There are 36 outcomes when rolling a pair of dice.
The ones adding up to 8 are: (2,6),(3,5),(4,4),(5,3),(6,2).
So the probability of rolling a sum of 8 is 5/36.
With three (independent) random throws, the probability of having a sum of 8 at least once equals 1-probability of no 8 at all.
P(not 8)=1-5/36=31/36
For this to happen three times, we use the multiplication rule
P(not 8 three times)=(31/36)^3=29791/46656=
Therefore
P(sum of 8 at least once)=1-29791/46656=16865/46656
[=0.3615 approx.]
Step-by-step explanation:
Absolute value of a number gives the positive quantity of the number.
For instance,
![|3|=3, |-3|=3](https://tex.z-dn.net/?f=%7C3%7C%3D3%2C%20%7C-3%7C%3D3)
From this example, therefore we know that
![\text{If }|x|=a,\text{ then }x=a\text{ or } -a.](https://tex.z-dn.net/?f=%5Ctext%7BIf%20%7D%7Cx%7C%3Da%2C%5Ctext%7B%20then%20%7Dx%3Da%5Ctext%7B%20or%20%7D%20-a.%20)
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In this question,
![-2|2.2x-3.3|=-6.6](https://tex.z-dn.net/?f=-2%7C2.2x-3.3%7C%3D-6.6)
![|2.2x-3.3|=3.3](https://tex.z-dn.net/?f=%7C2.2x-3.3%7C%3D3.3)
![2.2x-3.3=3.3\text{ or } -3.3](https://tex.z-dn.net/?f=2.2x-3.3%3D3.3%5Ctext%7B%20or%20%7D%20-3.3)
![2.2x=3.3+3.3 \text{ or }-3.3+3.3](https://tex.z-dn.net/?f=2.2x%3D3.3%2B3.3%20%5Ctext%7B%20or%20%7D-3.3%2B3.3)
![2.2x=6.6\text{ or } 0](https://tex.z-dn.net/?f=2.2x%3D6.6%5Ctext%7B%20or%20%7D%200)
![x=3\text{ or } 0](https://tex.z-dn.net/?f=x%3D3%5Ctext%7B%20or%20%7D%200)
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