316_F07_HW3_Sol - IOE 316 Fall 2007 Homework 3 Solutions...

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Unformatted text preview: IOE 316 Fall 2007 Homework 3 Solutions Due November 13, 2007 1. (25 points) Coin 1 comes up heads with probability 0.7 and coin 2 with probability 0.5. A coin is continually flipped until it comes up tails, at which time that coin is put aside and we start flipping the other one. (a) (10 points) What portion of flips use coin 1? (b) (5 points) If we start the process with coin 2, what is the expected number of flips until we go back to flipping coin 2? (c) (10 points) If we start the process with coin 1, what is the probability that coin 2 is used on the sixth flip? SOLUTION: (a) Let state 0 represent flipping coin 1 and state 1 represent flipping coin 2. Then the transition probability matrix for this system is P = . 7 0 . 3 . 5 0 . 5 . To find the steady state distribution for this system, we need to solve the following system of equations. . 7 + 0 . 5 1 = . 3 + 0 . 5 1 = 1 + 1 = 1 The solution to this system of equations is given by = 0 . 625, 1 = 0 . 375. Thus, 62.5% of the flips use coin 1. (b) The expected number of transitions between successive flips of coin 2 can be found using the formula m 11 = 1 1 = 8 3 . An alternative solution proceeds as follows. If we start with coin 2, then on the sec- ond flip, we use coin 2 with probability 0.5 and coin 1 with probability 0.5. Thus, the probability of returning after just one flip is 0.5. If we use coin 1, then the number of flips until returning to coin 2 is a geometric random variable with p = 0 . 3. Thus, the number of flips expected to return to coin 2 from coin 1 is 1- p p (1- p ) = 10 3 . Putting these two observations together, the expected number of flips to return to coin 2 is . 5(1) + 0 . 5( 10 3 + 1) = 8 3 . 1 (c) The coin used on the sixth flip is determined by the fifth flip. The five-step transition probability matrix is P 5 = . 6251 0 . 3749 . 6248 0 . 3752 ....
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316_F07_HW3_Sol - IOE 316 Fall 2007 Homework 3 Solutions...

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