1. Limited time only! Sign up for a free 30min personal 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!

Normal Random variable Q

  1. Nov 15, 2012 #1

    CAF123

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    1000 independent rolls of a fair die will be made. Compute an approximation to the probability that the number 6 will appear between 150 and 200 times inclusively. If the number 6 appears exactly 200 times, find the probability that the number 5 will appear less than 150 times.

    3. The attempt at a solution
    I have the first part correct. For the second part, I said: If X is the number of times a 6 appears and Y is the number of times a 5 appears, then we want P(Y<150|X=200). I simplified this to P(X=200 and Y<150)/(P(X=200), from which I could compute P(X=200). Since X and Y are not independent, I can't split the intersection up. I thought about rearranging the numerator into something involving a conditional probability
    (i.e like P(X=200 and Y<150) = P(X=200|Y<150)P(Y<150)) but I don't see how this helps.) I also noticed that this intersection looked like a joint RV, but I don't have the density function to do the necessary integration.
    Any advice on how to compute P(X=200 and Y<150)?
     
  2. jcsd
  3. Nov 15, 2012 #2
    If you know the number 6 appears 200 times, what would you say about the way in which the outcomes of the remaining 800 rolls are distributed?
     
  4. Nov 15, 2012 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    There are no densities and integrations involved here---only discrete probability mass functions, and summations (although later you can make a continuous approximation and THEN get a density).

    Besides the suggestion that Yuu Suzumi has made, you can also look at this as involving a trinomial distribution with three outcomes per trial: 5, 6 or 'other'. In N independent trials, what is the probability that you get k5 5s, k6 6s and N-k5-k6 'others'?

    RGV
     
  5. Nov 15, 2012 #4

    CAF123

    User Avatar
    Gold Member

    @Yuu I know that we have n independent Bernoulli trials => we have a binomial(800,1/6). Is this what you mean? How does this help? I recognised that in part a) of the question we had binomial(1000,1/6) which since n is large, approximates to a normal RV.
    @RGV I haven't come across the trinomial distribution. This problem comes from a chapter following content on the uniform, normal, exponential distributions.
     
  6. Nov 15, 2012 #5
    Be careful! When the six is removed, how many numbers share the 800 trials?
     
  7. Nov 15, 2012 #6

    CAF123

    User Avatar
    Gold Member

    Ok, so I should have binomial(800,1/5). I see how the condition on there being 200 sixes comes into play now. I would be able to compute the probability if there was exactly 150 fives. How to do it for less than 150 fives? I know P( x < 5), say is P(0) + P(1) + P(2)..+ P(4), but this is not efficient for larger numbers.
     
  8. Nov 15, 2012 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Originally, you said you had the first part correct (that is, the probab. of having between 150 and 200 sixes). How did you do that computation? Why can't you do something similar for the second part?

    RGV
     
  9. Nov 15, 2012 #8

    CAF123

    User Avatar
    Gold Member

    Many thanks. I didn't think they wanted an approximation for the second part, but this gives me the correct answer. One question: to get the correct answer, I had to do P(x<149.5) rather than P(x<150.5). I understand we add some small number on to allow a continuity correction, but why do we take the 1/2 away rather than add it on (in this case)? In general, can we add any small increment on?
     
  10. Nov 15, 2012 #9

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Because for a discrete random variable taking the values {0,1,2,...,800}, the event {X < 150} is the same as {X <= 149}, which you change to {X <= 149.5} in the continuous approximation. If you put had {X <= 150.5} you would have been be looking at {X <= 150}, not {X <= 149}.

    RGV
     
  11. Nov 15, 2012 #10

    CAF123

    User Avatar
    Gold Member

    ok, thanks. Does it have to be a half that we add on(or is this just a convenience)?
     
  12. Nov 15, 2012 #11

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No, there is a good reason. Say we have p(k) for k = 1,2,3,... and we are approximating this by a continuous probability density f(x). We have
    [tex] p(1) \approx \int_{1/2}^{3/2} f(x) \, dx,\\
    p(2) \approx \int_{3/2}^{5/2} f (x) \, dx, \\
    \vdots \\
    p(k) \approx \int_{k - 1/2}^{k + 1/2} f(x) \, dx,[/tex]
    so
    [tex] p(a) + \cdots + p(b) \approx \int_{a-1/2}^{b+1/2} f(x) \, dx .[/tex]

    In words, p(k) equals the area of a rectangle of height p(k) and with 1, which is almost the same as the area under the f(x) curve for x from k-1/2 to k+1/2. A sum of the probabilities equals the sum of the areas.

    RGV
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Normal Random variable Q
  1. Random variables (Replies: 5)

Loading...