To determine the probability distribution for who gets the car, we need to calculate the probability that each child (Peyton, Cameron, Dakota) gets the car when a single, fair, six-sided number cube (die) is rolled.
1. Identify the favorable outcomes for each child:
- Peyton: Peyton gets the car if a 1 or 2 is rolled, so there are 2 favorable outcomes.
- Cameron: Cameron gets the car if a 3 or 4 is rolled, so there are 2 favorable outcomes.
- Dakota: Dakota gets the car if a 5 or 6 is rolled, so there are 2 favorable outcomes.
2. Total number of possible outcomes:
- Since it is a fair six-sided die, there are 6 possible outcomes in total (1 through 6).
3. Calculate the probability for each child:
- Probability of Peyton getting the car:
[tex]\[
P(\text{Peyton}) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{2}{6} = \frac{1}{3}
\][/tex]
- Probability of Cameron getting the car:
[tex]\[
P(\text{Cameron}) = \frac{2}{6} = \frac{1}{3}
\][/tex]
- Probability of Dakota getting the car:
[tex]\[
P(\text{Dakota}) = \frac{2}{6} = \frac{1}{3}
\][/tex]
4. Convert probabilities to decimal form:
- The probability [tex]\(\frac{1}{3}\)[/tex] in decimal form is approximately 0.3333.
Thus, the probability distribution for Peyton, Cameron, and Dakota each getting the car is approximately [tex]\((0.333, 0.333, 0.333)\)[/tex].
Given the options, the correct probability distribution for who gets the car is:
\begin{tabular}{|c|c|c|}
\hline Peyton & Cameron & Dakota \\
\hline[tex]$\frac{1}{3}$[/tex] & [tex]$\frac{1}{3}$[/tex] & [tex]$\frac{1}{3}$[/tex] \\
\hline
\end{tabular}
