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. 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. b. 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. put each persons name on a separate piece of paper in a bag. randomly draw the winning name. d. 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.

Explanation:

Step1: Calculate probabilities for option A

When flipping a coin twice, there are \(2\times2 = 4\) possible outcomes: (H,H), (H,T), (T,H), (T,T).
The probability that Peter wins \(P(P)=\frac{1}{4}\), the probability that Roddy wins \(P(R)=\frac{1}{4}\), and the probability that Scott wins \(P(S)=\frac{2}{4}=\frac{1}{2}\).

Step2: Calculate probabilities for option B

When rolling a number - cube (with 6 faces), for Peter: \(P(P)=\frac{2}{6}=\frac{1}{3}\), for Roddy: \(P(R)=\frac{2}{6}=\frac{1}{3}\), for Scott: \(P(S)=\frac{2}{6}=\frac{1}{3}\).

Step3: Calculate probabilities for option C

Since there are 3 names in the bag, the probability of drawing each name is \(\frac{1}{3}\).

Step4: Calculate probabilities for option D

When rolling a number - cube, there are 3 even numbers (2, 4, 6) and 3 odd numbers (1, 3, 5). The probability that Peter wins \(P(P)=\frac{3}{6}=\frac{1}{2}\), the probability that Roddy wins \(P(R)=\frac{3}{6}=\frac{1}{2}\), and the probability that Scott wins \(P(S) = 0\).

Answer:

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