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    Two fair dice are rolled and the difference between the two score isn’t recorded (largest-smallest)
    Attachment 702380 (Sorry the image is upside down)

    This experiment is repeated 10 times. Find the probability that the recorded number is 0 on more than 3 occasions.
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    (Original post by Alexia_17)
    Two fair dice are rolled and the difference between the two score isn’t recorded (largest-smallest)
    Attachment 702380 (Sorry the image is upside down)

    This experiment is repeated 10 times. Find the probability that the recorded number is 0 on more than 3 occasions.
    If the recorded number is zero, that is equivalent to saying that the dice have the same number on them. What is the probability of this, and how can you use this to find the probability that this occurs on more than three occasions?
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    (Original post by Integer123)
    If the recorded number is zero, that is equivalent to saying that the dice have the same number on them. What is the probability of this, and how can you use this to find the probability that this occurs on more than three occasions?
    How would I find out if it occurs on more than three occasions out of the 10 repeated experiments?
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    (Original post by Alexia_17)
    How would I find out if it occurs on more than three occasions out of the 10 repeated experiments?
    Use the random variable X having a binomial distribution with parameter n=10 and p, which you should calculate. Then P(X>3)=1-P(X<=3)
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    (Original post by Integer123)
    Use the random variable X having a binomial distribution with parameter n=10 and p, which you should calculate. Then P(X>3)=1-P(X<=3)
    Would p be 6/36 from the table I made (Sorry the image Ian upside down)?

    If so, would I do:
    10C3x(6/36)^3x(30/36)^7
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    (Original post by Alexia_17)
    Would p be 6/36 from the table I made (Sorry the image Ian upside down)?

    If so, would I do:
    10C3x(6/36)^3x(30/36)^7
    Yeah p is 1/6. It may be easier to think of it as: irrespective of the first die, the second die has a chance of 1/6 of being the same. Alternatively, there are 6*6=36 ways of throwing two dice and 6 outcomes have a pair of (1,1), etc.

    Well what you have calcuated there is P(X=3). We want to find 1-P(X<=3) so if you add P(X=3) to the sum of P(X=k) for k=0,1,2, and subtract that result from 1, you have the desired probability.

    However, you may wish to simplify things using the probability tables (but these may not give exact values, and will be rounded).
 
 
 
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