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Homework Help: Simple probability that I can't get out

  1. Jan 30, 2010 #1

    rock.freak667

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    Well it's been about 4 years since I've done these types of questions, so I need some help in checking if I am overthinking it or if I am just straight wrong.

    I did simplify the wording a bit.

    1. The problem statement, all variables and given/known data
    a)Given 2 events, A & B, write in set notation
    1)the event that at most A or B occurs
    2) The event that exactly A or B occurs

    3. The attempt at a solution

    n=intersect, u=union

    1) P(AuB)=P(A)+P(B)-P(AuB)

    2) Not sure


    1. The problem statement, all variables and given/known data
    b)There are n people in a room, what is the probability that at least 2 have the same birthday?

    3. The attempt at a solution

    P(no 2 having the same birthday)+P(at least 2 having the same birthday)=1

    P(no 2 having the same birthday)=(365/365)*(364/365)

    P(no 3 having the same birthday)=(365/365)*(364/365)*(363/365)

    P(no 4 having the same birthday)=(365/365)*(364/365)*(363/365)*(362/365)
    .
    .
    .
    P(no 'n' having the same birthday)=(365/365)*(364/365)*(363/365)*(362/365)*(361/365)*...*(365-n+1)/365)

    It looks like the numerator should be 365! and the denominator, 365n, but I am not sure what happens to the 'n' in the numerator.


    1. The problem statement, all variables and given/known data
    c)There's an offer of a choice of 4 designs, 3 different heating systems, a garage or carport, and a patio or screened porch. How many different plans aer available?


    3. The attempt at a solution

    I think it would be 4*3*2*2 = 48

    1. The problem statement, all variables and given/known data
    d)A fair coin is tossed until a head appears for the first time. The tosses are independent.
    Give the sameple space and what is the probability that the first head appears when the toss number is odd.


    3. The attempt at a solution

    S={H,TH,TTH,TTTH,TTTTH,TTTTTH,...}

    Is that how it woud look?

    For the odd numbered toss, it would be

    P(H),P(TTH),P(TTTTH),...
    = 1/2,1/8,1/32,1/128,...

    which forms a GP with first term a=1/2 and common ratio r =1/4. So the sum to infinity is a/1-r

    =(1/2)(1-0.25)=(1/2)*(4/3)=2/3
     
  2. jcsd
  3. Jan 30, 2010 #2

    Dick

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    I think you simplified the wording a bit too much on the first one. I'm not sure I understand what e.g. "at most A or B occurs". Do you mean at most one of A or B occurs? That would mean the only events that don't count is where A and B both occur. For the birthday problem, what's the probability that NONE of the n people have the same birthday? c) and d) look ok to me.
     
  4. Jan 30, 2010 #3

    rock.freak667

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    sorry I left out the word one. It should be


    the event that at most one of A or B occurs
    The event that exactly one of A o B occurs


    Also if one person has a birthday, then the probability that the rest do not have that birthday is 364/365.
     
  5. Jan 30, 2010 #4

    Dick

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    Ok, then your whole event space, call it E, can be divided into four nonoverlapping regions, E-(AuB), A-B, B-A and AnB, right? Which combination is the answer to each part?
     
  6. Jan 30, 2010 #5

    Dick

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    Not exactly. If they all have DIFFERENT birthdays, then the first one can choose any day, the second has 364 days to choose from, the third 363, etc etc.
     
  7. Jan 30, 2010 #6

    rock.freak667

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    I think at most one of A or B would be E(AuB), and exactly one of A or B would by E(B) or E(A) ?

    Sorry I am a bit confused with it.


    Yeah, that is what I was trying with my attempt, I just didn't subtract it from 1.

    1-(365!/365n)
     
  8. Jan 31, 2010 #7

    Dick

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    The are asking for sets, not probabilities. I am thinking of E as the set of ALL events, so A and B are subsets of E. I don't know E(A) would mean. If you mean E-(AuB) for the 'at most' part, then sounds ok. Wouldn't 'exactly one' be (AuB)-(AnB)?
     
  9. Jan 31, 2010 #8

    Dick

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    If that's what you mean, then 1-365!/365^n is a pretty sloppy way to write it. It doesn't mean what you are describing.
     
  10. Jan 31, 2010 #9

    rock.freak667

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    Then would I not have to subtract the complement of of AuB? Unless the entire event is just A and B, which I don't think I can assume.



    so if (365n-365!)365n is incorrect, how do I go about doing it correctly?
     
  11. Jan 31, 2010 #10

    Dick

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    Sure, you can subtract complement of AuB. What makes you think that would change anything? On the second one, take n=1. That's 1-365!/365. 365! is MUCH larger then 365. So that expression is negative. 1-365!/365^n is not a correct expression for your answer.
     
  12. Jan 31, 2010 #11

    rock.freak667

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    Well I am seeing that I can write the equations in my first post as factorials.


    P(no 2 having the same birthday)=(365/365)*(364/365)=365!/363!3652

    P(no 3 having the same birthday)=(365/365)*(364/365)*(363/365)=365!/362!3653

    which can be extended to 'n' such that

    P(no 'n' having the same birthday)=365!/(365-n)!365n= 365Pn/365n

    So P(at least two having the same birthday)=1- 365Pn/365n

    I think this should always give a number less than 1.
     
  13. Jan 31, 2010 #12

    Dick

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    That looks better.
     
  14. Jan 31, 2010 #13

    rock.freak667

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    so it's correct then? thanks!
     
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