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

start with a line segment pq that we will copy. mark a point r that will be one endpoint of the new line segment. set the compasses' point on the point p of the line segment to be copied. adjust the compasses' width to the point q. the compasses' width is now equal to the length of the line segment pq.

step-by-step explanation:

$25

step-by-step explanation:

25 - 4.50 = 20.5

20.5 / 2 = 10.25

10.25 + 10 = 20.25

20.25 - 12.50 = 7.75

therefore the answer is $25 as shown as the first number.