第 4 / 23 页
非代码面试题
显示 20 / 443 道匹配题目
答题状态：未尝试未正确已正确
ID题目领域难度题型进度权限
420Variance of the k-th Uniform Order StatisticLet X 1, \ldots, X n be iid Uniform (0,1). Derive a closed-form expression for Var (X (k) ) for 1 \le k \le n.概率困难derivation未尝试面试订阅423Variance of the Range of Uniform Order StatisticsLet X 1, \ldots, X n be iid Uniform (0,1) and let R = X (n) - X (1) . Derive Var (R) as a function of n.概率中等derivation未尝试免费424Expected Gap Between the Two Largest UniformsLet X 1, \ldots, X 6 be iid Uniform (0,1). Find the expected gap between the largest and second-largest values: E[X (6) - X (5) ].概率中等数值题未尝试免费430Characterization of Memorylessness and the Residual Life ParadoxPart (a): Let X be a continuous, positive random variable satisfying P(X > s + t \mid X > s) = P(X > t) for all s, t 0. Prove that X must be exponentially distributed. Part (b): A lightbulb's lifetime L has CDF F(t) = 1 - 1 2 e -t - 1 2 e -3t for t 0 (a mixture of Exp (1) and Exp (3)). You arrive at a uniformly random time and observe the bulb currently in use. Let R be the residual lifetime of that bulb. Show that E[R] > E[L] and compute both values. Explain why memorylessness breaks down and causes this paradox.概率困难derivation未尝试面试订阅432Asymmetric Penalties in an Exponential RaceTwo independent alarms go off at Exp (4) and Exp (6) times respectively. If alarm 1 fires first you pay \3; if alarm 2 fires first you pay \5. After the first alarm fires, the remaining alarm is reset (memoryless restart) and you pay an additional \1 when it fires. Find the expected total payment.概率中等数值题未尝试免费434Second Failure in a Memoryless Component ArrayA system has 4 independent components, each with lifetime Exp (2). When a component fails, it is removed and the remaining components continue operating. By memorylessness, surviving components' residual lifetimes are still Exp (2). Find the expected time until the second component fails.概率中等数值题未尝试免费438Machine Replacements via Memoryless MinimumA factory runs 3 identical machines with independent lifetimes Exp (1). When any machine fails, it is instantly replaced with a new identical machine. All non-failed machines continue running (their residual lifetimes remain Exp (1) by memorylessness). Find the expected number of machine replacements in the time interval [0, 10].概率中等数值题未尝试免费439Sequential Elimination RaceThree players with independent lifetimes X 1 \sim Exp (1), X 2 \sim Exp (2), X 3 \sim Exp (4) compete. The first to "die" is eliminated, then the two survivors continue (by memorylessness, their residual lifetimes are fresh exponentials with the same rates). The second to die is eliminated, and the last survivor wins. (a) Find the probability that the elimination order is X 3, X 1, X 2 (i.e., player 3 dies first, then player 1, then player 2). (b) Find the expected total time until only one player remains.概率困难multi part未尝试面试订阅441Minimum of Three Identical ExponentialsLet X 1, X 2, X 3 be independent, each Exp (4). Find the distribution of M = \min(X 1, X 2, X 3) and compute E[M]. Then verify: given that M > 2, use memorylessness to find E[M \mid M > 2].概率简单数值题未尝试免费443Series System Replacement Costs via Competing ExponentialsA machine has two critical components in series: component A with lifetime Exp (3) and component B with lifetime Exp (5), independent of each other. When either fails, the entire machine stops, the failed component is replaced (cost \20 for A, \50 for B), and both components restart fresh (the survivor restarts by memorylessness, the replacement is new). Find the expected replacement cost per unit time in the long run.概率中等数值题未尝试免费444Full Ordering Probability for Four Competing ExponentialsFour independent exponential random variables X 1 \sim Exp (1), X 2 \sim Exp (2), X 3 \sim Exp (3), X 4 \sim Exp (6) represent task completion times. Using iterated applications of the memoryless property, find P(X 4 < X 3 < X 2 < X 1) — the probability that the tasks complete in the specific order 4, 3, 2, 1.概率困难数值题未尝试面试订阅448Threshold Exceedance for the Minimum of Two ExponentialsLet X \sim Exp (2) and Y \sim Exp (3) be independent. Define M = \min(X, Y) and fix a threshold c = 1. (i) Find P(M > 1). (ii) Given M > 1, find E[M - 1 \mid M > 1] and P(X < Y \mid M > 1) (the probability that X is the minimum, given both survived past 1).概率中等数值题未尝试免费449Memoryless Message Relay ChainA message must traverse a chain of relay nodes to reach its destination. Each node independently takes Geom (1/3) attempts to successfully forward the message to the next node. However, on each attempt, there is an independent probability 1/5 that the node permanently fails, destroying the message. If the node fails, the message is lost. If the chain has 2 nodes, find: (i) The probability the message reaches the destination (traverses both nodes). (ii) The expected total number of attempts across both nodes, given the message reaches the destination.概率中等数值题未尝试免费450Head Start in an Exponential RaceLet X \sim Exp ( ) and Y \sim Exp ( ) be independent. Player A finishes at time X and player B finishes at time Y + c where c > 0 is a head start for player A (player B starts c time units later). (a) Derive P(X < Y + c) — the probability that A finishes before B. (b) Show that as c 0, the result recovers the standard competing-exponentials formula. (c) Evaluate for = 3, = 2, c = 1 and interpret how the head start affects A's winning probability.概率困难multi part未尝试面试订阅452Sample Mean of Exponential Service TimesA server processes 100 independent requests. Each request takes Exp (1) time (mean 1 second). Let T = 1 100 \sum i=1 100 T i be the average processing time. **(a)** State what the Law of Large Numbers guarantees about T as n . **(b)** Using the CLT, approximate P( T > 1.2). You may use \Phi(2) \approx 0.9772.概率简单数值题未尝试免费453Call Center Overflow via Poisson CLTA call center receives calls according to a Poisson process with rate = 4 calls per minute. The center operates for an 8-hour shift (480 minutes). The center can handle at most 2000 calls per shift before service quality degrades. Using a suitable normal approximation, estimate the probability that the total number of calls in a single shift exceeds 2000. You may use the following: \Phi(1.83) \approx 0.9664.概率中等数值题未尝试免费455Geometric Mean of Random Gains via LLN and CLTLet X 1, X 2, \ldots be i.i.d.\ with P(X i = 1) = \tfrac 1 2 and P(X i = 0) = \tfrac 1 2 . Define the geometric-mean-like quantity Y n = (\prod i=1 n (1 + X i) ) 1/n . **(a)** Find \lim n Y n almost surely. **(b)** For n = 200, use the CLT to approximate P(Y 200 > 1.45). You may use: \ln 2 \approx 0.6931, \Phi(1.02) \approx 0.8461.概率困难derivation未尝试免费458Empirical Frequency Accuracy via the CLTA biased die shows a six with probability p = 1/3. You roll it n = 900 times independently and record p = ( number of sixes )/n. **(a)** State what the Law of Large Numbers guarantees about p as n . **(b)** Using the CLT, approximate P(| p - 1/3| < 0.02). You may use \Phi(1.27) \approx 0.8980.概率中等数值题未尝试免费462Estimating Pi by Monte Carlo and the Law of Large NumbersTo estimate , you draw n = 10 , 000 points (X i, Y i) independently and uniformly on the unit square [0,1] 2. Define Z i = 1 (X i 2 + Y i 2 \le 1), and let = 4 Z where Z = 1 n \sum i=1 n Z i. **(a)** Explain why E[ ] = and why almost surely. **(b)** Using the CLT, find an approximate 95\% confidence interval for given that the observed Z = 0.7854. You may use \Phi(1.96) \approx 0.975 and \approx 3.1416.概率简单数值题未尝试免费466Election Poll Margin of Error via the CLTA pollster surveys n = 1 , 600 voters to estimate the proportion p supporting a candidate. Suppose the true proportion is p = 0.5. Using the CLT, find the probability that the sample proportion p falls within 0.02 of the true value, i.e., approximate P(| p - 0.5| < 0.02). You may use \Phi(1.60) \approx 0.9452.概率简单数值题未尝试免费