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)

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