Jake, Daniel, Timmy, Amy, and Pam are going on a fishing trip. They are taking a pick-up truck which only holds 3 33 people inside the truck. The other 2 22 people will have to ride in the back of the truck. No one wants to ride in the back. They decide to let fate determine who has to ride in the back by using this week's winning lottery ticket, which is about to be announced. The winning lottery ticket is a list of 5 55 numbers. Each number is a different random integer from 1 11 to 1000 10001000. Match each method for deciding who rides in the back with the correct assessment of its fairness. (Consider a system to be fair when the probabilities of each event are equal.) Method 1 Method 1start text, M, e, t, h, o, d, space, end text, 1: Of the five numbers in the ticket, if all 5 55 numbers are even, 0 00 of them are even, or exactly 1 11 is even, Jake rides in back. If 1 11 of them is even or 2 22 of them are even, Daniel rides in back. If 2 22 are even or 3 33 are even, Timmy rides in back. If 3 33 are even or 4 44 are even, Amy rides in back. If there are 4 44 even numbers, all 5 55 numbers are even, or none of them are even, Pam rides in back. Method 2 Method 2start text, M, e, t, h, o, d, space, end text, 2: Jake is assigned the first number that is read. Daniel is assigned the second number. Timmy gets the third number. Amy gets the fourth number. Pam gets the fifth number. The two people with the smallest numbers ride in the back. Method 3 Method 3start text, M, e, t, h, o, d, space, end text, 3: If the first number is higher than the second number, Jake rides in back. If the second number is higher than the first number, Daniel rides in back. Of the final 3 33 numbers, if the third number is highest, Timmy rides in back, if the fourth number is highest, Amy rides in back, and if the fifth number is highest, Pam rides in back.