##### Section 3.4 Example 4f Sheldon Ross

An infinite sequence of independent trials is to be performed. Each trial results in a success with probability p and a failure with probability 1 − p. What is the probability that

(a) at least 1 success occurs in the first n trials;

(b) exactly k successes occur in the first n trials;

(c) all trials result in successes?

**Answer**

In order to determine the probability of at least 1 success in the first n trials, it is easiest to compute first the probability of the complementary event: that of no successes in the first n trials. If we let E_{i }denote the event of a failure on the ith trial, then the probability of no successes is, by independence,

P(E_{1}E_{2} · · · E_{n}) = P(E_{1})P(E_{2}) · · · P(En) = \((1 - p)^n\)

Hence, the answer to part **(a)** is \(1 - (1 - p)^n\).

**(b) **Consider any particular sequence of the first \(n\) outcomes containing \(k\) successes and \(n-k\) failures. Each one of these sequences

will, by the assumed independence of trials, occur with probability \(p^k(1-p)^{n-k}\)

There are \(\binom{n}{k}\) such sequences , the desired probability in part (b) is

\(P\{exactly\ \ k \ \ success\} = \binom{n}{k} p^k(1-p)^{n - k}\)

**(c)**

The probability of the first n trials all resulting in success is given by

\(P(E_1^cE_2^c....E_n^c) = p^n\)