Need Help with Basic Problem

[riazkj4]

How do you solve this?

For a certain discrete random variable on the non-negative integers, the probability
function satisfies the relationships
P(0) = P(1),
P(k + 1) =(1/k) P(k), k = 1, 2, 3, · · ·

Find P(0).

[dcg_phil]

P(0) = P(1)
P(2) = P(1+1)=1/1*P(1)=P(0)
P(3) = P(2+1)=1/2*P(2)=1/2*P(0)
P(4) = P(3+1)=1/6*P(0)
.
.
.
P(n)=P(n-1+1) = 1/(n-1)!*P(0)
1= P(0)[1+1+1/2+1/6+...+1/(n-1)!]

1+1+1/2+1/6+...+1/(n-1)! = taylor's series with x = 1

1=P(0)*e^1
P(0) = 0.3679

[riazkj4]

Thanks very much, are Taylor and Maclurin series on the initial P/1 exam?

[Puoya]

NO, just memorize the identies the authors of the manuals highlight or bold.