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*}