Question

mr. piper is driving peter, roddy, and scott home from school. all of them want to ride in the front seat. how can he make a fair decision about who rides in the front seat?
select all of the correct answers.
a. roll a number cube. if the number is even, peter wins. if the number is odd, roddy wins. if it’s any other number, scott wins.
b. flip a coin twice. if both tosses are heads, peter wins. if both tosses are tails, roddy wins. if one is heads and one is tails, scott wins.
c. put each person’s name on a separate piece of paper in a bag. randomly draw the winning name.
d. roll a number cube. if it lands on 1 or 2, peter wins. if it lands on 3 or 4, roddy wins. if it lands on 5 or 6, scott wins.

Explanation:

To determine a fair decision, each person should have an equal probability of winning.

Step1: Analyze Option A

A number cube has 6 faces (1 - 6). Even numbers: 2, 4, 6 (3 outcomes), odd numbers: 1, 3, 5 (3 outcomes). But "any other number" is impossible (all numbers are even or odd), so Scott has 0 probability. Not fair.

Step2: Analyze Option B

Flipping a coin twice has 4 outcomes: HH, HT, TH, TT. HH (Peter): 1 outcome, TT (Roddy): 1 outcome, HT/TH (Scott): 2 outcomes. Probabilities: Peter = 1/4, Roddy = 1/4, Scott = 2/4 = 1/2. Not equal. Not fair.

Step3: Analyze Option C

Putting 3 names (Peter, Roddy, Scott) in a bag and drawing one. Each has a 1/3 chance. Fair.

Step4: Analyze Option D

Number cube: 1 - 6. Peter: 1,2 (2 outcomes), Roddy: 3,4 (2 outcomes), Scott: 5,6 (2 outcomes). Each has 2/6 = 1/3 chance. Fair.

Answer:

C. Put each person’s name on a separate piece of paper in a bag. Randomly draw the winning name.
D. Roll a number cube. If it lands on 1 or 2, Peter wins. If it lands on 3 or 4, Roddy wins. If it lands on 5 or 6, Scott wins.
B. Flip a coin twice. If both tosses are heads, Peter wins. If both tosses are tails, Roddy wins. If one is heads and one is tails, Scott wins. (Wait, no, earlier analysis for B was wrong? Wait, no: HH (1), TT (1), HT and TH (2). So Peter: 1/4, Roddy: 1/4, Scott: 2/4 = 1/2. So B is not fair. Wait, my mistake. Let's re - check:

Wait, the problem says "select all correct". Let's re - evaluate:

Option A: Scott has 0 chance. Eliminate.

Option B: Outcomes: HH, HT, TH, TT. Peter: HH (1/4), Roddy: TT (1/4), Scott: HT, TH (2/4 = 1/2). So probabilities are not equal. So B is not fair.

Option C: 3 names, each has 1/3 chance. Fair.

Option D: 1 - 6. Peter: 1,2 (2/6 = 1/3), Roddy: 3,4 (2/6 = 1/3), Scott: 5,6 (2/6 = 1/3). Fair.

So correct options are C and D.

Wait, let's re - check Option B again. The problem says "select all correct". Let's see: when flipping a coin twice, the possible outcomes are HH, HT, TH, TT. So:

• Peter: HH (probability 1/4)
• Roddy: TT (probability 1/4)
• Scott: HT or TH (probability 2/4 = 1/2)

Since 1/4 ≠ 1/2, the probabilities are not equal, so B is not a fair method.

Option A: A number cube has numbers 1 - 6. Even numbers: 2,4,6 (3 numbers), odd numbers: 1,3,5 (3 numbers). The option says "if it's any other number, Scott wins" but there are no other numbers (all are even or odd), so Scott's probability is 0. Not fair.

Option C: 3 names, each on a paper. Drawing one: each has probability 1/3. Fair.

Option D: Number cube: 1 - 6. Peter: 1,2 (2 numbers), Roddy: 3,4 (2 numbers), Scott: 5,6 (2 numbers). So each has 2/6 = 1/3 probability. Fair.

So the correct options are C and D.