An ordinary (fair) coin is bossed 3 times. Outcomes are thus triples of \"heads\

Question

An ordinary (fair) coin is bossed 3 times. Outcomes are thus triples of "heads" (h) "tails"(t) which we write For each outcome. Let R be the random variable counting the number of heads in each outcome for example, if the outcome is ttt, then R (ttt) = 0. Suppose that the random variable X is defined in terms or R as follows X = 2 R - 2 R - 2 R2 - 1. The values of X are thus. Calculate the probability distribution functions of X, I e the function px (x). First fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row.

Explanation / Answer

Solution: The only possible values of x are: x = -13, x = -5 and, x = -1 There are no other possible values for x. So the top row of the table will simply have the values -13, -5, and -1 placed in each box respectively Now count how many outcomes occur in the filled out table given above. There are 8 total possible outcomes. Out of those 8 outcomes, the is only one way to get a -13 (and it's when you get hhh) So P(X = -13) = 1/8 (the fraction 1 over 8) So the value that goes in the first box in row 2 is the fraction 1/8 ----------- Now count the number of -5s that occur. There are three instances of -5. So P(X = -5) = 3/8 So the number that goes in the second box of row 2 is the fraction 3/8 ----------- Finally, you can either count that there are 4 instances of -1 or you can use the idea that the sum of the probabilities must be equal to 1. In any event, you'll then get P(X = -1) = 4/8 = 1/2 So the number that goes in the last box in the third row is the fraction 1/2

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