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Probablility of a gambling machine

  1. Aug 22, 2014 #1
    1. The problem statement, all variables and given/known data
    A gambling machine shows one out of five colours, red , orange , yeloow, green and blue. When the machine is working properly, every colour has an equal chance of appearing the colour. And the colour shown at one instant is independent of the colour shown earlier. Calculate the probablity that the machine shows different colours for 5 consecutive times.

    the ans is 24/625.

    but my working is 0.2^5 = 1/3125

    why cant i do in this way? by the way? what is the porper working of getting the ans?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 22, 2014 #2
    Can you work out the probabability that the machine shows red, orange, yellow, green and blue in that order? Do you still think ## 0.2^5 ## can be the right answer?

    If the machine is going to show different colours for 5 consecutive times, does it matter what colour it shows the first time? How many choices of coulour can it show the second time?
     
  4. Aug 22, 2014 #3
    sorry. i still cant get what do you mean . can you explain in other words?
     
  5. Aug 22, 2014 #4

    phinds

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    Let the first choice be arbitrary. Any color. Then how many choices would you have for the second color if you have to avoid duplicating any color?
     
  6. Aug 22, 2014 #5

    Orodruin

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    If the machine has shown red on the first try, what is the probability that it will show a different color (any color different from red) in the second try?
    If the machine has shown red on the first try and blue on the second try, what is the probability that it will show a color different from red and blue on the third?
    Take it from there.

    Your expression ##0.2^5## is the probability of obtaining a particular sequence (say red, blue, yellow, orange, green), but there are other sequences that also show all five colors.
    Edit: You can actually use this to solve the problem as well by computing the number of different sequences that fulfill your condition of not repeating any color.
     
  7. Aug 22, 2014 #6
    WHY NOT 1/(5x4x3x2x1)= 1/120 ....?
    after a certain colour(from n choices) is chosen for the first attempt , i have n-1 choices for the subsequent attempt , where n = 5,4 ,3,2 ,1
     
  8. Aug 22, 2014 #7

    Ray Vickson

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    This answer is incorrect. The number 12 = 5! is the number of permutations of 5 different things, so 1/120 is the probability of a PARTICULAR permutation. However, that is not what the question asks for. In your question you are trying to count only a subset of all possible outcomes---those that actually DO constitute a permutation of the 5 numbers. In other words, you want to discard those outcomes in which one or more of the numbers is repeated more than once, and look only at the ones that constitute permutations.
     
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