Answer :
Let's go through the problem step-by-step to fill in the blanks correctly.
### Step 1: Determine the type of events for Ryan's selection
Ryan selects two tiles without replacement. This means that when he selects the second tile, the total number of tiles in the bag has decreased by one, and possibly the number of vowels as well. Therefore, these selections are dependent events.
When Ryan selects his tiles, selecting the first tile and selecting the second tile are dependent events.
### Step 2: Calculate the probability that both tiles selected by Ryan will be vowels
Given:
- Total number of tiles = 26 (each letter of the alphabet).
- Number of vowels = 5.
The probability that the first tile Ryan selects is a vowel is:
[tex]\[ \text{Probability of selecting a vowel (first tile)} = \frac{5}{26} \approx 0.1923 \][/tex]
If the first tile selected is a vowel, there are now 25 tiles left, and 4 of those are vowels. The probability that the second tile selected is also a vowel, under this condition, is:
[tex]\[ \text{Probability of selecting a vowel (second tile, after the first is a vowel)} = \frac{4}{25} = 0.16 \][/tex]
Therefore, the probability that both of Ryan's tiles will be vowels is:
[tex]\[ \text{Total probability (Ryan)} = \frac{5}{26} \times \frac{4}{25} = 0.0308 \][/tex]
### Step 3: Calculate the probability that both tiles selected by Curtis will be vowels
Curtis selects a tile, replaces it, and then selects another tile. This means each selection is independent.
The probability that the first tile Curtis selects is a vowel is:
[tex]\[ \text{Probability of selecting a vowel (first tile)} = \frac{5}{26} \approx 0.1923 \][/tex]
Since the tile is replaced, the total number of tiles and the number of vowels remain the same for the second selection. Thus, the probability that the second tile Curtis selects is also a vowel is:
[tex]\[ \text{Probability of selecting a vowel (second tile)} = \frac{5}{26} \approx 0.1923 \][/tex]
Therefore, the probability that both of Curtis's tiles will be vowels is:
[tex]\[ \text{Total probability (Curtis)} = \frac{5}{26} \times \frac{5}{26} \approx 0.037 \][/tex]
### Step 4: Compare the probabilities
The probability that both of Ryan’s tiles are vowels is approximately [tex]\( 0.0308 \)[/tex] and the probability that both of Curtis’s tiles are vowels is approximately [tex]\( 0.037 \)[/tex]. Therefore, we can conclude that:
[tex]\[ 0.0308 \text{ (Ryan's probability)} < 0.037 \text{ (Curtis's probability)} \][/tex]
So, the probability that the two tiles Ryan selects will both be vowels is less than the probability that the two tiles Curtis selects will both be vowels.
### Final Solution
Combining all the steps and correct answers:
When Ryan selects his tiles, selecting the first tile and selecting the second tile are dependent events.
The probability that the two tiles Ryan selects will both be vowels is less than the probability that the two tiles Curtis selects will both be vowels.
### Step 1: Determine the type of events for Ryan's selection
Ryan selects two tiles without replacement. This means that when he selects the second tile, the total number of tiles in the bag has decreased by one, and possibly the number of vowels as well. Therefore, these selections are dependent events.
When Ryan selects his tiles, selecting the first tile and selecting the second tile are dependent events.
### Step 2: Calculate the probability that both tiles selected by Ryan will be vowels
Given:
- Total number of tiles = 26 (each letter of the alphabet).
- Number of vowels = 5.
The probability that the first tile Ryan selects is a vowel is:
[tex]\[ \text{Probability of selecting a vowel (first tile)} = \frac{5}{26} \approx 0.1923 \][/tex]
If the first tile selected is a vowel, there are now 25 tiles left, and 4 of those are vowels. The probability that the second tile selected is also a vowel, under this condition, is:
[tex]\[ \text{Probability of selecting a vowel (second tile, after the first is a vowel)} = \frac{4}{25} = 0.16 \][/tex]
Therefore, the probability that both of Ryan's tiles will be vowels is:
[tex]\[ \text{Total probability (Ryan)} = \frac{5}{26} \times \frac{4}{25} = 0.0308 \][/tex]
### Step 3: Calculate the probability that both tiles selected by Curtis will be vowels
Curtis selects a tile, replaces it, and then selects another tile. This means each selection is independent.
The probability that the first tile Curtis selects is a vowel is:
[tex]\[ \text{Probability of selecting a vowel (first tile)} = \frac{5}{26} \approx 0.1923 \][/tex]
Since the tile is replaced, the total number of tiles and the number of vowels remain the same for the second selection. Thus, the probability that the second tile Curtis selects is also a vowel is:
[tex]\[ \text{Probability of selecting a vowel (second tile)} = \frac{5}{26} \approx 0.1923 \][/tex]
Therefore, the probability that both of Curtis's tiles will be vowels is:
[tex]\[ \text{Total probability (Curtis)} = \frac{5}{26} \times \frac{5}{26} \approx 0.037 \][/tex]
### Step 4: Compare the probabilities
The probability that both of Ryan’s tiles are vowels is approximately [tex]\( 0.0308 \)[/tex] and the probability that both of Curtis’s tiles are vowels is approximately [tex]\( 0.037 \)[/tex]. Therefore, we can conclude that:
[tex]\[ 0.0308 \text{ (Ryan's probability)} < 0.037 \text{ (Curtis's probability)} \][/tex]
So, the probability that the two tiles Ryan selects will both be vowels is less than the probability that the two tiles Curtis selects will both be vowels.
### Final Solution
Combining all the steps and correct answers:
When Ryan selects his tiles, selecting the first tile and selecting the second tile are dependent events.
The probability that the two tiles Ryan selects will both be vowels is less than the probability that the two tiles Curtis selects will both be vowels.
