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A Color Sequence Experiment

A Color Sequence Experiment Suppose that we have a special box - each time we press a button on the box, it prints out a sequence of colors, in order - it prints four colors at a time. Suppose the box follows the following Probabilities for each Color Sequence: The Model.

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A Color Sequence Experiment

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  1. A Color Sequence Experiment Suppose that we have a special box - each time we press a button on the box, it prints out a sequence of colors, in order - it prints four colors at a time. Suppose the box follows the following Probabilities for each Color Sequence: The Model

  2. Suppose we define the event E={Blue(B) is printed in the 2nd or 3rd slot}. Compute the probability for event E, and show me how you did it. Also, interpret the probability for event E. The only color sequence meeting the definition of E is the sequence BBBB. So, we write Pr{E} = Pr{BBBB} = .10 = 10%. This means that in long runs of box-prints, that approximately 10% of the prints will show as BBBB.

  3. Suppose we define the event F={Yellow is printed at least once in the sequence}. Compute the probability for event F, and show me how you did it. Also, interpret the probability for event F. The event F merely requires that Yellow be present. Yellow is present in the following color sequences: YYYY, BYRG and RYYB. So we write Pr{F} = Pr{exactly one of YYYY, BYRG or RYYB is printed}= Pr{ YYYY }+Pr{ BYRG }+Pr{ RYYB } = .30+.15+.15 = .60 = 60% In long runs of box-prints, approximately 60% of prints will contain at least one Yellow in the color sequence.

  4. Suppose we define the event G={Green(G) is printed in the 2nd slot}. Compute the probability for event G, and show me how you did it. Also, interpret the probability for event G. The event G requires that Green be present in the 2nd slot. This requirement is met in the following color sequences: BGGB, RGGR. So we write Pr{G} = Pr{exactly one of BGGB or RGGR is printed} = Pr{ BGGB }+Pr{ RGGR = .25+.05 = .30 = 30% In long runs of box-prints, approximately 30% of prints will show green in the 2nd slot.

  5. Suppose we define the event H={Red is not the 1st color}. Compute t|he probability for the event H, and use the Complementary Rule. Also, interpret the probability for H. The event notH requires that the sequence lead off with Red. This requirement is met in the following color sequences: RGGR, RYYB. Pr{notH} = Pr{Red is the first color } = Pr{exactly one of RGGR or RYYB is printed } = Pr{ RGGR }+Pr{ RYYB } = .05 + .15 = .20 = 20%. So, Pr{H} = 1 - Pr{notH} = 1 - .20 = .80 = 80% So in long runs of box-prints, approximately 80% of color sequences will not show Red as the first color in the sequence.

  6. Suppose we define the event I={Blue(B) is not the 4th color}. Compute the probability for the event I, and use the Complementary Rule. Also, interpret the probability for I. The event notI requires that the sequence end with Blue. This requirement is met in the following color sequences: BBBB, BGGB and RYYB. Pr{notI} = Pr{Blue is the 4th color } = Pr{exactly one of BBBB, BGGB or RYYB is printed } = Pr{BBBB}+Pr{ BGGB }+Pr{ RYYB } = .10 + .25 + .15 = 50%. So, Pr{I} = 1 - Pr{notI} = 1 - .50 = .50 = 50% So in long runs of box-prints, approximately 50% of color sequences will not show Blue as the last color in the sequence.

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