Bob is asked to construct a probability model for rolling a pair of fair dice. He lists the outcomes as 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Because there are 11 outcomes, he reasoned, the probability of rolling a three must be one eleventh

. What is wrong with Bob's reasoning?

Answer and Step-by-step explanation:

For 2 dice, we have 36 possibles outcomes:

D₁ = {1,2,3,4,5,6}

D₂ = {1,2,3,4,5,6}

U = ({1,1}; {1,2}; {1,3}; ...; {2,1}; {2,2}; ...; {6,6})

This way, it's 6 for D₁ and 6 for D₂, 6.6 = 36

The possible 3 that might appear when rolling a pair of dice would be

{1,2} and {2,1}, so, 2.

The probability of rolling 3 is 2/36 = 1/18