Question

You roll 2 dice. What is the probability that the sum of the dice is greater than 6 and 1 die shows a 2? A 6 X 6 table of dice outcomes will help you to answer this question

Answer 1

\[\left[\begin{matrix} -& 1 & 2 & 3 & 4 & 5 & 6 \\ 1& 2 & 3 & 4 & 5 & 6 & 7 \\ 2& 3 & 4 & 5 & 6 & 7 & 8\\3& 4 & 5 & 6 & 7 & 8 & 9 \\4& 5 & 6 & 7 & 8 & 9 & 10\\5& 6 & 7 & 8 & 9 & 10 & 11\\6& 7 & 8 & 9 & 10 & 11 & 12\end{matrix}\right]\]

Answer 2

I answered your question...

Answer 3

the first row is the values for the 1st die, the first column is the values for the 2nd die

Answer 4

ow my eyes!!

Answer 5

the inside consists of the sum of dice, so find all values greater than 6, then of those, take the ones with a value of 2 one of the die

Answer 6

the chart confuses me :(

Answer 7

skittles this question has already been answered in my page look aty my page im takeing the same course

Answer 8

shows a 2 are these;

1 2 3 4 5 6
1 x
2 x x x x x x
3 x
4 x
5 x
6 x

now we need to account for those that are greater than 6


1 2 3 4 5 6
1
2 x x
3
4
5 x
6 x

if i did it right

Answer 9

@curry. im doing this for apex to make up for algebra 2 semester 2.
@amister. so i just add all them up and i ge the answer?

Answer 10

The sum of numbers can be either even or odd....
The probability of having a sum as odd is 1/2.

Probability of none of the dices showing a four is (5/6)*(5/6) = 25/36
Thus probability of at least one four is 1 - 25/36 = 11/36

Total probability = 1/2 + 11/36 = 29/36

Answer 11

yep; there are 4 ways to get a sum greater than 6 AND one of them is a 2

4/36 = 1/9 right?

Answer 12

got it. thx amistre. new questoin coming up

Answer 13

another way to do this is just create the sample space and weed out the offending options:

11 21 31 41 51 61
12 22 32 42 52 62
13 23 33 43 53 63
14 24 34 44 54 64
15 25 35 45 55 65
16 26 36 46 56 66

delete everything that doesnt have 2

21
12 22 32 42 52 62
23
24
25
26

now delete everything that sums to 6 or less



52 62
25
26

and thats it; 4 left out of 36