Miguel is playing a game in which a box contains four chips with numbers written on them. Two of the chips have the number 1, one chip has the number 3, and the other chip has the number 5. Miguel must choose two chips, and if both chips have the same number, he wins $2. If the two chips he chooses have different numbers, he loses $1 (–$1). Let X = the amount of money Miguel will receive or owe. Fill out the missing values in the table. (Hint: The total possible outcomes are six because there are four chips and you are choosing two of them.) Xi 2 –1 P(xi) What is Miguel’s expected value from playing the game?

Miguel's choose the best way to win the 2 dollars by pulling the two chips with the number 1. In total, there are four balls, so his probability of winning is:

[tex]\to \frac{2}{4} \times \frac{1}{3}= \frac{1}{6}[/tex]

The chances of losing a dollar add up to that amount, so 5/6.

Price predicted = [tex]\frac{1}{6} \times (2) + \frac{5}{6} \times (-1)[/tex]

[tex]=\frac{1}{6} \times 2 + \frac{5}{6} \times -1\\\\=\frac{1}{3} - \frac{5}{6} \\\\=\frac{2-5}{6}\\\\=\frac{-3}{6}\\\\=- \frac{1}{2}\\[/tex]

The final answer is "[tex]-\frac{1}{2}[/tex]"