##### 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?

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 Ei denote the event of a failure on the ith trial, then the probability of no successes is, by independence,

P(E1E2 · · · En) = P(E1)P(E2) · · · 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

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