To solve for [tex]\( P(1 \text{ red}) \)[/tex], we need to understand the different outcomes and their frequencies from the experiment of drawing and replacing marbles. The table provided shows the outcomes and their respective frequencies after the experiment has been conducted 100 times.
First, let's summarize the outcomes:
- Red, Red: 19
- Red, Blue: 32
- Blue, Blue: 21
- Blue, Red: 28
The total number of trials is the sum of the frequencies of all outcomes:
[tex]\[
\text{Total trials} = 19 + 32 + 21 + 28 = 100
\][/tex]
Next, we need to determine the number of trials in which exactly one red marble was drawn. The relevant outcomes for this scenario are:
- "Red, Blue": 32 trials
- "Blue, Red": 28 trials
Sum the frequencies for these outcomes:
[tex]\[
\text{Number of trials with 1 red marble} = 32 + 28 = 60
\][/tex]
Now, we can calculate the probability [tex]\( P(1 \text{ red}) \)[/tex] by dividing the number of trials with exactly one red marble by the total number of trials:
[tex]\[
P(1 \text{ red}) = \frac{\text{Number of trials with 1 red marble}}{\text{Total trials}} = \frac{60}{100} = 0.6
\][/tex]
Thus, the probability of drawing exactly one red marble, [tex]\( P(1 \text{ red}) \)[/tex], is [tex]\( 0.6 \)[/tex] or [tex]\( 60\% \)[/tex].
