p(k) denotes the probability of exactly k successes in n independent trials of a random experiment in which the probability of a success is p. q=1-p How do you show that : p(k+1) = (n-k)p , k=0,1,2,...,n-1 p(k) (k+1)q
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Originally Posted by Pink Lady p(k+1) = (n-k)p , k=0,1,2,...,n-1 p(k) (k+1)q Can you tell me what the above is. I really have no idea what is could possibly mean.
p(k) = $\displaystyle {n\choose k}p^kq^{n-k}$ so I guess you need to play with p(k+1) = $\displaystyle {n\choose k+1}p^{k+1}q^{n-k-1}$ But I'm not too sure what this.... (n-k)p , k=0,1,2,...,n-1 p(k) (k+1)q.... is
It was meant to be written as follows: p(k+1) / p(k) = (n-k)p / (k+1)q , when k=0,1,2,3,...,n-1 and q=1-p Sorry about the confusion
Well, that follows directly from what I wrote. Just divide the two and use x!/(x-1)!=x twice. AND it's k=0,1,..,n not n-1.
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