MATH
121616949-math.225

# 121616949-math.225 - 9.8 Probability 211 topping the list...

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9.8 Probability 211 topping the list at about 6 million. The sum of all rolls would be 1 million times 2 plus 2 million times 3, and so on, and dividing by 36 million we would get the average: ¯ x = (2 · 10 6 + 3(2 · 10 6 ) + · · · + 7(6 · 10 6 ) + · · · + 12 · 10 6 ) 1 36 · 10 6 = 2 10 6 36 · 10 6 + 3 2 · 10 6 36 · 10 6 + · · · + 7 6 · 10 6 36 · 10 6 + · · · + 12 10 6 36 · 10 6 = 2 P (2) + 3 P (3) + · · · + 7 P (7) + · · · + 12 P (12) = 12 X i =2 iP ( i ) = 7 . It is apparent that there is nothing special about the 36 million in this calculation. No matter what the number of rolls, once we simplify the average, we get the same 12 X i =2 iP ( i ). A variable, say X , that can take certain values, each with a corresponding probability, is called a random variable ; in the example above, the random variable was the sum of the two dice. If the possible values for X are x 1 , x 2 , . . . , x n , then the expected value of the random variable is E ( X ) = n X i =1 x i P ( x i ). In many applications of probability, the number of possible values of a random vari- able is very large, perhaps even infinite. To deal with the infinite case we need a different
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