W(n)=? 2 i=1 ? 2 j=1 ? 2 k=1 ................? 2 m=1 w1w2w3...wm ...........(1)
ID: 1917645 • Letter: W
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W(n)=? 2 i=1 ? 2 j=1 ? 2 k=1 ................? 2 m=1 w1w2w3...wm ...........(1) Drive the binomial distribution in the follwing algabric way,which doesn't involve any explicit combinatorial analysis .One is again interested in finding the probability W(n) of n successes out of a total of N independent trials.Let w1=p denote thr probability of a success ,w2=1-p-q the corresponding probability of a failure.Then W(n) can be obtianed by writing Here each terms contains N factors and is the probability of a particular combination success and failures.The sum over all combinations is then to be taken only over those terms involving w1 exactly n times ,i.e,only over those terms involving w1^n. By rearranging the sum (1) ,show that the unrestriched sum can be written in the form: W(n) = (w1+w2)^N Expanding this by binomial theorem ,show the sum of all terms in (1) involving w1 ^n ,i.e, the desired probability W(n) ,is the simply gives by the one binomial expension term which involves w1^nExplanation / Answer
Hi, An expression like (n) nCr (r) will be used to find the number of combinations of r out of n things occurring. So 7 nCr 5 would tell how many combinations of any 5 of 7 things could be chosen. We will use this in calculating our probabilities. When choosing from 4 red and three blue cards, the probability of having any 5 of 7 cards drawn from the complete set of 7 cards to be red is: 7 nCr 5 (4/7)^5 (3/7)^2 This is a combination of any 5 of the 7 times the probability of getting red raised to the fifth for 5 reds times the probability of getting blue raised to the second for 2 blues. We will use this pattern for your problems. a) Probability of no successes. (n) nCr (0) (p)^0 (1-p)^(n - 0) = (1 - p)^n b) Probability of at least one success. The probability of at least one success = 1 - probability of no success, so the probability of at least one success is: 1 - (1 - p)^n c) Probability of at most one success. This would be the probability of either 0 or 1 success. The probability of no successes was already calculated in part a), so now we need to find the probability of 1 success to add onto that earlier answer. The probability of 1 success is: (n) nCr (1) (p)^1 (1-p)^(n - 1) = n*p*(1-p)^(n - 1) So the probability of at most one success is: (1 - p)^n + n*p*(1-p)^(n - 1) d) Probability of at least two successes. The probability of at least two successes = 1 - probability of at most one success. That would be: 1 - [(1 - p)^n + n*p*(1-p)^(n - 1)]
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