Question

Suppose you have a three-sided die, the die is loaded
so that the probability of 1 or 2 coming out is the same and equal to
one
4 while the third side has a probability of 1
two
. If it is launched
twice the die is X the random variable that returns the sum of the
obtained results, write the table for the distribution function of
probability p for the random variable X.

Answer 1

There is a three sided die with P(1) = 1/4 , P(2) = 1/4 , and P(3) = 1/2.

Die is rolled twice and the sum of the resultant number is noted. The possible Outcomes are:

(1,1) , (1,2) (1,3)

(2,1) (2,2) (2,3)

(3,1) (3,2) (3,3)

Case 1 : Outcome is (1,1)

Sum is noted as 2(1+1).

P(1,1) = P(1on 1st roll)*P(1 on 2nd roll of die)

= (1/4)*(1/4)

= 1/16

Case 2: Outcome is (1,2)

Sum is noted as 3(1+2).

P(1,2) = P(1on 1st roll)*P(2 on 2nd roll of die)

= (1/4)*(1/4)

= 1/16

Case 3: Outcome is (1,3)

Sum is noted as 4(1+3).

P(1,3) = P(1on 1st roll)*P(3 on 2nd roll of die)

= (1/4)*(1/2)

= 1/8

Case 4 : Outcome is (2,1)

Sum is noted as 3(2+1).

P(2,1) = P(2on 1st roll)*P(1 on 2nd roll of die)

= (1/4)*(1/4)

= 1/16

Case 5 : Outcome is (2,2)

Sum is noted as 4(2+2).

P(2,2) = P(2 on 1st roll)*P(2 on 2nd roll of die)

= (1/4)*(1/4)

= 1/16

Case 6 : Outcome is (2,3)

Sum is noted as 5(2+3).

P(2,3) = P(2 on 1st roll)*P(3 on 2nd roll of die)

= (1/4)*(1/2)

= 1/8

Case 7 : Outcome is (3,1)

Sum is noted as 4(3+1).

P(3,1) = P(3 on 1st roll)*P(1 on 2nd roll of die)

= (1/2)*(1/4)

= 1/8

Case 8 : Outcome is (3,2)

Sum is noted as 5(3+2).

P(3,2) = P(3 on 1st roll)*P(2 on 2nd roll of die)

= (1/2)*(1/4)

= 1/8

Case 9 : Outcome is (3,3)

Sum is noted as 6(3+3).

P(3,3) = P(3 on 1st roll)*P(3 on 2nd roll of die)

= (1/2)*(1/2)

= 1/4

Thus , the table for the distribution function is given as :

X : Sum	P(x)
2	1/16 ( Case 1 )
3	2/16 ( Case 2 + 4)
4	5/16 ( Case 3 + 5 + 7)
5	4/16 (Case 6 + 8)
6	4/16 ( Case 9)

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