1. Probability

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

2. 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.

3. p(k) = ${n\choose k}p^kq^{n-k}$

so I guess you need to play with

p(k+1) = ${n\choose k+1}p^{k+1}q^{n-k-1}$

But I'm not too sure what this.... (n-k)p , k=0,1,2,...,n-1 p(k) (k+1)q.... is

4. 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