Question

(3) For the random experiment “roll a balanced die 3 times”, what’s the total count of the sample space? What are the chances that the three numbers add up to 5? What are the chances that there are two even numbers and one odd number among the 3 rolls? What are the chances that there are more even numbers than odd numbers among the 3 rolls?

Answer 1

3-a)since there are 6 outcomes on each roll:

total count of the sample space =6*6*6 =216

3-b)

number of ways 3 numbers can be added to 5 =6 (outcomes are (1,1,3),(1,3,1),(3,1,1),(2,2,1),(2,1,2),(1,2,2))

therefore chances that the three numbers add up to 5 =6/216 =1/36

3-c)since probability of even number =probability of odd number on a die =1/2

P( two even numbers and one odd number among the 3 rolls ) =(₃C₂)*(1/2)²*(1/2) =3/8

3-d)

P(more even number than odd) =P(2 even numbers)+P(3 even numbers) =(₃C₂)*(1/2)²*(1/2) +(₃C₃)*(1/2)³*(1/2)⁰ = 3/8+1/8=1/2