# Bernoulli(p) to Binomial(n,p)?

## Homework Statement

So this is the problem, from Pitman: "In Bernoulli (p) trials let V_n be the number of trials required to produce either n successes or n failures, whichever comes first. Find the distribution of V_n."

## Homework Equations

P(k successes in n trials) = (n choose k) * p^k * q^(n-k) 
(Sorry, tried LaTeX'ing it, it created problems with the superscript)
P(A or B) = P(A) + P(B) - P(AB) 

## The Attempt at a Solution

Okay, so here is my reasoning. I know that if I were only looking for the number of successes, this would be a binomial(n,p) distribution . Since I cannot have n successes and n failures at the same time, the P(AB) term in  drops out, so I thought I would add Bin(n,p) and Bin(n,1-p).

Q1: Is my reasoning right?
Q2: Here is my solution:

P(Vn = k) = Bin(n,p) + Bin(n,1-p)
= (n choose k) * (p)^k * (1-p)^(n-k) + (n choose k) * (1-p)^k * (p)^(n-k)
= (n choose k) [(1-p)^n * p^n]

I don't think this is the right answer because it does not make sense when I ask this question: Say I am looking for V_1. Then the answer must be 1. I am wrong somewhere, and I cannot figure it out. Where am I wrong? Any insight welcome.

Last edited: