Let's solve the problem step by step.
First, let's define some terms:
- Ryan draws two tiles without replacement. This means that after Ryan draws the first tile, he does not put it back in the bag before drawing the second tile.
- Curtis draws two tiles with replacement. This means that after Curtis draws the first tile, he puts it back in the bag before drawing the second tile.
Next, we need to determine the probability for each of these draws.
### Ryan's Draw (Without Replacement)
1. Probability for the First Tile to be a Vowel:
There are 5 vowels in the alphabet (A, E, I, O, U) out of the 26 letters.
The probability of drawing a vowel first is:
[tex]\[
\frac{5}{26} \approx 0.1923
\][/tex]
2. Probability for the Second Tile to be a Vowel given the First Tile was a Vowel:
After drawing the first vowel, there are now 4 vowels left in a total of 25 tiles.
The probability of drawing a vowel next is:
[tex]\[
\frac{4}{25} = 0.16
\][/tex]
3. Combined Probability for Both Tiles to be Vowels (Multiplication Rule):
We multiply the probabilities of each individual event happening:
[tex]\[
0.1923 \times 0.16 \approx 0.0308
\][/tex]
Therefore, the probability that the two tiles Ryan selects will both be vowels is approximately \(0.0308\).
### Curtis' Draw (With Replacement)
1. Probability for the First Tile to be a Vowel:
As before, the probability of drawing a vowel first is:
[tex]\[
\frac{5}{26} \approx 0.1923
\][/tex]
2. Probability for the Second Tile to be a Vowel (Same as First, Due to Replacement):
Since Curtis puts the first tile back, the probability of drawing a vowel again remains:
[tex]\[
\frac{5}{26} \approx 0.1923
\][/tex]
3. Combined Probability for Both Tiles to be Vowels (Multiplication Rule):
We again multiply the probabilities of each individual event happening:
[tex]\[
0.1923 \times 0.1923 \approx 0.037
\][/tex]
Therefore, the probability that the two tiles Curtis selects will both be vowels is approximately \(0.037\).
### Completing the Statements
1. When Ryan selects his tiles, selecting the first tile and selecting the second tile are dependent events. (Because the outcome of the first draw affects the second)
2. The probability that the two tiles Ryan selects will both be vowels is approximately 0.0308.
3. The probability that the two tiles Curtis selects will both be vowels is approximately 0.037.
