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Unformatted text preview: Lecture 8. Random Variables (continued), Expected Value, Variance Example 4. A biased coin will land on heads with probability p . Let us flip such a coin k times. An outcome is a sequence k of H s and T s. Now, the outcomes are not equally probable. We get all H s with probability p k , and all T s with probability (1- p ) k . A moment of thought shows that the probability of an outcome depends on k , p and the number x of heads, but not on order in which the x heads and k- x tails occur. Indeed, the probability of any such outcome is p x (1- p ) k- x . Now, there are ( k x ) ways to place x heads in a sequence of length k , so the probability of flipping x heads in a sequence of k flips is: b ( x, k, p ) = parenleftbigg k x parenrightbigg p x (1- p ) k- x . If we define X to be the random variable that counts the number of heads in a sequence of k flips of a biased coini.e., X ( ) = the number of heads in then P ( X = x ) = b ( x, k, p ). Problems 4. (See Mathematica Notebook 8.) a) In a Mathematica notebook, define a Mathematica function that calculates b ( x, k, p ), the probability of getting x heads, in k flips of a coin that has probability p of landing on heads. (You should view b ( x, k, p ) as a descriptive...
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