Question

question 3
1 pts
suppose the 9 letters forming the word cafeteria were each placed on a ticket and put in a box. if two letters were drawn from the box (without replacement between draws), what is the probability that you end up with at least one vowel (aeiou)?
(hint: what is the complement of at least one vowel?)
0.167
0.198
0.691
0.833
0.722
0.802
question 4
1 pts
event a: earning an a on all three statistics tests
event b: earning an a in your statistics class
are these two events dependent or independent?
dependent
independent
question 5
1 pts

Explanation:

Question 3

Step 1: Identify vowels and consonants in "CAFETERIA"

The word "CAFETERIA" has 9 letters. Vowels (A, E, I, O, U) in it: C, A, F, E, T, E, R, I, A. So vowels are A, E, E, I, A (5 vowels) and consonants are C, F, T, R (4 consonants).

Step 2: Find the complement event (no vowels)

The complement of "at least one vowel" is "no vowels" (both drawn are consonants). The number of ways to choose 2 consonants from 4 is \( C(4,2) \), and the number of ways to choose 2 letters from 9 is \( C(9,2) \).

The combination formula is \( C(n,k)=\frac{n!}{k!(n - k)!} \). So \( C(4,2)=\frac{4!}{2!2!}=\frac{4\times3}{2\times1}=6 \), and \( C(9,2)=\frac{9!}{2!7!}=\frac{9\times8}{2\times1}=36 \).

The probability of no vowels is \( \frac{C(4,2)}{C(9,2)}=\frac{6}{36}=\frac{1}{6}\approx0.1667 \).

Step 3: Calculate the probability of at least one vowel

Using the complement rule: \( P(\text{at least one vowel}) = 1 - P(\text{no vowels}) \). So \( 1 - \frac{1}{6}=\frac{5}{6}\approx0.833 \)? Wait, no, wait. Wait, consonants are 4? Wait, wait, let's re - check the letters. "CAFETERIA" letters: C, A, F, E, T, E, R, I, A. So vowels: A, E, E, I, A (that's 5 vowels: positions 2,4,6,8,9). Consonants: C (1), F (3), T (5), R (7) → 4 consonants. So choosing 2 consonants: \( C(4,2) = 6 \), total ways \( C(9,2)=36 \). So \( P(\text{no vowels})=\frac{6}{36}=\frac{1}{6}\approx0.1667 \). Then \( P(\text{at least one vowel}) = 1 - 0.1667\approx0.833 \). Wait, but let's check again. Wait, maybe I miscounted vowels. Wait, A, E, E, I, A: that's 5 vowels. Consonants: C, F, T, R: 4. So yes. So the probability is \( 1-\frac{C(4,2)}{C(9,2)} = 1-\frac{6}{36}=1 - \frac{1}{6}=\frac{5}{6}\approx0.833 \).

Wait, but let's check the options. The options include 0.833. So that's the answer.

Event A is earning an A on all three statistics tests, and Event B is earning an A in the statistics class. Earning an A in the class is likely dependent on earning A's on the tests (since the tests are part of the class grade). So the outcome of Event A affects the outcome of Event B, meaning the events are dependent.

Answer:

0.833

Question 4