These are kinds of problems that Electrical Computer Engineers normally encounter in 3rd year.
Probability Theory and Random Processes: Ensemble model of randomness. Conditional probability, and independence, and Bayes' theorem. Random variables, probability mass and probability density. Expected values. Collections of random variables, joint and marginal probability, correlation and regression. Confidence intervals. Random processes, stationarity and ergodicity, power spectral density. Poisson process, birthdeath process and queues.
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1) Steps for installing firestarter (fire wall) on Ubuntu 5.10 2) Shell Scripting with Linux/Unix 3) Shell Scripting with Linux/Unix ( Part 2) 4) Mathematical Software ( 2D Graphs Plot in Mathcad) 5) Computer Security ( Bell LaPadula Model) 7) Artificial Intelligence ( Algorithms) 8) Critical Essay and Analysis Part 1. 9) Critical Essay and Analysis Part 2
 1) Ten hunters are waiting for ducks to fly by. When a flock of ducks flies overhead, the hunters fire at the same time, but each chooses his target at random, independently of the others. If each hunter independently hits his target with probability p, compute the expected number of ducks that escape unhurt when a flock of size 10 flies overhead. Solution: Let Xi=1 if the ith duck escapes unhurt and 0 otherwise for i=0,1,2,3...10. The expected number of ducks escape can be expressed as E[X1+X2+...Xn]=E[X1]+E[X2]+...+E[Xn] To compute E[Xi]=P(Xi=1) we note that the number of hunters will, independently hit the ith duck with probability p/10 and therefore using basic combination and permutation we can see: P(Xi=1)=(1p/10)^10 and E[X]= 10(1p/10)^10
2) Two hunters shoot at a deer which is hit by exactly one bullet. If the first hunter hits his target with probability 0.3 and the second with probability 0.6, what is the probability the second hunter killed the deer? Solution: Let H2 be the event that hunter 2 shot the deer and Hc 2 the event that hunter 2 did not shoot the deer. Let B be the event that only one hunter shot the deer. Then P(B) = .3 *0.4 + 0.7 *0.6 =0.12 + 0.42 = 0.54 and the conditional probability P(H2B) = .42/(.12+.42) = 7/9
