We have a loaded die and the numbers 2 and 3 are twice as li

We have a loaded die and. the numbers 2 and 3 are twice as likely as any other number to occur. Define an appropriate probability measure that describes our assumption. Alternately, if we scratch away the numbers 2 and 4 from the original fair die and turn it into a second and a third 5. what is the appropriate probability measure.

Solution

7)

a) In a fair die.

P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1/6

Since in a loaded die, the number 2 and 3 are twice as likely to occur as any other number

Let P(1) = P(4) = P(5) = P(6) = x

then P(2) = P(3) = 2x

Using the first formula

x + 2x + 2x + x + x + x = 1

8x = 1

x = 1/8

Hence probability of occuring 1,4,5,6 = 1/8

probabiity of occuring 2 or 3 = 1/4

b)

In a fair die.

P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1/6

Since in a loaded die, the number 2 and 4 are removed and changed into a 5

Hence P(2) = P(4) = 0

P(5) = P(2) + P(4) + P(5) = 1/6 + 1/6 + 1/6 = 1/2

Hence probability of occuring 1,3,6 = 1/6

probabiity of occuring 2 or 4 = 0

probability of occuring 5 = 1/2