since the three rolls are independent.

  55. [Examples] Three dice are rolled; the probability of at least one 6 is 91/21691/21691/216 .What is wrong with the reasoning "each die gives a 6 with probability 1/61/61/6 , so at least one 6 has probability 3⋅(1/6)=1/23 \cdot (1/6) = 1/23⋅(1/6)=1/2 "?

    Probabilities of events may be added only if the events are exclusive, and "6 on the first roll", "6 on the second roll", "6 on the third roll" are not. The sum counts outcomes with two or three 6's two or three times, so 1/21/21/2 is larger than the correct 91/21691/21691/216 .Example: with 12 dice the same reasoning gives 12⋅(1/6)=212 \cdot (1/6) = 212⋅(1/6)=2 , a probability larger than 1.

  56. [Examples] People randomly take seats. What are the two common ways to find the probability of a given seating?

    Count outcomes Count the desired outcomes and all outcomes, and take their ratio. Example: 3 people in a row, middle-height person in the middle seat: 2 of the 3!=63! = 63!=6 orderings, so 1/31/31/3 . Assign seats one at a time Multiply the probabilities of success at each stage (with the rules for "and" and "or"). Example: 5 people in order of height in a row of 5 seats: 15⋅14⋅13⋅12⋅11\frac15 \cdot \frac14 \cdot \frac13 \cdot \frac12 \cdot \frac1151​⋅41​⋅31​⋅21​⋅11​ . Solving both ways is a ch…

  57. [Examples] Three people randomly take three chairs in a line. What is the probability that the person of middle height gets the middle seat?

    1/3\boldsymbol{1/3}1/3 Assign the middle-height person first: each of the 3 seats has probability 1/31/31/3 .Example: with people 1, 2, 3 (tallest to shortest), the orderings 123 and 321 of the 6 have 2 in the middle: 2/6=1/32/6 = 1/32/6=1/3 .

  58. [Examples] Three people (tallest TTT , middle-height MMM , shortest) randomly take three chairs in a line.Find P(M in the middle seat)P(M \text{ in the middle seat})P(M in the middle seat) by assigning the tallest person TTT first.

    13⋅0+23⋅12=13\boldsymbol{\frac13 \cdot 0 + \frac23 \cdot \frac12 = \frac13}31​⋅0+32​⋅21​=31​ TTT is in the middle with probability 1/31/31/3 (then MMM cannot be); otherwise ( 2/32/32/3 ) MMM gets the middle seat with probability 1/21/21/2 (one of the two remaining seats).Formally: P(Mmid)=P(Tmid)P(Mmid∣Tmid)+P(Tnot mid)P(Mmid∣Tnot mid)P(M_{\text{mid}}) = P(T_{\text{mid}})P(M_{\text{mid}} | T_{\text{mid}}) + P(T_{\text{not mid}})P(M_{\text{mid}} | T_{\text{not mid}})P(Mmid​)=P(Tmid​)P(Mmid​∣Tmid…

  59. [Examples] What intuitive method does Morin recommend instead of writing out sums of products of conditional probabilities?

    Imagine a large number of trials and count how many of them succeed. Example: 600 trials of three people taking three seats, P(middle-height person M gets the middle seat)P(\text{middle-height person } M \text{ gets the middle seat})P(middle-height person M gets the middle seat) :In 13⋅600=200\frac13 \cdot 600 = 20031​⋅600=200 the tallest is in the middle (failure).Of the other 400, half (200) have MMM in the middle, so P=200/600=1/3P = 200/600 = 1/3P=200/600=1/3 .

  60. [Examples] Five people randomly take five chairs in a line. What is the probability that they sit in order of decreasing height from left to right?

    1/5!=1/120\boldsymbol{1/5! = 1/120}1/5!=1/120 Only 1 of the 5!=1205! = 1205!=120 arrangements works.One at a time: the tallest gets the leftmost seat with probability 15\frac1551​ , then the second tallest the next seat with 14\frac1441​ , and so on: 15⋅14⋅13⋅12⋅11\frac15 \cdot \frac14 \cdot \frac13 \cdot \frac12 \cdot \frac1151​⋅41​⋅31​⋅21​⋅11​ .

  61. [Examples] P(A and B)=P(A)⋅P(B∣A)P(A \text{ and } B) = P(A) \cdot P(B|A)P(A and B)=P(A)⋅P(B∣A) . What is the extension to five events, P(A and B and C and D and E)P(A \text{ and } B \text{ and } C \text{ and } D \text{ and } E)P(A and B and C and D and E) ?

    P(A)⋅P(B∣A)⋅P(C∣A,B)⋅P(D∣A,B,C)⋅P(E∣A,B,C,D)\boldsymbol{P(A) \cdot P(B|A) \cdot P(C|A, B) \cdot P(D|A, B, C) \cdot P(E|A, B, C, D)}P(A)⋅P(B∣A)⋅P(C∣A,B)⋅P(D∣A,B,C)⋅P(E∣A,B,C,D) Each factor is conditioned on all the earlier events (" A,BA, BA,B " means " AAA and BBB ").Example: five people in order of height in a row: 15⋅14⋅13⋅12⋅11=1120\frac15 \cdot \frac14 \cdot \frac13 \cdot \frac12 \cdot \frac11 = \frac{1}{120}51​⋅41​⋅31​⋅21​⋅11​=1201​ .

  62. [Examples] Five people randomly take five chairs arranged in a circle. What is the probability that they sit in order of decreasing height going clockwise (starting anywhere)?

    5/120=1/24\boldsymbol{5/120 = 1/24}5/120=1/24 5 of the 5!=1205! = 1205!=120 arrangements work: the tallest can take any of the 5 seats, and then the others' seats are determined.One at a time: the tallest can sit anywhere, then 14⋅13⋅12⋅11=124\frac14 \cdot \frac13 \cdot \frac12 \cdot \frac11 = \frac{1}{24}41​⋅31​⋅21​⋅11​=241​ .

  63. [Examples] Three girls and three boys randomly take six chairs in a line. What is the probability that the girls get the three leftmost seats?

    3! 3!6!=36720=120\boldsymbol{\dfrac{3!\,3!}{6!} = \dfrac{36}{720} = \dfrac{1}{20}}6!3!3!​=72036​=201​ 3!3!3! orders of the girls on the left times 3!3!3! orders of the boys on the right, out of 6!6!6! arrangements.One girl at a time: 36⋅25⋅14=120\frac36 \cdot \frac25 \cdot \frac14 = \frac{1}{20}63​⋅52​⋅41​=201​ .

  64. [Examples] Three girls and three boys randomly take six chairs in a line. The probability that the girls get the three leftmost seats is 3! 3!/6!3!\,3!/6!3!3!/6! .Which binomial coefficient does 3! 3!/6!3!\,3!/6!3!3!/6! suggest, and what is the solution behind it?

    1/(63)\boldsymbol{1 / \binom{6}{3}}1/(36​) : choose the set of 3 seats for the girls; the (63)=20\binom63 = 20(36​)=20 choices are equally likely and exactly one is the three leftmost seats.

  65. [Examples] A drawer has two blue and two red socks. You randomly pick two. What is the probability of a matching pair?

    1/3\boldsymbol{1/3}1/3 After the first sock, 1 of the 3 remaining socks has its color.Example: counting gives 2 matching pairs (blue-blue, red-red) out of (42)=6\binom42 = 6(24​)=6 pairs.

  66. [Examples] A drawer has four blue and two red socks. You randomly pick two. What is the probability of a matching pair?

    (42)+(22)(62)=6+115=715\boldsymbol{\dfrac{\binom42 + \binom22}{\binom62} = \dfrac{6 + 1}{15} = \dfrac{7}{15}}(26​)(24​)+(22​)​=156+1​=157​ 6 blue pairs and 1 red pair among the 15 possible pairs.One sock at a time: 46⋅35+26⋅15=1430=715\frac46 \cdot \frac35 + \frac26 \cdot \frac15 = \frac{14}{30} = \frac{7}{15}64​⋅53​+62​⋅51​=3014​=157​ (first blue then blue, or first red then red).

  67. [Examples] Six dice are rolled. What is the probability that each of the numbers 1 through 6 appears exactly once?

    6!/66=5/324≈1.5%\boldsymbol{6!/6^6 = 5/324 \approx 1.5\%}6!/66=5/324≈1.5% Of the 666^666 ordered outcomes, the successful ones are the 6!6!6! permutations of the six numbers.Rolling in succession: 1⋅56⋅46⋅36⋅26⋅161 \cdot \frac56 \cdot \frac46 \cdot \frac36 \cdot \frac26 \cdot \frac161⋅65​⋅64​⋅63​⋅62​⋅61​ (each roll must avoid the numbers already seen), the same fraction.

  68. [Examples] Six dice are rolled. What is the probability of three pairs (three different numbers that each appear twice)?

    (63)(62)(42)(22)66=20⋅15⋅6⋅166=25648≈3.9%\boldsymbol{\dfrac{\binom63 \binom62 \binom42 \binom22}{6^6} = \dfrac{20 \cdot 15 \cdot 6 \cdot 1}{6^6} = \dfrac{25}{648} \approx 3.9\%}66(36​)(26​)(24​)(22​)​=6620⋅15⋅6⋅1​=64825​≈3.9% (63)\binom63(36​) ways to pick the three numbers; then (62)\binom62(26​) ways to pick which two dice show the first number, (42)\binom42(24​) for the second, (22)\binom22(22​) for the third.Multiplying probabilities roll by roll is messy here: many different sequences lead…

  69. [Examples] A coin is flipped 5 times. What is the probability P(k)P(k)P(k) of exactly kkk Heads?

    P(k)=(5k)/25\boldsymbol{P(k) = \binom{5}{k} / 2^5}P(k)=(k5​)/25 (5k)\binom5k(k5​) ways to choose which kkk flips show Heads, out of 252^525 ordered outcomes.Example: P(0),…,P(5)=132,532,1032,1032,532,132P(0), \dots, P(5) = \frac{1}{32}, \frac{5}{32}, \frac{10}{32}, \frac{10}{32}, \frac{5}{32}, \frac{1}{32}P(0),…,P(5)=321​,325​,3210​,3210​,325​,321​ .

  70. [Examples] A coin is flipped nnn times; P(k)=(nk)/2nP(k) = \binom{n}{k}/2^nP(k)=(kn​)/2n is the probability of exactly kkk Heads.Why is P(0)+P(1)+⋯+P(n)=1P(0) + P(1) + \dots + P(n) = 1P(0)+P(1)+⋯+P(n)=1 , physically and mathematically?

    Physically: the number of Heads must be some kkk . Mathematically: ∑k(nk)=2n\sum_k \binom{n}{k} = 2^n∑k​(kn​)=2n . Example: 1+5+10+10+5+1=32=251 + 5 + 10 + 10 + 5 + 1 = 32 = 2^51+5+10+10+5+1=32=25 .

  71. [Examples] You are dealt 7 cards from a 52-card deck and form the best 5-card hand. What is the probability of being able to form a Royal flush (10, J, Q, K, A of one suit)?

    4⋅(472)(527)=4,324133,784,560≈0.0032%\boldsymbol{\dfrac{4 \cdot \binom{47}{2}}{\binom{52}{7}} = \dfrac{4{,}324}{133{,}784{,}560} \approx 0.0032\%}(752​)4⋅(247​)​=133,784,5604,324​≈0.0032% 4 choices of suit for the Royal flush, then (472)\binom{47}{2}(247​) choices of the other two cards from the remaining 47.For a 5-card hand it is 4/(525)≈0.00015%4/\binom{52}{5} \approx 0.00015\%4/(552​)≈0.00015% , exactly 21 times smaller.

  72. [Examples] In a 5-card poker hand, what is the probability of a "full house" of suits (three cards of one suit and two of another)?

    4⋅(133)⋅3⋅(132)(525)=267,6962,598,960≈10.3%\boldsymbol{\dfrac{4 \cdot \binom{13}{3} \cdot 3 \cdot \binom{13}{2}}{\binom{52}{5}} = \dfrac{267{,}696}{2{,}598{,}960} \approx 10.3\%}(552​)4⋅(313​)⋅3⋅(213​)​=2,598,960267,696​≈10.3% 4 choices for the suit appearing 3 times and (133)\binom{13}{3}(313​) for its cards; 3 choices for the suit appearing twice and (132)\binom{13}{2}(213​) for its cards.

  73. [Examples] Why is a "full house" of suits (3 cards of one suit, 2 of another) about 70 times more likely than an actual full house (3 cards of one value, 2 of another)?

    There are only 4 suits but 13 values, so it is much harder for five cards to involve only two values than only two suits. ≈10.3%\approx 10.3\%≈10.3% against 3,744/2,598,960≈0.14%3{,}744 / 2{,}598{,}960 \approx 0.14\%3,744/2,598,960≈0.14% ; the exact ratio is 71.5.

  74. [Examples] In a 5-card poker hand, what is the probability that all cards belong to at most two suits?

    (42)(265)−8(135)(525)=384,3842,598,960≈14.8%\boldsymbol{\dfrac{\binom42 \binom{26}{5} - 8 \binom{13}{5}}{\binom{52}{5}} = \dfrac{384{,}384}{2{,}598{,}960} \approx 14.8\%}(552​)(24​)(526​)−8(513​)​=2,598,960384,384​≈14.8% (42)\binom42(24​) pairs of suits, (265)\binom{26}{5}(526​) hands from the 26 cards of each pair, minus the flushes counted more than once.

  75. [Examples] Counting 5-card hands from at most two suits as (42)(265)\binom42 \binom{26}{5}(24​)(526​) ( (42)\binom42(24​) pairs of suits, then 5 of their 26 cards) overcounts.Why, and what is the correction?

    A flush (all one suit) is counted 3 times instead of once, so subtract 2(135)2 \binom{13}{5}2(513​) for each of the 4 suits: 8(135)8\binom{13}{5}8(513​) . Example: an all-hearts hand is in the heart/diamond, heart/spade and heart/club sets of hands.

  76. [Examples] Into which three types (by cards per suit) do 5-card hands from at most two suits split, and how many hands has each type?

    (5, 0): a flush 4(135)=5,1484 \binom{13}{5} = 5{,}1484(513​)=5,148 (4, 1) 4(134)⋅3(131)=111,5404 \binom{13}{4} \cdot 3 \binom{13}{1} = 111{,}5404(413​)⋅3(113​)=111,540 (3, 2): a suit full house 4(133)⋅3(132)=267,6964 \binom{13}{3} \cdot 3 \binom{13}{2} = 267{,}6964(313​)⋅3(213​)=267,696 Total 384,384384{,}384384,384 hands; the (3, 2) type is about 2/32/32/3 of them.

  77. [Four classic problems] The Birthday Problem. How many people must be in a room for the probability that at least two share a birthday to exceed 1/21/21/2 ? (365 days, no leap years.)

    23\boldsymbol{23}23 With 23 people the probability is 50.7%50.7\%50.7% ; with 22 it is 47.6%47.6\%47.6% .

  78. [Four classic problems] In the Birthday Problem, what is the probability PnnoP^{\text{no}}_nPnno​ that nnn people all have different birthdays?

    Pnno=1⋅364365⋅363365⋯365−(n−1)365\boldsymbol{P^{\text{no}}_n = 1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365} \cdots \dfrac{365 - (n-1)}{365}}Pnno​=1⋅365364​⋅365363​⋯365365−(n−1)​ Place the birthdays one at a time: the kkk -th person must avoid the k−1k - 1k−1 days already taken. This is the "art of not": Pnyes=1−PnnoP^{\text{yes}}_n = 1 - P^{\text{no}}_nPnyes​=1−Pnno​ .Example: n=2n = 2n=2 gives 364365\frac{364}{365}365364​ , the probability that a second person avoids the first one's birthda…

  79. [Four classic problems] nnn people all have different birthdays with probability 1⋅364365⋅363365⋯365−(n−1)3651 \cdot \frac{364}{365} \cdot \frac{363}{365} \cdots \frac{365 - (n-1)}{365}1⋅365364​⋅365363​⋯365365−(n−1)​ .Why is the last factor 365−(n−1)365\frac{365 - (n-1)}{365}365365−(n−1)​ and not 365−n365\frac{365 - n}{365}365365−n​ , and how can you check it?

    When the nnn -th person is placed, n−1n - 1n−1 days are taken. Check with small nnn . Example: n=2n = 2n=2 : the last factor is 364365\frac{364}{365}365364​ , correct. Morin advises such a check for any expression with a parameter like nnn .

  80. [Four classic problems] About how likely is a shared birthday among 30 people, and among 50?

    30 people: ≈70.6%\approx 70.6\%≈70.6% ; 50 people: ≈97.0%\approx 97.0\%≈97.0% n10202330506070100Pnyes11.7%41.1%50.7%70.6%97.0%99.4%99.92%99.99997%\begin{array}{c|cccccccc} n & 10 & 20 & 23 & 30 & 50 & 60 & 70 & 100 \\ \hline P^{\text{yes}}_n & 11.7\% & 41.1\% & 50.7\% & 70.6\% & 97.0\% & 99.4\% & 99.92\% & 99.99997\% \end{array}nPnyes​​1011.7%​2041.1%​2350.7%​3070.6%​5097.0%​6099.4%​7099.92%​10099.99997%​​

  81. [Four classic problems] Why do many people find the Birthday Problem's answer of 23 people unbelievable?

    They answer a different question: how many other people are needed for someone to have my birthday. No one of nnn people has a given birthday with probability (364/365)n(364/365)^n(364/365)n , which drops below 1/21/21/2 only at n=253n = 253n=253 .

  82. [Four classic problems] The Game-Show Problem (Monty Hall problem). A prize is behind one of three doors, goats behind the other two. The host, who knows where the prize is, announces that after you pick a door he will open one of the other two and reveal a goat. He does so and offers a switch to the remaining door. Should you switch?

    Yes: switching wins with probability 2/32/32/3 , staying with 1/31/31/3 .

  83. [Four classic problems] Game-Show Problem: three doors, one prize. You pick a door; the host (who knows where the prize is, and announced this) opens another door with a goat and offers a switch to the remaining door.Why does switching win with probability 2/32/32/3 ?

    You pick door 1. The equally likely cases are PGG, GPG, GGP; switching wins in GPG and GGP. (P = prize, G = goat.) In GPG and GGP the host must open the other goat door, so the unopened door has the prize. Switching loses only if your first pick had the prize, probability 1/31/31/3 .

  84. [Four classic problems] Game-Show Problem: three doors, one prize. You pick a door; the host (who knows where the prize is, and announced this) opens another door with a goat and offers a switch.What is wrong with "two doors remain and the prize was placed randomly, so each has probability 1/21/21/2 "?

    Revealing a goat breaks the symmetry between the two remaining doors. One is the door you picked; the other is one of the two you didn't pick. Two possibilities need not have equal probabilities.

  85. [Four classic problems] Game-Show Problem: three doors, one prize. You pick a door; the host (who knows where the prize is, and announced this) opens another door with a goat and offers a switch.A friend who knows the procedure enters after the goat is revealed and sees two unopened doors. For the friend, each has probability 1/21/21/2 . Why isn't it 1/21/21/2 for you too?

    You have information the friend lacks: which door you picked and which door the host chose to leave closed. Probabilities depend on information, so they can differ between people.Example: I flip a coin and see Heads without showing you: P(Heads)P(\text{Heads})P(Heads) is 1/21/21/2 for you and 1 for me.

  86. [Four classic problems] Game-Show Problem: three doors, one prize. You pick a door; the host (who knows where the prize is, and announced this) opens another door with a goat and offers a switch.What is wrong with "your door has probability 1/31/31/3 , the other door has 1/31/31/3 too, since a goat is always behind one of the other two doors, so switching doesn't matter"?

    The reveal does give information: it tells you which of the other two doors has no prize. The two probabilities 1/31/31/3 also don't add up to 1. The part "your own door wins 1/31/31/3 of the time" is correct.

  87. [Four classic problems] Game-Show Problem with 1000 doors: one prize, you pick a door, then the host (as announced) opens 998 other doors with goats. What is your chance of winning if you switch?

    999/1000\boldsymbol{999/1000}999/1000 You lose by switching only if your first pick had the prize. The host shows you which of the other 999 doors has the prize (when one of them has it).

  88. [Four classic problems] Game-Show Problem: three doors, one prize. You pick a door; the host opens another door with a goat and offers a switch.Why does it matter that the host announced beforehand that he would reveal a goat? Give the strategy for three possible host behaviours.

    Host always reveals a goat Switch (wins 2/32/32/3 ). Host opens a random other door, which happened to show a goat Switching doesn't matter ( 1/21/21/2 each). Host reveals a goat only if your door has the prize Don't switch (staying wins with certainty). Without the announcement you don't know which rule the host follows, so no strategy can be stated.

  89. [Four classic problems] How does Morin suggest settling an argument about the Game-Show Problem?

    Play the game many times with the switching and the non-switching strategy. Example: the "host" rolls a die without showing it: 1 or 2 puts the prize behind door 1, 3 or 4 behind door 2, 5 or 6 behind door 3.

  90. [Four classic problems] The Prosecutor's Fallacy. In a city of one million, 1 person in 10,000 fits a description of a criminal. Police arrest someone solely because he fits it. What is the probability that he is the criminal?

    1/1001/1001/100 , so he is innocent with probability 99%. 106/104=10010^6 / 10^4 = 100106/104=100 people fit the description, and only one of them is guilty.

  91. [Four classic problems] In a city of one million, 1 person in 10,000 fits a description of a criminal; a man is arrested only because he fits it.The prosecutor says: (1) only 1 in 10,000 fits the description; (2) so an innocent person very unlikely fits it; (3) so the defendant is very unlikely innocent. Where is the error?

    Statement 3 does not follow from 2: it confuses P(D∣I)P(D|I)P(D∣I) with P(I∣D)P(I|D)P(I∣D) (D = fits description, I = innocent). Statements 1 and 2 are true: P(D∣I)=99/999,999≈1/10,000P(D|I) = 99/999{,}999 \approx 1/10{,}000P(D∣I)=99/999,999≈1/10,000 . But P(I∣D)=99/100P(I|D) = 99/100P(I∣D)=99/100 .

  92. [Four classic problems] In a city of one million, 1 person in 10,000 fits a description; exactly one person is guilty, and he fits it. D = fits description, I = innocent.Why is P(D∣I)P(D|I)P(D∣I) tiny while P(I∣D)P(I|D)P(I∣D) is close to 1?

    Both have the same numerator BBB (innocent and fitting), but different denominators: P(D∣I)=BB+AP(D|I) = \frac{B}{B + A}P(D∣I)=B+AB​ , P(I∣D)=BB+CP(I|D) = \frac{B}{B + C}P(I∣D)=B+CB​ . A=999,900A = 999{,}900A=999,900 innocent people not fitting, B=99B = 99B=99 innocent fitting, C=1C = 1C=1 guilty fitting: P(D∣I)=99999,999P(D|I) = \frac{99}{999{,}999}P(D∣I)=999,99999​ , P(I∣D)=99100P(I|D) = \frac{99}{100}P(I∣D)=10099​ .

  93. [Four classic problems] In the Boy/Girl Problem, which assumptions of general randomness does Morin make?

    Every child is a boy or a girl with probability 1/21/21/2 each, independently, and the parent you meet and any choice the parent makes are random. So nothing is special about the person you meet, and there are no correlations such as identical twins.

  94. [Four classic problems] Boy/Girl Problem (a). A random person says: "I have two children. At least one of them is a boy." What is the probability that the other child is also a boy?

    1/3\boldsymbol{1/3}1/3 Families BB, BG, GB, GG (older child first) are equally likely. "At least one boy" rules out only GG, and of BB, BG, GB only BB works.Example: of 1000 two-child parents, 250 are GG and are discarded; of the other 750, 250 are BB: 250/750250/750250/750 .

  95. [Four classic problems] For two-child families, why are there four equally likely possibilities (BB, BG, GB, GG) and not three (BB, BG, GG)?

    The older and the younger child are each a boy or a girl with probability 1/21/21/2 , so BG (older boy) and GB (older girl) are different cases. Example: like HH, HT, TH, TT for two coin flips, where one Head and one Tail has probability 1/21/21/2 .

  96. [Four classic problems] Boy/Girl Problem (b). A random person says: "I have two children. The older one is a boy." What is the probability that the other child is also a boy?

    1/2\boldsymbol{1/2}1/2 Of BB, BG, GB, GG (older child first), only BB and BG remain, and BB is one of them.

  97. [Four classic problems] Boy/Girl Problem (c). A random person says: "I have two children, one of whom is this boy standing next to me." What is the probability that the other child is also a boy?

    1/2\boldsymbol{1/2}1/2 8 equally likely cases: 4 families (older child first) times which child is out on the walk (bold). The 4 boxed cases have a boy present; in 2 of them the other child is a boy.

  98. [Four classic problems] Two children. (a) "At least one is a boy": other is a boy with probability 1/31/31/3 . (b) "The older one is a boy" and (c) "one of them is this boy here": 1/21/21/2 .What quick argument gives 1/21/21/2 in (b) and (c), and why does it fail in (a)?

    (b) and (c) are statements about one particular child, so they say nothing about the other child, who is a boy with probability 1/21/21/2 . In (a) the statement is about the two children together, so "the other child" is not even well defined.

  99. [Four classic problems] Two children; the 8 equally likely cases are BB, BG, GB, GG (older first) times which child is present. (a) "At least one is a boy" gives 1/31/31/3 ; (c) "one of them is this boy here" gives 1/21/21/2 .Which cases does adding "and here he is" to (a) rule out, and why does that raise the probability?

    It rules out the BG and GB cases in which the girl is the child present. (a) leaves 6 cases, 2 with the other child a boy; (c) leaves 4 cases, 2 with the other child a boy.

  100. [Four classic problems] In the Boy/Girl Problem, how does the answer to (a) "at least one is a boy" change if a parent who has a girl always says "at least one is a girl", and says "at least one is a boy" only otherwise?

    It becomes 1: only BB parents make the "boy" statement. To avoid this, ask: "Do you have exactly two children? If so, is at least one a boy?"