Tell me exactly what you need. Do you have specific probability problems that you need help with? Or are you looking for something to read online?

check this site..very useful - http://www.statsoft.com/

first law of and Probability and Stastistics if something has 1% chance of going wrong then it probably will if u get your head around this then i will post other theories and laws for u to read

" if something has 1% chance of going wrong then it probably will " what is 100% perfect? remember probablity can never be more than 1.

boston girl ji, I am very new to those topics and i need to take 3 credit. I mean anything on web that help me to understand and solve the problrms. By the way, I am PhD Student on Bioterrorism.

Some good links: http://mathforum.org/probstat/probstat.html http://www.mathpages.com/home/iprobabi.htm Here is an interesting example: ----------------------------------------------------------------- Optimizing Your Wife ----------------------------------------------------------------- If a man can expect to meet exactly N eligible women in his life, what strategy should he use to maximize his chances of choosing the very best one? We typically make some idealized assumptions when solving this type of problem, namely, that he meets the women in random order of "goodness", that there is an unambiguous ordering of the women, and that the man is concerned ONLY to maximize his chances of choosing the VERY best of the N candidates, placing no value at all on choosing the second-best versus the worst. Under these conditions the man's strategy can obviously never be to select a candidate that isn't the best he has met so far. Also, at each stage his only information is (1) how many women he has evaluated so far, and (2) whether the current women is the best so far. At some stage, his strategy must be to select the current woman if she is the best so far, and clearly if he has reached this stage the optimum strategy can't subsequently be to pass on a "best so far" candidate. Thus, his optimum strategy for selecting the very best of a sequence of N candidates is to pass on the first j-1 candidates (where j is a number to be determined), and then select the next "best so far" that he encounters. This strategy will result in choosing the very best woman if and only if there is one and only one "best so far" in the sequence from j to N. Of course, the probability that the overall best woman is the kth woman in the sequence is just 1/N. Also, the probability that that best woman of the first k-1 women would be one of the first j-1 women is (j-1)/(k-1). Hence, the probability of selecting the optimum woman at the kth round, for k in the range j to N based on some particular value of j, equals the product / 1 \ / j-1 P{N,j} = ( --- )( ----- ) \ N / \ k-1 / So, for this value of j, the probability of selecting the optimum woman out of all N candidates based on this strategy (pass on the first j-1, then choose the next "best so far") is the sum of the above expression for k ranging from j to N. Thus we have /j-1\ N / 1 P{N,j} = ( --- ) SUM ( ----- ) \ N / k=j \ k-1 / If N is fairly large, the summation on the right is a large span of the simple harmonic series, i.e., sum of the inverses of consecutive integers. Recalling that the sum of 1/m for m=1 to s is asymptotically equal to ln(s) + gamma (where gamma=0.57 is Euler constant) it follows that the above probability approaches /j-1\ / N P{N,j} = ( --- ) ln( ----- ) \ N / \ j-2 / as N and j increase. To maximize this probability, we differentiate with respect to j and set the result to zero, which gives j-1 / N --- = ln( ----- ) j-2 \ j-2 / Thus, as j increases, the left side approaches 1, and we can take the exponential of both sides to give N e = --- j-2 Consequently, for a large number of candidates, the optimum strategy for maximizing the chances of selecting the very best woman from N sequential candidates is to pass on the first N/e women (approximately) and then select the next "best so far" that is encountered. With this strategy the probability of success approaches 1/e. (Incidentally, it's been called to my attention that the above formula could, in certain circumstances, be used in a Bayesian way to estimate the number N of candidates that a man could have expected to encounter over his entire life, based on knowledge of having already met the very best woman. The "Amanda Rule" states that if a man knows, by some means, that the kth woman he has encountered is actually the very best, then a Bayesian estimate for the number N of women he would have expected to meet overall is roughly e*k. Of course, if he has already determined that the kth woman is the very best, he presumably has no interest in the remaining k(e-1) candidates, so the formula is of only academic interest.) The above is the traditional answer, and it gives a surprisingly good probability of finding the single best woman from N candidates even as N increases without limit. However, as noted previously, this solution essentially treats the second best woman as no more suitable than the least suitable. In other words, the strategy is entirely focused on maximizing the probability of selecting the very best candidate, without distinguishing between the utilities of any other outcomes. A more pragmatic criterion for choosing a strategy might be to maximize the expected "goodness" of the selection based on some weighting of the outcomes. This certainly affects the answer, as can be seen in the case of N=5 with a linear weighting where 0 is the worst and 4 is the best. The results for a strategy of "passing" on the first k candidates and then choosing the next "best so far" (or the last candidate if necessary) are as shown below probability expected of selecting goodness of the best the selected k candidate candidate ---- ----------- -------------- 0 24/120 240/120 = 2.000 1 50/120 348/120 = 2.900 2 52/120 336/120 = 2.800 3 41/120 297/120 = 2.475 4 24/120 240/120 = 2.000 This shows that to maximize your chances of selecting the very best candidate you should "pass" on the first two, whereas to maximize the expected goodness of the selected candidate you should only "pass" on the first one. By the way, the numerators of the exact probabilities of selecting the best of N candidates using a strategy of k "passes" are as shown below, based on denominators of N!. N k --- -------------------------------- 3 2 3 2 4 6 11 10 6 5 24 50 52 41 24 6 120 274 308 271 206 120 7 720 1764 2088 1950 1640 1237 720 Is there is simple way of generating more rows of this table (aside from just counting permutations)? Do these coefficients have any other applications?

If you want to see the ungarbled text version, open the following link http://www.mathpages.com/home/kmath018.htm

The probability of Prachanda selling Nepal?

Thanks all, but I need more about basic probability and statistics. Please do suggest.

येसो तथ्यान्क सास्त्र को जोखना हेर्दा खेरि हाम्रो राहुल् भाइ लाई अब बिस्तारै amrika मन पर्ने सम्भाबना देखिन्छ । येसो प्रोबाबिलिटि स्टाटिस्टिक्स् को बारे मा पनि सानो मुक्तक् कबिता बाचन् गरम् न त। सपरिवार् लाइ दशै को मङल्मय शुवकामना!

धन्यवाद भुन्टे जी, तर मलाई लाग्छ कि मलाई अ्एरिका फाप्दैन । यहाँ आउनु अघी अर्कै कुरो आए पछि अर्कै भयो । मान्छे हरु को देश होइन यो, लाश हरु मान्छे को अभिनय गर्ने उदेकलाग्दो सुनसान मशान जस्तो । भएको जागीर छोडेर बुढेसकाल म डाक्तरी पढ्न आइएछ । चोकटा खान गएकी बुढी--------- जस्तो भा छ । तै पनि यसो तथ्यांक शास्त्र र संभब्यता अध्ययन बारे कसै लाई केही जानकारी छ भने आभारी हुने छु ।

rahulvai, I am not sure what sorts of info u r looking for, prob and stat is a very broad topic, if you are looking for some specific topic, then just go into one of the math help sites like mathforum, or ams.org has some info too. It all depends on what you are looking for. If you are looking for some math homework help, the site below is very helpful. It has problems on topolpgy, graph theory, real variable and stats too. http://at.yorku.ca/cgi-bin/bbqa? forum=homework_help_2001;task=show_msg;msg=0078 or better yet, go to your school lib, check out some basic prob and stats book and go from there!! sorry wasnt much of a help!

Rahulvai, dherai birakta lagdo kura garnu bhayo... sansar ka sabai prani haru lai hamro sancho ma dhalna sakindaina ni. pachhi bistarai yehi kai chal chalan ramro lagne chha tapai lai pani..dashai ko tehi suvakamana chha ani homro nepal ki chhori bhrikuti le fone magera kina fone nagarya bhanera bhute dai le gunaso gardai hunthyo bhanera sunai dinu hola For statistics, there r tons of resources in web. they will be good for general background. but for grade, just follow what your instructor has asked to read books, articles, softwares, etc.

Dhanayabaad Bhnute ji, Sallah ra Sujhaw ko laagi. Instructor chai Chinese paryo, mulaa le bhaneko ni bujhinna ani feri I hate cChinese people more n more Kharaab than Dhoti bhai.