Answer: -0.50 dollars
This means Miguel expects to lose on average $0.50 (aka 50 cents) per game.
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Explanation:
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.
