Dealing with weighted dice:

Question

For this problem, assume that all the odd numbers are equally likely to come up, all the even numbers are equally likely to come up, and the odd numbers are two times as likely to come up as the even numbers.
(1) What is the value of w1?


(2) What is the value of w4?


(3) What is the value of w2?

Answer 1

Are you talking about rolling a dice here? 

If it is a dice then each of the numbers in a dice has the probability of 1/6 since each of the 6 outcomes are equally likely. 

 

But in this case the odd numbers are two times as likely to occur as the even numbers so, if x is an outcome for an even number and y is an outcome for an odd number

y = 2x

but the probability of odd and even numbers add up to 1

so 

x + y = 1

x + 2x = 1

3x = 1

x = 1/3

y = 2/3

 

Since there are 3 even outcomes each equally likely to occur, w2 or w4 = 1/3*1/3 = 1/9

w1 = 2/3*1/3 = 2/9