Question

1. janna has white, blue, and pink shirts. she has black, white, and beige pants. she has black and brown shoes. if she chooses one shirt, one pair of pants, and one pair of shoes, how many unique outfits can she make?

a. 6
b. 8
c. 9
d. 18

1. darwin has a spinner and a fair number cube, as shown.

spinner and number cube images omitted
if darwin spins the spinner and rolls the cube at the same time, what is the probability of the spinner landing on the number 1 and the number cube landing on a 6?
a. \\(\frac{1}{14}\\)
b. \\(\frac{1}{30}\\)
c. \\(\frac{1}{12}\\)
d. \\(\frac{1}{18}\\)

1. mike, nancy, orlando, and paul are running in a race. in how many ways can the four friends finish the race?

a. 4
b. 6
c. 12
d. 24

Explanation:

Question 4 (Jamie's Outfits)

Step 1: Count number of shirts

Jamie has 3 shirts (white, blue, pink).

Step 2: Count number of pants

Jamie has 3 pants (black, white, orange).

Step 3: Count number of shoes

Jamie has 2 shoes (black, brown).

Step 4: Apply multiplication principle

Total outfits = Shirts × Pants × Shoes = \( 3 \times 3 \times 2 \)
\( 3 \times 3 = 9 \), \( 9 \times 2 = 18 \)? Wait, no, wait the options have 18 as D, but wait let me recheck. Wait, the problem says "black, white, and orange pants" – 3 pants. Shirts: white, blue, pink – 3 shirts. Shoes: black and brown – 2 shoes. So 3×3×2=18. But wait the options: A.6, B.8, C.9, D.18. So D is 18. Wait, but maybe I misread. Wait the original problem: "She has white, blue, and pink shirts (3), black, white, and orange pants (3), black and brown shoes (2). So 3×3×2=18. So answer D.

Question 5 (Darwin's Probability)

Step 1: Probability spinner lands on 1

Spinner has 5 sections (1,2,3,4,5), so probability \( P(\text{spinner}=1) = \frac{1}{5} \)

Step 2: Probability cube lands on 6

Fair number cube (die) has 6 faces, so probability \( P(\text{cube}=6) = \frac{1}{6} \)

Step 3: Multiply probabilities (independent events)

\( P(\text{spinner}=1 \cap \text{cube}=6) = \frac{1}{5} \times \frac{1}{6} = \frac{1}{30} \)
Wait the options: A. \( \frac{1}{15} \), B. \( \frac{1}{30} \), C. \( \frac{1}{11} \), D. \( \frac{1}{18} \). So B is \( \frac{1}{30} \).

Question 6 (Race Finish Order)

Step 1: Recognize permutation of 4 people

Number of ways to arrange \( n \) distinct objects is \( n! \) (n factorial). Here \( n = 4 \) (Mike, Nancy, Orlando, Paul).

Step 2: Calculate \( 4! \)

\( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
Options: A.4, B.6, C.12, D.24. So D is 24.

Question 4 Answer: D. 18

Question 5 Answer: B. \( \frac{1}{30} \)

Question 6 Answer: D. 24

Answer:

Step 1: Recognize permutation of 4 people

Number of ways to arrange \( n \) distinct objects is \( n! \) (n factorial). Here \( n = 4 \) (Mike, Nancy, Orlando, Paul).

Step 2: Calculate \( 4! \)

\( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
Options: A.4, B.6, C.12, D.24. So D is 24.

Question 4 Answer: D. 18

Question 5 Answer: B. \( \frac{1}{30} \)

Question 6 Answer: D. 24