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If you play roulette, are you going to break-even? Maybe sometimes, but not if you play forever, because the expected value is negative. The expected value finds the average outcome of random events.
Let \(X\) be a uniform random variable over \(\{0, 1, \dots, 6\},\) and \(Y\) a random variable over \(\{0, 1, \dots, 4\}\) with \(P(Y=y) = \frac{y}{10}.\) If \(X\) and \(Y\) are independent, what is \( E[XY] ?\)
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If \(X\) and \(Y\) are independent random variables such that \( E[X] = 12\) and \( E[Y] = 17 ,\) what is \( E[ (X-4)( Y- 5) ] ?\)
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If \(X\) and \(Y\) are independent random variables such that \( E[X] = 11,\) \( E[Y] = 20 ,\) and \( E[ (X-c)( Y- 9) ] = 88,\) what is \(c?\)
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If \(X\) and \(Y\) are independent random variables such that \( E[X] = 15\) and \(E[ (X-4)( Y- 9) ] = 99,\) what is \( E[Y] ?\)
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If \(X, \ Y, \ Z,\) and \(W\) are independent random variables such that \( E[X] = 6,\) \( E[Y] = 5 ,\) \( E[Z] = 4 ,\) and \( E[W] = 11 ,\) what is \[ E[ (X-3)( Y- 3)( Z- 2)( W- 3) ] ?\]
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