1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Binomial Dist.

  1. Jan 8, 2014 #1
    Hello all,

    I just have a question which covers binomial distribution.

    Sally is a goal shooter. Assume each attempt at scoring a goal is independent, in the long term her scoring rate has been shown as 80% (i.e. 80% success rate).

    Question:

    What's the probability, (correct to 3 decimal places) that her first 4 attempts at scoring a goal are successful, given that exactly 6 of her first 8 attempts are successful?

    I've solved the problem the following way.

    X~ Bi(8,0.8)

    Pr(X=6) = 0.8^6*0.2^2 = 0.0105

    Pr(X=4) = 0.8^4*0.2^4 = 0.0007


    Pr(X=4 | X=6) = Pr(X=4 & X=6) = Pr(X=4) = 0.0007 = 0.067
    ------------------- ---------- --------- <--------divided signs
    Pr(X=6) Pr(X=6) 0.0105

    Note:{'&' means Intersection}, the answer given = 0.214. Is that correct?

    I would like your thoughts.

    Note: The previous two parts of the same question are (where the above is part iii):
    (i) chance of getting 8 out of 8 shoots as goals.
    (ii) change of getting exactly 6 out of 8 shoots as goals.
    ref: MathsQuest(Yr-12)
     
  2. jcsd
  3. Jan 8, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Need the reasoning.

    ##p^n(1-p)^{N-n}## is the probability of n successes out of N trials.
    It is not the probability that those n successes are the first n of them.

    The probability P(X=6) you gave is the probability that any six are successes right?

    If I read you correctly:
    The probability P(X=4) you gave is the probability that any four of the trials are successes.
     
    Last edited: Jan 8, 2014
  4. Jan 10, 2014 #3

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Pr(X=4 | X=6) is zero, because if X=6 it is never equal to 4.

    Some food for thought: Does this part of the question even require knowing what Sally's probability of scoring each individual goal is?
     
  5. Jan 10, 2014 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Umm.. backwards?
    ##p^n(1-p)^{N-n}## is the probability of n successes out of N trials, and those being the first n.
    ##^NC_np^n(1-p)^{N-n}## is the probability of n successes out of N trials, those being any n.
     
  6. Jan 10, 2014 #5

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    @haruspex:
    I picked the right issue but got it backwards - thanks.
    <annoyed with self>

    Advise to OP would be to be more careful about defining events.
    Off the wording in post #1,
    It looks like X is the number of trials that are successes
    So X=x would be the event that exactly x trials were successes.
    P(X=x) equation given was for the event that a particular x trials were a success.
    P(X=x|X=y) equation... was for different X again?

    By the first definition of X: P(X=x|X=y)=0 unless x=y - so that's not what is intended!

    How about:
    X is the event that the 1st n out of N trials are successes - the remaining N-n trials being anything.
    Y is the event that any m out of the N trials are successes
    X[itex]\small\cap[/itex]Y would be the event that there are m successes in N trials, in which the 1st n trials are successes.
    ... or something.
     
    Last edited: Jan 10, 2014
  7. Jan 28, 2014 #6
    sorry for not getting back sooner. Thanks for responses,

    To both of the above comments.
    Correct. The intersection is equal to 0.

    If we consider the problem as a tree diagram, there are 6 branchs of X=6 given the first 4 shots are successfull.
    and given;
    Pr(x=6) = 0.2936 (exactly 6 isn't the same as only 6, my bad).

    Therefore:
    (6*0.8^6*.2^2)/0.2936 = 0.214
     
  8. Jan 28, 2014 #7

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That's the right answer.
    The easiest way is to to turn it into: pick 6 objects from 8 in a line; what's the probability you pick the first 4? A: 4C4*4C2/8C6
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted