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Can someone explain this to me? full answer written, probabilty

  1. Nov 9, 2006 #1
    Hello everyone, i'm having troubles grasping the answer to this question. I'm studying for the exam so the answers are given in full by the professor.

    Here is the question:
    Two faces of six sided die are painted red, two are painted blue , and two are painted yellow. The die is rolled 3 times, and the colors that appeared face up on the first, second and thrid rollsa re recorded.

    Consider the situation described above:
    a. Find the probabily of the event that exactly one of the colors that appears face up is red.

    b. find the probabily of the event that at least one of the colors that appears face up is red.


    Answer:
    a: Exactly one color being red can happen in 12 ways. (One can list them all and count them. Or one can use
    the multiplication rule from the next section to calculate how many 3-letter sequences will show up in the list:
    there are 3 positions in which the 'R' can occur, and the two remaining positions can be filled with B or Y in 2x2 ways. So there will be 3 x 2 x 2 = 12, 3 lettered sequences in the list.

    So the probabilty is 12/(3x3x3) = 4/9 .

    I'm confused on why there are 3 postions in which R can occur, is it because if you roll it 3 times, then that means its possible to get R, every single time you roll the die?

    I'm trying to apply the multipcation rule, so i need to break this down into a step by step process in which its covering all the cases.

    So would my first step be:
    On roll 1, you get a red face
    step #2.
    On roll 2, you get either a blue or yellow
    step #3.
    On roll 3, you get either a blue or yellow

    I don't see how this can be applied to the mulicpation rule becuase you could end up with more than 1 choice depending on what happened before hand, can someoen explain?

    I can see how you can get how many possible total combinations in 3 rolls, like
    step 1:
    Roll the die, 3 things could pop up
    step 2:
    roll the die, 3 things could pop up
    step 3:
    roll the die, 3 things could pop up
    so the total number of ways that Red, blue or yellow will appear is 3x3x3 = 27

    b: At least one color being red can happen in 19 ways. (Again, one can list them all and count them, or one
    can calculate that 2^3 = 8 of the 3^3 = 27 possible outcomes involve only blue and yellow, so the remaining
    27 − 8 = 19 outcomes include at least one red.) Therefore, the probability of at least one red is 19/27, which
    is about 70.4%.

    So for the at least 1 case, can you always use this rule, find out 2^3, excluding the 1 your looking for in this case Red face comes up. Then find the total number of possible outcomes if all 3 where rolled which is 3^3, then subtract the 2 and you'll be left with how many times red will occur?


    Thanks any help would be great.

    EDIT: a simple example of the multipcation rule that i understand is the following:
    Suppose a computer installation has 4 input/output units (a, b, c, and d) and 3 CPU's x,y, z. Any input/ouput can be paired with any central processing unit. How many ways are there to pair an input/ouptu unit with a central processing unit?

    Step 1: Choose input/ouput unit.
    step 2: choose the CPU

    Well in step 1, there are 4 ways to choose the input/output unit, either a, b, c, or d
    Step is there are 3 ways to choose the cpu, either x, y or z,

    so the total number of ways is 4x3 = 12.

    I get this but i don't see how the top problem can be transfored into this.


    Thanks
     
    Last edited: Nov 9, 2006
  2. jcsd
  3. Nov 10, 2006 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes. Since you are only interested in the probability of "exactly one Red", represent any other color by "O". The three possible ways to get "exactly one red" are ROO (red on first throw but not on next two), ORO, (non-red on first throw, red on second, non-red on third), and OOR (red only on third throw). Since there are 2 out of 6 faces= 2/6= 1/3 that are red and 4 out of 6 faces= 4/6= 2/3 that are not, you could calculate the probability of "exactly one red" in two different ways:
    The probability of "ROO", in that order, is 1/3(2/3)(2/3)= 4/27.

    The probability of "ORO", in that order, is 2/3(1/3)(2/3)= 4/27.
    The probability of "OOR", in that order, is 2/3(2/3)(1/3)= 4/27. The sum of those 4/27+ 4/27+ 4/27= 12/27= 4/9.

    Or, since the individual probabilites are the same, 4/27, just multiply that by 3: 3(4/27)= 4/9.

    Again, use "other" to be "either blue or yellow". Then you have "red", "other", "other".

    Yes, that is true.

    b: At least one color being red can happen in 19 ways. (Again, one can list them all and count them, or one
    can calculate that 2^3 = 8 of the 3^3 = 27 possible outcomes involve only blue and yellow, so the remaining
    27 − 8 = 19 outcomes include at least one red.) Therefore, the probability of at least one red is 19/27, which
    is about 70.4%.[/quote]
    Enumerating all possibilities is the 'hard' way. Better is to calculate:

    "Probabilty of exactly one red" which you have already done: 4/9.

    "Probability of exactly two red". The possible orders are now "RRO", "ROR", "ORR", again 3. (Since 3C1= 3C2= 3.) Now the probability of each of those is (1/3)(1/3)(2/3)= 2/27 so the probability of getting exactly 2 red is 3(2/27)= 2/9.

    "Probability of exactly three red". Now the only possible order is "RRR" (3C3= 1) and the probability of that is (1/3)(1/3)(1/3)= 1/27.

    The probability that one of those will occur is their sum: 4/9+ 2/9+ 1/27= 12/27+ 6/27+ 1/27= 19/27.

    If I read this correctly, yes, you can do that but, again, it is better to think in terms of two possible outcomes: "red" and "anything other than red". It simplifies the calculations!


    It's not simply a "choice" problem, it is a "permutations" problem. "How many different ways can you get exactly one red and two others?" is the same as "How many permutations are there of 3 things if two are identical". A "permutation" problem can be transformed in to a "counting" problem. To see how that is much better than enumerating all the cases, look at "how many ways can you get roll this die 6 times and get exactly 3 red?". Again, I can imagine writing "R" each time it comes up red, "O" each time it comes up something other than red and the question becomes "how many ways can I order 3 "R"s and 3 "O"s. That's a total of 6 letters. Suppose, first, they were all different. Then I would have 6 choices for the first letter, 5 for the second, 4 for the third, etc. A total of 6(5)(4)(3)(2)(1)= 6! different order. But, in fact, 3 of the letters are identical: "R" and so I don't want to count as different just switching the "R"s. There are 3(2)(1)= 3! ways to to do that and my 6! is counting each of those as different. I need to divide by 3! to allow for that. Similarly, there are 3(2)(1)= 3! ways to switch the "O"s while leaving the "R"s alone. We need to divide by another 3! to allow for that. The actual number of "permutations" of 6 thing if 3 are the same and the other 3 are the same is 6!/(3!3!)= (6(5)(4)(3)(2)(1))/(3(2)(1))(3(2)(1))= (6(5)(4))/(3(2)(1))= (5)(4)= 20. If you really want to, you can write out all ways of writing 3 "R"s and 3 "O"s and see if you get 20 different ways!
    Since the probabilty of RRROOO is (1/3)(1/3)(1/3)(2/3)(2/3)(2/3) = 8/729 and each different order just commutes those fractions, the probability of rolling this die 6 times and getting exactly 3 red is (20)(8/729)= 160/729.
     
    Last edited: Nov 10, 2006
  4. Nov 10, 2006 #3
    Thanks so much for the detailed responce it helped alot!
     
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