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!

Discrete Probability Theory

  1. Feb 2, 2007 #1
    1. The problem statement, all variables and given/known data
    A die is loaded so that the numbers 2 through 6 are equally likely to appear but that 1 is 3 times as likely as any other number to appear. What is the probability of getting an even number?

    2. Relevant equations

    This is where I become lost - I'm not sure how to get the outcome.

    3. The attempt at a solution

    P(2) + P(4) + P(6) = 1/8 + 1/8 + 1/8 = I think the answer is 3/8 but can't tell you how I got that. I don't understand the P(1) + P(2) + P(3) thing. Could someone help me to understand this?
    Last edited: Feb 2, 2007
  2. jcsd
  3. Feb 2, 2007 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    What P(1)+P(2)+P(3) thing?

    You know that there are 6 outcomes

    P(2)=P(3)=..=P(6), P(1)=3P(2), and P(1)+P(2)+..P(6)=1.

    So from this you can find P(2), and hence P(2 or 4 or 6)=P(2)+P(4)+P(6).
  4. Feb 2, 2007 #3
    That's just it - I don't understand this. I guess I see where P(2)*P(3) = P(6) but where I get lost is P(1)=3P(2), and P(1)+P(2)+..P(6)=1.
  5. Feb 2, 2007 #4
    Load a die and die cast yourself.
  6. Feb 2, 2007 #5
    Geez - thanks for the great input there! Some of us aren't math wizz's.
  7. Feb 2, 2007 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    The montecarlo method!

    You "see" the one part I don't see, and are lost on the easy ones!

    The problem says "1 is 3 times as likely as any other number to appear". If the probability a 2 will occur is P(2) and 1 is "3 times more likely", then P(1)= 3P(1). And, of course, since some number must occur the probability some number must occur is 1. That is, the sum of all the individual probabilities is 1. That's generally true for any probability distribution- the sum of probabilities of all "elementary" events is 1.

    If you let p be the common probability of 2- 6, P(2)= P(3)= P(4)= P5)= P6) and P1)= 3p so you must have 3p+ p+ p+ p+ p+ p= 1. Solve that for p and the rest is easy.

    But what in the world makes you think that P(2)*P(3)= P(6)? They are not, P(2)= P(3)= P(6). Are you thinking that 2*3= 6? That has nothing to do with probabilities.
  8. Feb 3, 2007 #7
    So P(2) + P(4) + P(6) = 1/8 + 1/8 + 1/8?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Discrete Probability Theory