Question

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

Answer 1

2 occurred 3times

$\\ \sf\longmapsto S=6\times 3=18$

$\\ \sf\longmapsto |S|=18$

Now

$\\ \sf\longmapsto E=No\:of\:times\: 2\:occured$

$\\ \sf\longmapsto |E|=3$

we know

$\boxed{\sf P(E)=\dfrac{|E|}{|S|}}$

$\\ \sf\longmapsto P(E)=\dfrac{3}{18}$

$\\ \sf\longmapsto P(E)=\dfrac{1}{6}$