Answer:
Sum of 8 = 13.8888... ≈ 14
Two 6's = 2.7777... ≈ 3
Sum of 8 and two 6's = 0.3858... ≈ 0
Sum of 8 or two 6's = 16.6666... ≈ 17
Step-by-step explanation:
Create a sample space for the sum of rolling two dice (where x are the die):
[tex]\begin{array}{|c||c|c|c|c|c|c|}\cline{1-7} \vphantom{\dfrac12}x& \textbf{1} &\textbf{2} &\textbf{3} &\textbf{4} &\textbf{5} &\textbf{6}\\\cline{1-7} \textbf{1}&2 &3 & 4& 5&6 &7 \\\cline{1-7} \textbf{2}& 3& 4& 5& 6& 7&8 \\\cline{1-7} \textbf{3}&4 &5 & 6&7 & 8&9 \\\cline{1-7} \textbf{4}&5 &6 &7 & 8&9 &10 \\\cline{1-7} \textbf{5}& 6 &7 & 8&9 &10 & 11\\\cline{1-7} \textbf{6}&7 & 8&9 &10 & 11&12\\\cline{1-7} \end{array}[/tex]
Therefore, there are 36 possible outcomes.
Let Event A be getting a sum of 8.
Let Event B be getting two 6's.
From inspection of the the sample space diagram:
[tex]\sf P(A) = \dfrac{5}{36}[/tex]
[tex]\sf P(B) = \dfrac{1}{36}[/tex]
Therefore:
[tex]\implies \sf P(A)\;and\;P(B)=\dfrac{5}{36} \times \dfrac{1}{36}=\dfrac{5}{1296}[/tex]
[tex]\implies \sf P(A)\;or\;P(B)=\dfrac{5}{36} +\dfrac{1}{36}=\dfrac{6}{36}=\dfrac{1}{6}[/tex]
Given the two dice are tossed 100 times.
Number of times you will get a sum of 8 from 100 tosses of two dice:
[tex]\implies 100 \times \dfrac{5}{36}=13.8888...[/tex]
Number of times you will get a two 6's from 100 tosses of two dice:
[tex]\implies 100 \times \dfrac{1}{36}=2.7777...[/tex]
Number of times you will get a sum of 8 and two 6's from 100 tosses of two dice:
[tex]\implies 100 \times \dfrac{5}{1296}=0.385802...[/tex]
Number of times you will get a sum of 8 or two 6's from 100 tosses of two dice:
[tex]\implies 100 \times \dfrac{1}{6} = 16.6666...[/tex]
