Question

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 1

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