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25 February, 00:47

A six-sided die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on that face. For example, a six is three times as probable as a two.

What is the probability of getting either a 5 or a 2 in one throw?

Hint: Let m be a distribution function for this experiment. If a six is three times as probable as a two, then in terms of the distribution function, m (6) = 3. m (2)

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  1. 25 February, 00:51
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    P (2U5) = 7/21 = 1/3

    the probability of getting either a 5 or a 2 in one throw is 1/3

    Step-by-step explanation:

    Given that; the probability of each face turning up is proportional to the number of dots on that face

    P (1) = 1*P (1)

    P (2) = 2*P (1)

    P (3) = 3*P (1)

    P (4) = 4*P (1)

    P (5) = 5*P (1)

    P (6) = 6*P (1)

    P (T) = 21*P (1)

    Where;

    P (x) is the probability of getting number x on the dice.

    P (T) is the total probability of obtaining any number

    N (x) is the number of possible number x in terms of the distribution function.

    P (x) = N (x) / N (T) ... 1

    And since P (T) is constant, and P (T) is proportional to N (T) then,

    P (x) is directly proportional to N (x)

    So, equation 1 becomes;

    P (x) = N (x) / N (T) = P (x) / P (T) ... 2

    The probability of getting either a 5 or a 2 in one throw

    P (2U5) = (P (2) + P (5)) / P (T)

    Substituting the values of each probability;

    P (2U5) = (2P (1) + 5P (1)) / 21P (1)

    P (2U5) = 7P (1) / 21P (1)

    P (1) cancel out, to give;

    P (2U5) = 7/21 = 1/3

    the probability of getting either a 5 or a 2 in one throw is 1/3
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