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Probability of getting an ace in case of a loaded die

  1. Nov 21, 2017 #1
    1. The problem statement, all variables and given/known data

    upload_2017-11-21_17-15-19.png
    2. Relevant equations


    3. The attempt at a solution
    12)a)
    a) Since both events are independent : P( 33) = 9 / 441

    b) Let’s have

    4: throwing 4

    Not 4 : not throwing 4

    P ( 4 and not throwing 4 ) = P( 4) P ( not throwing 4) = (17*4)/441


    c) Let’s consider the sample event S for the above problem:

    S : {(6,4),(4,6),(5,6),(6,5),(5,5),(6,6)}

    Probability of getting 5 and 5 under given condition is : P(55/S)

    Using Bayes’ theorem,

    P(55/S) = P(55 and S)/ P(S)


    As S contains (5,5) as a sample point, P(55 and S) = P(55)

    Is it always true that P (A and B) = P (A),when A is a subset of B ?


    P(55/S) = ## \frac { \frac {5 *5 } {21*21 } } {(2*6*4 +2*5*6 +25+36)/441 } ##

    = ## \frac { 25 } { 169 } ##


    d) I think what the question says is:

    I throw the dice n times, what is the probability that I will get an ace once?



    The dice throwing n times and getting an ace once is equivalent to that I have n boxes and I have to put one ace in any of these n boxes. The probability of getting an ace is 1/21 in one throw. So, the probability of getting one ace in n throws is n/21.


    Now, ## \frac { n } {21 } \geq ½

    \\ n \geq 11 ##
    e) S = {(2,5),(2,6), (4,5),(4,6), (6,5),(6,6)}

    P(S) = 132/441

    Is this correct ?
     
  2. jcsd
  3. Nov 21, 2017 #2
    A) Simplfiy the 9/441
    B) Good
    C) Good
    D) The probability of not getting an ace in 1 throw is 20/21. In N throws, the probability of no ace is (20/21)^N.
    E) Simplify 132/441
     
  4. Nov 21, 2017 #3
    The probability of getting not " no ace" i.e. at least one ace in N throws is:
    1- ## {\frac{20}{21}}^N##.
    Now, 1- ## {\frac{20}{21}}^N \geq 0.5
    \\ N = 15##
    But, this will tell me probabilty of getting at least one ace, not exacctly one ace.
     
  5. Nov 21, 2017 #4

    BvU

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    Hey, didn't my reply come through ? Weird. Your logic in d) is the same as claiming you have 100% chance of throwing a six in six throws with an unloaded die.

    In the problem wording 'an ace' means 'not zero aces'
     
  6. Nov 21, 2017 #5

    Ray Vickson

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    Let ##N## be the number of tosses you need to make to get your first ace. That follows a geometric distribution with "success" probability ##p = 1/21##:
    $$P(N = n) = p (1-p)^{n-1}, n = 1,2, \ldots,$$
    and
    $$P(N > n) = (1-p)^n, n=0,1,2, \ldots .$$
    In other words, probability of needing more than ##n## tosses to your first ace is ##(20/21)^n##, because the first ##n## tosses must all end in a non-ace.

    Be careful If you Google "geometric distribution", because some sources (such as http://mathworld.wolfram.com/GeometricDistribution.html ) give the "alternative" form, which is the number of tosses before your first ace. That is essentially ##N_b = N-1##, and has distribution ##P(N_b = n) = p (1-p)^n, n=0,1,2, \ldots##
     
    Last edited: Nov 21, 2017
  7. Nov 21, 2017 #6

    haruspex

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    Are you implying that @Pushoam 's response is incorrect somewhere, either in logic or result? It looks fine to me, and a bit simpler than thinking terms of throws until the first Ace.
    And it produces the same answer, yes?
     
  8. Nov 21, 2017 #7

    Ray Vickson

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    I'm not implying anything; I am just responding to his/her statement "But, this will tell me probabilty of getting at least one ace, not exactly one ace" in Post #3. The probability of getting at least one ace is exactly what is needed in this problem; the probability of getting exactly 1 ace in ##n## tosses will not answer the original question.
     
  9. Nov 21, 2017 #8

    haruspex

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    Ok, thanks for clarifying.
     
  10. Nov 23, 2017 #9
    Thanks for clarifying.

    I am not getting the quote, reply, post a new thread button. What should I do now?
    Please help me.
    Pushoam
     
  11. Nov 23, 2017 #10

    haruspex

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    Have you tried reloading the page?
     
  12. Nov 23, 2017 #11
    Yes, l did. It seems as if I have got banned.
    How can I get to know whether it is temporary or permanent?
    And why did it happen?
    I want to use my account.
    For how long will it remain banned?
    Do they not inform before baning?
    To whom should l request to get the permission?
     
  13. Nov 23, 2017 #12

    berkeman

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    I explained it to you in my PM a couple yours ago today. Do not create alternate accounts to try to get around the 10-day temporary ban. You have accumulated too many infraction points for many problem posts, and that lead to the temporary ban. Thread is closed.
     
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