## Section10.3Bernoulli Trials

Suppose we have a jar with $$7$$ marbles, four of which are red and three are blue. A marble is drawn at random and we record whether it is red or blue. The probability $$p$$ of getting a red marble is $$4/7\text{;}$$ and the probability of getting a blue is $$1-p=3/7\text{.}$$

Now suppose the marble is put back in the jar, the marbles in the jar are stirred, and the experiment is repeated. Then the probability of getting a red marble on the second trial is again $$4/7\text{,}$$ and this pattern holds regardless of the number of times the experiment is repeated.

It is customary to call this situation a series of Bernoulli trials. More formally, we have an experiment with only two outcomes: success and failure. The probability of success is $$p$$ and the probability of failure is $$1-p\text{.}$$ Most importantly, when the experiment is repeated, then the probability of success on any individual test is exactly $$p\text{.}$$

We fix a positive integer $$n$$ and consider the case that the experiment is repeated $$n$$ times. The outcomes are then the binary strings of length $$n$$ from the two-letter alphabet $$\{S,F\}\text{,}$$ for success and failure, respectively. If $$x$$ is a string with $$i$$ successes and $$n-i$$ failures, then $$P(x)=\binom{n}{i}p ^i(1-p)^{n-i}\text{.}$$ Of course, in applications, success and failure may be replaced by: head/tails, up/down, good/bad, forwards/backwards, red/blue, etc.

###### Example10.12.

When a die is rolled, let's say that we have a success if the result is a two or a five. Then the probability $$p$$ of success is $$2/6=1/3$$ and the probability of failure is $$2/3\text{.}$$ If the die is rolled ten times in succession, then the probability that we get exactly four successes is $$C(10,4)(1/3)^4 (2/3)^{6}\text{.}$$

###### Example10.13.

A fair coin is tossed $$100$$ times and the outcome (heads or tails) is recorded. Then the probability of getting heads $$40$$ times and tails the other $$60$$ times is

\begin{equation*} \binom{100}{40}\left(\frac{1}{2}\right)^{40}\left(\frac{1}{2}\right)^{60} =\frac{\binom{100}{40}}{2^{100}}. \end{equation*}
###### Discussion10.14.

Bob says that if a fair coin is tossed $$100$$ times, it is fairly likely that you will get exactly $$50$$ heads and $$50$$ tails. Dave is not so certain this is right. Carlos fires up his computer and in few second, he reports that the probability of getting exactly $$50$$ heads when a fair coin is tossed $$100$$ times is

\begin{equation*} \frac{12611418068195524166851562157}{158456325028528675187087900672} \end{equation*}

which is $$.079589\text{,}$$ to six decimal places. In other words, not very likely at all. Xing is doing a modestly more complicated calculation, and he reports that you have a $$99$$% chance that the number of heads is at least $$20$$ and at most $$80\text{.}$$ Carlos adds that when $$n$$ is very large, then it is increasingly certain that the number of heads in $$n$$ tosses will be close to $$n/2\text{.}$$ Dave asks what do you mean by close, and what do you mean by very large?