By Y. Wallach
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9 Solution In this ”double-or-nothing” type game, there are only two possible payoﬀs. The ﬁrst is zero dollars, which happens when we lose 6 straight bets, and the second payoﬀ is 64 dollars which happens unless we lose 6 straight bets. So the PMF of Y is ⎧ y=0 ⎨ (1/2)6 = 1/64 6 1 − (1/2) = 63/64 y = 64 (1) PY (y) = ⎩ 0 otherwise The expected amount you take home is E [Y ] = 0(1/64) + 64(63/64) = 63 So, on the average, we can expect to break even, which is not a very exciting proposition. 10 Solution By the deﬁnition of the expected value, n E [Xn ] = x x=1 n = np x=1 n x p (1 − p)n−x x (1) (n − 1)!
0 otherwise (1) (b) The child will throw the frisbee more than four times iﬀ there are failures on the ﬁrst 4 trials which has probability (1 − p)4 . 4096. 5 Solution Each paging attempt is a Bernoulli trial with success probability p where a success occurs if the pager receives the paging message. (a) The paging message is sent again and again until a success occurs. Hence the number of paging messages is N = n if there are n − 1 paging failures followed by a paging success. That is, N has the geometric PMF (1 − p)n−1 p n = 1, 2, .
To simulate the replacement of the jth device by the ultrareliable version by replacing the jth column of W by the column vector R in which a device has failure probability q/2. Lastly, for each column replacement, we count the number N of working devices. 2) ans = 93 89 91 92 90 93 From the above, we see, for example, that replacing the third component with an ultrareliable component resulted in 91 working devices. The results are fairly inconclusive in that replacing devices 1, 2, or 3 should yield the same probability of device failure.