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Statistics Probability Densities

  1. Jan 20, 2013 #1
    1. It is known that the probability of being able to log on to a computer from a remote terminal at any given time is .7. Let X denote the number of attempts that must be made to gain access to the computer.
    a) Find the first four terms of the density table.
    b) Find a closed form expression for f(x).
    c) Find P[X = 6].


    3. I got this:
    a) x P[X = x]
    -------------
    1 0.7
    2 0.49
    3 0.343
    4 0.2401
    b) f(x) = (0.7)^x where x = 1, 2, 3, 4, ...
    0 elsewhere
    c) f(x) = (0.7)^6
    = 0.117649

    I'm not sure if I did it correctly.
     
  2. jcsd
  3. Jan 20, 2013 #2

    LCKurtz

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    Remember that for the first successful attempt to be the second try, you must fail on the first try. That changes things..
     
  4. Jan 20, 2013 #3
    Hmmm why do you have to fail on the first try? Couldn't you be successful on the first two tries?
     
  5. Jan 20, 2013 #4

    haruspex

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    Because if you succeed the first time, you don't need to try again. You're only logging on once.
     
  6. Jan 20, 2013 #5
    Oh so my table instead should be
    1 0.7
    2 0.21 (.3)*(0.7)
    3 0.063
    4 0.0189

    I think?
     
  7. Jan 20, 2013 #6

    LCKurtz

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    That's better.
     
  8. Jan 20, 2013 #7
    Thank You!! Could you also help me on this problem?

    The basic storage unit of a digital computer is a "bit". A bit is a storage position that can be designated as either on (1) or off (0) at any given time. In converting picture images to a form that can be transmitted electronically, a picture element, called a pixel is used. Each pixel is quantized into gray levels and coded using a binary code. For example, a pixel with four gray levels can be coded using two bits by designating the gray levels by 00, 01, 10, and 11.
    (a) How many gray levels can be quantized using a four bit code?
    (b) How many bits are necessary to code a pixel quantized to 32 gray levels?

    I'm not sure how to start it. I believe it has something to do with permutations and combinations.
     
  9. Jan 20, 2013 #8

    Dick

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    Not really. It has more to do with powers of two and how many binary digits you need to encode them. Sure you aren't overthinking this? You can encode 4 gray levels with 2 bits because you have two choices 0 and 1 for each bit. Since you have 2 bits you have 2*2 choices.
     
  10. Jan 20, 2013 #9
    Oh!! So for
    a) Since I have 2 choices and 4 bits, I would have 2 * 4 = 8 gray levels.
    b) 32/2 = 16 bits

    Is this correct?
     
  11. Jan 20, 2013 #10

    Dick

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    Nope. If you had 3 bits you would have two choices for EACH bit and three bits. That's 2*2*2=8 total choices. With four bits you should have more. Think again.
     
  12. Jan 20, 2013 #11
    Ah! So this is what I got,
    a) 2 ^ 4 = 16
    b) 2 ^ x = 32
    x = 5 bits
     
  13. Jan 20, 2013 #12

    Dick

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    Yes, that's it.
     
  14. Jan 20, 2013 #13
    Thank You!!
     
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