QUESTION 1- Two balanced 4 sided-dice, each with faces labeled by 1, 2, 3 and4. if Ali plays a game where he tosses two balanced 4 sided-dice He wins 3 points if the sum is 6. He wins 2 points if the sum is greater than 6. He loses 1 point if the sum is less than 6. Find the probability distribution sum. 2- A dice is loaded in such away that each odd number is twice as likely to occur as each even number. find P(G) , where G is the event that a number greater than 3 occurs on a single roll of dice.

Kindly check explanation

Explanation:

Given that:

Two Four-sided dice numbered : 1, 2, 3, 4

Sample space = 4² = 16

3 points wonif sum = 6

2 points won If sum > 6

1 point lost if sum < 6

SAMPLE SPACE

___ 1 2 3 ___4

1 __2345

2__3456

3__4567

4__5678

Probability = (number of required outcomes / total possible outcomes)

P(6) = 3 / 16 = 0.1875

P(>6) = 3 / 16 = 0.1875

P(<6) = 10 / 16 = 0.625

X32 - 1

P(X)0.18750.18750.625__

XP(X)0.56250.375 - 0.625__

Sum of the distribution :

0.1875 + 0.1875 + 0.625 = 1

Σ XP(X) = (0.5625 + 0.375 + (-0.625))

Σ XP(X) = 0.5625 + 0.375 - 0.625

Σ XP(X) = 0.3125

2.) A dice is loaded in such away that each odd number is twice as likely to occur as each even number. find P(G) , where G is the event that a number greater than 3 occurs on a single roll of dice.

Since each odd number is twice as likely to occur as each even number

Odd numbers on a dice = 1, 3, 5

Hence, sample space = {1, 1, 2, 3, 3, 4, 5, 5, 6}

find P(G) , where G is the event that a number greater than 3 occurs on a single roll of dice.

Required outcomes = numbers greater than 3 =(4, 5, 5, 6)

P(G) = (number of required outcome / Total possible outcomes)

P(G) = 4 / 9