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!

Homework Help: Statistics- Rolling a loaded die

  1. May 7, 2013 #1
    1. The problem statement, all variables and given/known data

    A certain die is weighted such that probabilities of showing a 1, 2, 3, 4, 5, and 6 are

    A) If two such dice are thrown, and you are told that the sum of the two is 10 or larger. What is the probability that the result was a pair of 5's?

    B) How many times would you have to throw this die to have the probability of throwing a 2 exceed 40 percent?

    2. Relevant equations

    3. The attempt at a solution

    A) For this the possible outcomes for a sum of 10 or greater are:


    Getting rid of duplicates, since order doesn't matter there is a 1/4 chance it's double fives. I multiplied this by (8/34)(8/34) and got a probability of .01384

    B) For this I used the equation

    Probability = Successful outcomes/Total number of outcomes

    so .4 = (8/34)/n

    solving for n I get .588, which doesn't make sense.

    I'm just learning statistics, so if anything I tried to do offends you mathematically, I'm very sorry!
  2. jcsd
  3. May 7, 2013 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Don't apologize; we are not offended by errors. Besides, this is a probability question, not a statistics one.

    Anyway, in (a): eliminating duplicates is an error. For example, 6-4 and 4-6 both contribute equally to getting '10'. For (a) I get the answer 8/21 ≈ 0.38095.

    For (b), you want to find the smallest n so that the probability of getting at least one '2' in trials 1,2,...,n is = 0.40. I'll just give a hint to get you going. Look at the experiments as having only two outcomes: S (success = get a '2') or F (failure = not a '2'). Let p = 8/34 = 4/17 be the success probability per trial and q = 1-p = 13/17 be the failure probability per trial.

    What is the probability that your first S occurs in trial n? For that to happen, the first (n-1) trials must all give F. What is the probability for that to happen? The probability that the first S occurs on or before trial n is a sum of such probabilities.
    Last edited: May 7, 2013
  4. May 7, 2013 #3
    Would you mind explaining how you got part A? I can't figure out how you did it.
  5. May 7, 2013 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    If S is the sum we have
    [tex] P\{(5,5)|S \geq 10\} = \frac{P\{(5,5)\; \& \;S \geq 10 \}}{P\{S \geq 10\}} \\
    = \frac{P\{(5,5)\}}{ P\{S \geq 10 \}}[/tex]
    because {(5,5) & S ≥ 10} = {(5,5)} (since the event {(5,5)} is a subevent of {S ≥ 10}).
    So, you need to compute P{S ≥ 10}.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted