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.
so I guess you need to play with
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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