Asked by karol
Tracy has a set of 26 letter tiles, one for each letter in the alphabet, including 5 vowels and 21 consonants. She put them in a bag, and will pull out 3 tiles without looking.
Which statements are true? Select two that are correct.
A.
There are 10 ways to choose 3 vowels.
B.
There are 60 ways to choose 3 vowels.
C.
There are 2,600 ways to choose 3 letters.
D.
There are 15,600 ways to choose 3 letters.
E.
The probability that all 3 letters will be vowels is approximately 0.02308.
Which statements are true? Select two that are correct.
A.
There are 10 ways to choose 3 vowels.
B.
There are 60 ways to choose 3 vowels.
C.
There are 2,600 ways to choose 3 letters.
D.
There are 15,600 ways to choose 3 letters.
E.
The probability that all 3 letters will be vowels is approximately 0.02308.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To determine which statements about the letter tiles are true, let's evaluate each one based on the combinations and probability involved:
We have 5 vowels (A, E, I, O, U) and 21 consonants (the rest of the letters).
**A. There are 10 ways to choose 3 vowels.**
- To choose 3 vowels from 5, we use the combination formula:
\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
where \( n = 5 \) (vowels), and \( r = 3 \):
\[
\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10
\]
**This statement is true.**
**B. There are 60 ways to choose 3 vowels.**
- As calculated above, there are 10 ways to choose 3 vowels, not 60.
**This statement is false.**
**C. There are 2,600 ways to choose 3 letters.**
- To find ways to choose 3 letters from the total of 26, we use:
\[
\binom{26}{3} = \frac{26!}{3!(26-3)!} = \frac{26 \times 25 \times 24}{3 \times 2 \times 1} = \frac{15600}{6} = 2600
\]
**This statement is false.**
**D. There are 15,600 ways to choose 3 letters.**
- Using the combination calculation:
\[
\binom{26}{3} = 15600
\]
**This statement is true.**
**E. The probability that all 3 letters will be vowels is approximately 0.02308.**
- The total ways to choose 3 letters is 15600 (as calculated). The ways to choose 3 vowels is 10. So, the probability is:
\[
P(\text{3 vowels}) = \frac{\text{Ways to choose 3 vowels}}{\text{Total ways to choose 3 letters}} = \frac{10}{15600} = \frac{1}{1560} \approx 0.000641
\]
This is far less than 0.02308.
**This statement is false.**
The two correct statements are:
- **A. There are 10 ways to choose 3 vowels.**
- **D. There are 15,600 ways to choose 3 letters.**
We have 5 vowels (A, E, I, O, U) and 21 consonants (the rest of the letters).
**A. There are 10 ways to choose 3 vowels.**
- To choose 3 vowels from 5, we use the combination formula:
\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]
where \( n = 5 \) (vowels), and \( r = 3 \):
\[
\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10
\]
**This statement is true.**
**B. There are 60 ways to choose 3 vowels.**
- As calculated above, there are 10 ways to choose 3 vowels, not 60.
**This statement is false.**
**C. There are 2,600 ways to choose 3 letters.**
- To find ways to choose 3 letters from the total of 26, we use:
\[
\binom{26}{3} = \frac{26!}{3!(26-3)!} = \frac{26 \times 25 \times 24}{3 \times 2 \times 1} = \frac{15600}{6} = 2600
\]
**This statement is false.**
**D. There are 15,600 ways to choose 3 letters.**
- Using the combination calculation:
\[
\binom{26}{3} = 15600
\]
**This statement is true.**
**E. The probability that all 3 letters will be vowels is approximately 0.02308.**
- The total ways to choose 3 letters is 15600 (as calculated). The ways to choose 3 vowels is 10. So, the probability is:
\[
P(\text{3 vowels}) = \frac{\text{Ways to choose 3 vowels}}{\text{Total ways to choose 3 letters}} = \frac{10}{15600} = \frac{1}{1560} \approx 0.000641
\]
This is far less than 0.02308.
**This statement is false.**
The two correct statements are:
- **A. There are 10 ways to choose 3 vowels.**
- **D. There are 15,600 ways to choose 3 letters.**
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