Math help



  • you have a hat with ten “numbers”.  there are 3 0s, 3 2s and 4 1s. you have to get a sum of 3. you get to draw two numbers out. you put the 1st number back in and then draw the second( so  results are indpentant) what is the chance of you getting that sum of 3 or better?

    i really want to know how to do this so i can use it for a differnt probability problem.


  • 2007 AAR League

    math is definently not my strong point


  • 2007 AAR League

    Ok, first of all, how can you get a sum of 3?  You need to draw 1 “one” and 1 “two” right?  You can’t draw 2 of the same.  You can’t draw any "zero"s.  So if you know you must draw a “one” and a “two” and you can draw them in any order.  So successful outcomes are as follows:
    FIRST you draw a “one” THEN you draw a “two”.  or
    FIRST you draw a “two” THEN you draw a “one”.

    p1 is 4/10.  p2 is 3/10.  Then p1 and p2 = 12/100.  Also p2 and p1 = 12/100

    so add them together and your probability is 24/100.  or 6/25.

    Maybe I made a mistake?  Anyone else?



  • I think it’s easier to think of all the combinations.

    There are 8 possible ways to draw numbers.  Just write them out.  Then look to see which ones equal three or more.  Out of all the possibilities, only three will get you the desired sum ( 1/2, 2/1, 2/2).

    But the likelihood of drawing a one is better than drawing a 0 or 2.

    So let’s see what happens if we draw 1 first.  That’s 4/10 probability.  We MUST draw a 2, which is a 3/10 probability.  That is 12/100, or 3/25, overall probability of happening (multiply the two probabilities).

    Let’s see what happens if we draw 2 first.  That’s a 3/10 probability.  You can draw a 2 or 1, which is a 7/10 probability.  That makes a 21/100 probability overall.

    Since either can happen, we find that you have a 33/100 probability of achieving the desired amount, or pretty much 1/3…

    (Sorry, rj - if I am right, and it has been a LONG time, then you are my inspiration…)


  • 2007 AAR League

    Jermo,

    I think you are correct if Cyan needs a sum of 3 or more:
    @cyan:

    what is the chance of you getting that sum of 3 or better?

    I was basing my solution on my understanding that he needed a sum of exactly 3:
    @cyan:

    you have to get a sum of 3

    Since it’s kind of ambiguous in the question, Cyan will need to clear that up.  Depending on the question, one of these solutions should satisfy him  😉



  • @rjclayton:

    Jermo,

    I think you are correct if Cyan needs a sum of 3 or more:
    @cyan:

    what is the chance of you getting that sum of 3 or better?

    I was basing my solution on my understanding that he needed a sum of exactly 3:
    @cyan:

    you have to get a sum of 3

    Since it’s kind of ambiguous in the question, Cyan will need to clear that up.  Depending on the question, one of these solutions should satisfy him  😉

    sorry i meant 3 or better. but i was hoping for an equation  or something to make it easier. like for premutations or combunations. thanks for the help, i tried what jermo said and it worked…


  • 2007 AAR League

    Well, as a combination I guess it is:
    [(4 choose 1) x (3 choose 1)] + [(3 choose 1) x (7 choose 1)]

    but since you’re only choosing 1 each time, there really isn’t a need to use combinations / permutations.

    Cheers.



  • I admit I didn’t know any shortcuts - I was just trying to break it down better so cyan could see the meat & potatoes of the thing…


  • 2007 AAR League

    @cyan:

    you have a hat with ten “numbers”.  there are 3 0s, 3 2s and 4 1s. you have to get a sum of 3. you get to draw two numbers out. you put the 1st number back in and then draw the second( so  results are indpentant) what is the chance of you getting that sum of 3 or better?

    i really want to know how to do this so i can use it for a differnt probability problem.

    33%

    You could just draw the matrix of 10x10 cells and label all the possible combinations then count the ones that work.


  • 2007 AAR League

    AN EQUATION!!!  😮

    Next thing you know you will be discussing the polarity of ZERO!!!  😛

    You defined 10 possible selections so the odds of any single selection is 1/10.

    You defined the possible selections as three 0’s, four 1’s and three 2’s.

    This means the odds of selecting a 0 is 3/10 or 30% in a single draw.  For a 1 it is 4/10 or 40% and for a 2 it is 3/10 or 30%.

    You then specify two draws with replacement between the draws.

    Finally, you requested the sum of the two draws equal or exceed “3”.

    The possible results of the two draws are (x,y) where x is the first draw and y is the second draw.

    These possible results are (0,0); (0,1); (0,2); (1,0); (1,1); (1,2); (2,0); (2,1); (2,2).

    The only results you are interested in are (1,2); (2,1) and (2,2).

    These are (0.4 * 0.3) + (0.3 * 0.4) + (0.3 *0.3)

    This equals 0.12 + 0.12 + 0.09 or 0.33 which is 33%.

    For a bunch of fancy pants math terms take a look at this:
    http://en.wikipedia.org/wiki/Dice#Probability


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