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Mathematics of Data Management - Probability distributiono

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data
    A spinner has eight equally-sized sectors, numbered 1 through 8. What is the probability that the arrow on the spinner will stop on a prime number?

    2. Relevant equations
    P(x) = 1 / n, out come of a uniform probability distribution
    P(x) = Probability distribution
    n = Total # of possible outcomes

    3. The attempt at a solution
    n = 8, since the total number of possible outcome range from 1 through 8
    P(x) = 1/n
    Prime numbers are 1, 3, 5, 7
    *Note that some textbooks don't include 1 as a prime number

    P(x) = 1/n
    = 1/8
    ** This is wrong but that's how it seems like it is suppose to be done

    P(x) = 4/8
    4 = total number of primes
    P(x) = 1/2 or 3/8 ( if you don't consider 1 as a prime number)
    ** this is the correct answer

    The second way of solving it is the correct answer but what i don't understand is that the formula is P(x) = 1/n, why did 1 become a 4, how would i know that i'm suppose to change the 1? and yes i have checked the 1 is not an L or an I...
    Last edited: Jan 17, 2012
  2. jcsd
  3. Jan 17, 2012 #2


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    Homework Helper

    the probability of any single number coming up is 1/8. As you have 4 (or 3) potential outcomes, the probability is 4/8 (or 3/8)
  4. Jan 17, 2012 #3
    But why is it 4/8 or 3/8? the formula is p(x) = 1/n
  5. Jan 17, 2012 #4


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    Homework Helper

    not sure if I'm understanding you

    say you have n, distinct evenly distributed outcomes (labelled 1 to n)

    the probability of getting an outcome x is p(x) = 1/n

    the probabilty of getting one of m outcomes is m/n

    If you want to break it right down, you know
    [tex] p(x)=\frac{1}{8} [/tex]

    [tex] p(1)=p(3)=p(5)=p(7)=\frac{1}{8} [/tex]

    you also know only a single number can appear at a time, they are mutually exclusive events
    [tex] p(i\cap j)=0 [/tex]

    then, as the intersection is zero (mutually exclusive)
    [tex] p(1\cup3 \cup 5\cup7)=p(1)+p(3)+p(5)+p(7)= \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}= \frac{4}{8}[/tex]
    Last edited: Jan 17, 2012
  6. Jan 17, 2012 #5
    My lord! Thank you that is just perfect!!!!!!!!!!!!!<3
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