Sure! Let's work through the problem step-by-step.
### Part A: Completing the table
We know that:
- The total number of days is [tex]\( 100 \)[/tex].
- On 50% of the days, Camila rides her bike.
- On 20% of the days, Camila goes running.
- Running and biking are mutually exclusive events, meaning she does not run and bike on the same day.
Let's fill in the table with this information.
1. Calculate the number of days Camila bikes and runs:
- Number of days Camila bikes: [tex]\( 100 \times 0.50 = 50 \)[/tex] days.
- Number of days Camila runs: [tex]\( 100 \times 0.20 = 20 \)[/tex] days.
2. Calculate the number of days she neither bikes nor runs:
- Number of days either activity occurs: [tex]\(50 + 20 = 70 \)[/tex] days.
- Number of days neither activity occurs: [tex]\(100 - 70 = 30 \)[/tex] days.
3. Fill in the table:
According to the information above:
- Days she runs: 20
- Days she bikes: 50
- Days she neither runs nor bikes: 30
Thus, the filled-in table should look like this:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline & \multicolumn{2}{|c|}{ Bikes } & \multicolumn{2}{|c|}{ Does not bike } & \multicolumn{2}{|c|}{ Total } \\
\hline Runs & 0 & 0 & 20 & 20 & 20 & [tex]$\checkmark$[/tex] \\
\hline Does not run & 50 & 50 & 30 & 30 & 80 & [tex]$\checkmark$[/tex] \\
\hline Total & 50 & [tex]$\checkmark$[/tex] & 50 & [tex]$\checkmark$[/tex] & 100 & [tex]$\checkmark$[/tex] \\
\hline
\end{tabular}
### Part B: Probability of Running OR Biking
The probability of Camila running or biking on a randomly selected day can be calculated by adding the probabilities since they are mutually exclusive:
[tex]\[ P(\text{Run or Bike}) = P(\text{Run}) + P(\text{Bike}) = 0.20 + 0.50 = 0.70 \][/tex]
Therefore, the probability of running or biking on any given day is [tex]\(0.7\)[/tex].
### Part C: Probability of Running Given She Bikes
Since running and biking are mutually exclusive (Camila cannot do both on the same day), the probability that she goes running given she has already gone on a bike ride is [tex]\(0\)[/tex]:
[tex]\[ P(\text{Run | Bike}) = 0 \][/tex]
This step-by-step solution adheres to the rules of mutually exclusive events, providing probabilities based on the given conditions.
