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IE111_F09_Exam2a_solutions

# IE111_F09_Exam2a_solutions - IE 111 9:00 Section Exam 2...

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IE 111 Solutions 9:00 Section Exam 2 Fall 2009 Question 1 Three flips of a biased coin are conducted in which the probability of a head is 0.4. Let X 1 = the number of heads in the first flip (i.e. flip number 1). Let X 2 = the number of heads in the last two flips (i.e. flip numbers 2 and 3). Let X 3 = the number of heads in the last flip (i.e. flip number 3). a) Find P(X 1 +X 2 <2) This happens exactly when there is 0 or 1 head on the three flips together. The probability is thus 0.6 3 3 × 0.6 2 × 0.4 = 0.648 . b) Find E(X 1 +X 2 +X 3 ) By the properties of the mean, we can write E(X 1 +X 2 +X 3 )=EX 1 +EX 2 +EX 3 , which we can compute by terms. EX 1 =0.4=EX 3 , and EX 2 =0.8, since X 2 is a binomial random variable with N=2, p=0.4, and the expected value is Np. So the some of the three terms is 1.6. c) Find P(X 1 +X 2 +X 3 <3) There are only 8 possibilities for the coin flips, so we can lsit all of them together with the value of X 1 +X 2 +X 3 and the probability of that combination.

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