(I'll give brainliest)
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. What is Miguel’s expected value from playing the game?

Let's consider each selection to be a different slot to fill. There are 4 chips to choose from for the first slot. Then we have 3 left for the second slot. That gives 4*3 = 12 permutations if order mattered. However, order doesn't matter, so we'll divide that by 2 to get 12/2 = 6.

There are 6 ways to select the two chips from a pool of four total.

Out of those 6 ways, there's only one way to win money: that's to select the two "1"s. So if X = 2, then P(X) = 1/6 to represent a 1/6 chance of winning $2. Otherwise, P(X) = 5/6. The two fractions 1/6 and 5/6 add to 1 to represent 100%

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Based on that, we can form this table:

[tex]\begin{array}{c|c}\textbf{X} & \textbf{P(X)}\\2 & 1/6\\-1 & 5/6\\\end{array}[/tex]

Let's add a third column where we multiply the two original columns

[tex]\begin{array}{c|c|c}\textbf{X} & \textbf{P(X)} & \textbf{X*P(X)}\\2 & 1/6 & 2/6\\-1 & 5/6 & -5/6\\\end{array}[/tex]

Adding the stuff in the third column gets us: (2/6)+(-5/6) = -0.50

He should expect to lose about 50 cents per game on average.