This question is based on the probability. Therefore, the probability that at least 1 show as a 2 is 0.22974 if 3 such number cubes are rolled.
Given:
In a certain board game a 12 sided number cube showing numbers 1-12 is rolled.
We need to determined the probability that at least 1 show as a 2, if 3 such number cubes are rolled.
It is given that total faces of = 12 labeled 1 - 12
Now calculating probability of showing a two ;
[tex]Probability = \dfrac{ Required\, outcome }{Total\, possible \,outcomes}[/tex]
[tex]Probability\, of\, showing\,a \,two=\dfrac{1}{12}[/tex]
Hence, probability of not showing a 2 is,
[tex]Probability\, of\, not\,showing\,a \,two=1-\dfrac{1}{12}[/tex]
Now, calculating probability that at least one of 3 cubes shows as a 2 =
1 - P(none of the cubes shows a 2)
[tex]P(none\, of\, the\, cubes\, shows\, a \,2) = \dfrac{11}{12} \times \dfrac{11}{12} \times \dfrac{11}{12} \\\\P(none\, of\, the\, cubes\, shows\, a \,2)=\dfrac{1331}{1728} = 0.770[/tex]
Probability that at least one of 3 cubes shows as a 2 = 1 - 0.770 = 0.22974
Therefore, the probability that at least 1 show as a 2 is 0.22974 if 3 such number cubes are rolled.
For more details, prefer this link:
https://brainly.com/question/11234923
