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Connor rolled a die 6 times. A 2 occurred three times. Why isn't the theoretical probability 3/6 instead of 1/6?​

Respuesta :

2 occurred 3times

[tex]\\ \sf\longmapsto S=6\times 3=18[/tex]

[tex]\\ \sf\longmapsto |S|=18[/tex]

Now

[tex]\\ \sf\longmapsto E=No\:of\:times\: 2\:occured[/tex]

[tex]\\ \sf\longmapsto |E|=3[/tex]

we know

[tex]\boxed{\sf P(E)=\dfrac{|E|}{|S|}}[/tex]

[tex]\\ \sf\longmapsto P(E)=\dfrac{3}{18}[/tex]

[tex]\\ \sf\longmapsto P(E)=\dfrac{1}{6}[/tex]

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