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!

Probability - array of balls

  1. Aug 4, 2014 #1


    User Avatar

    Here is my task:
    There are 4 white and 6 red balls in array. Find probability that there are three white balls in beginning of array.
    How to calculate m?
    Last edited: Aug 4, 2014
  2. jcsd
  3. Aug 4, 2014 #2


    User Avatar
    Science Advisor

    Where in the world did you get 70? There are total of 10 balls, 4 white and 6 red. How many different orders are there?

    If we require that there be exactly three white balls at the beginning (so the fourth ball must be red), then the last 6 balls are 1 white and 5 red. How many different orders are there?
  4. Aug 4, 2014 #3


    User Avatar

    Since there are 4 white and 6 red balls I should use formula for permutations with repetition to calculate all possible orders?
    Orders of interest are where first three balls are white and other 7 balls (6 red and 1 white) could be in any order so m would be $$\binom{4}{3} \frac{7!}{6!1!}= 28$$ and $$P(A) = \frac{28}{210}$$?
    Last edited: Aug 4, 2014
  5. Aug 4, 2014 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You need to decide whether you want to use permutations or combinations. I think permutations are a more natural description of the actual, physical situation, and also would probably be what you would prefer to use if you were writing a Monte-Carlo simulation of the experiment for running on a computer. So: number the balls from 1--10; the first 4 are white and the remaining 6 are red. The sample space consists of all the permutations of the numbers (1,2,...,10), and we tacitly assume all are equally likely.

    How many permutations are there altogether? How many permutations have numbers from 1--4 in the first three places? So, assuming we don't care what is in the 4th place, what would be the resulting probability? Next, if we assume we also want a non-white in position 4, how many permutations would there be now? What is the corresponding probability?

    Another way to do the problem would be via conditional probability arguments. If the events E1, E2, E3 correspond to having a number from 1--4 in positions 1,2 and 3, respectively, then
    [tex] P(E_1 \& E_2 \& E_3) = P(E_1) P(E_2 \& E_3 | E_1),\\
    P(E_2 \& E_3 | E_1) = P(E_3 | E_1 \& E_2) P(E_2|E_1)[/tex]
    What is ##P(E_1)?## What is ##P(E_2 | E_1),## etc.?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted