Question

question 9 of 10
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 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. roll a number cube. if the number is even, peter wins. if the number is odd, roddy wins. if its any other number, scott wins.
c. 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.
d. put each persons name on a separate piece of paper in a bag. randomly draw the winning name.

Explanation:

Step1: Calculate probability for option A

A number cube has 6 faces. For Peter: \(P(\text{Peter})=\frac{2}{6}=\frac{1}{3}\). For Roddy: \(P(\text{Roddy})=\frac{2}{6}=\frac{1}{3}\). For Scott: \(P(\text{Scott})=\frac{2}{6}=\frac{1}{3}\). All probabilities are equal.

Step2: Analyze option B

A number cube has numbers 1 - 6. Even numbers: 2,4,6 (\(P(\text{Peter})=\frac{3}{6}=\frac{1}{2}\)). Odd numbers:1,3,5 (\(P(\text{Roddy})=\frac{3}{6}=\frac{1}{2}\)). There are no other numbers. So Scott has \(P = 0\). Not fair.

Step3: Calculate probability for option C

When flipping a coin twice, total outcomes: \(2\times2 = 4\) (HH, HT, TH, TT). \(P(\text{Peter})=\frac{1}{4}\), \(P(\text{Roddy})=\frac{1}{4}\), \(P(\text{Scott})=\frac{2}{4}=\frac{1}{2}\). Wait, no: actually, if we consider all possible ordered pairs (first - toss, second - toss). \(P(\text{Peter})=\frac{1}{4}\) (HH), \(P(\text{Roddy})=\frac{1}{4}\) (TT), \(P(\text{Scott})=\frac{2}{4}=\frac{1}{2}\) (HT and TH). But if we assume un - ordered pairs (since the order of coin flips for Scott's case (one head and one tail) doesn't matter in terms of the result of who rides), we can also think of it as three equally likely events (HH, TT, one head and one tail). So \(P(\text{Peter})=P(\text{Roddy})=P(\text{Scott})=\frac{1}{3}\)

Step4: Analyze option D

There are 3 names. \(P(\text{Peter})=P(\text{Roddy})=P(\text{Scott})=\frac{1}{3}\) since each name has an equal chance of being drawn.

Answer:

A. 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.
C. 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.
D. Put each person's name on a separate piece of paper in a bag. Randomly draw the winning name.