Question

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)

Answer 1

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

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

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