Julie wants to randomly choose two out of three friends, Margarite, Anita, and Sasha, to go with her to a concert. In which method does one of Julie's friends have more than a fair chance of being chosen?

Julie rolls a number cube. If the number is 1 or 2. then Sasha does not go. If the number is 3 or 4, Margarite does

not go. If the number is 5 or 6. Anita does not go.

Julie flips three coins, one for each friend. If exactly two coins match, then the two corresponding friends go.

Otherwise, she continues to flip the three coins until exactly two coins match.

Julie has 5 slips of paper numbered 1 to 5. Each friend draws a slip at random, and the friends with the highest

and lowest numbers go.

Julie flips a coin. If it's tails, Margarite and Anita will go the concert. If it's heads, then Sasha will go to the concert.

Julie will then flip a coin to choose between Margarite and Anita.