Editing Bernoulli trial (section)

==Examples==

===Tossing coins===
Consider the simple experiment where a fair coin is tossed four times. Find the probability that exactly two of the tosses result in heads.

====Solution====
[[File:Coin flip outcomes.png|thumb|A representation of the possible outcomes of flipping a fair coin four times in terms of the number of heads. As can be seen, the probability of getting exactly two heads in four flips is 6/16 = 3/8, which matches the calculations.]]
For this experiment, let a heads be defined as a ''success'' and a tails as a ''failure.'' Because the coin is assumed to be fair, the probability of success is <math>p = \tfrac{1}{2}</math>. Thus, the probability of failure, <math>q</math>, is given by
:<math>q = 1 - p = 1 - \tfrac{1}{2} = \tfrac{1}{2}</math>.

Using the equation above, the probability of exactly two tosses out of four total tosses resulting in a heads is given by:
:<math>\begin{align}
P(2)
  &= {4 \choose 2} p^{2} q^{4-2} \\
  &= 6 \times \left(\tfrac{1}{2}\right)^2 \times \left(\tfrac{1}{2}\right)^2 \\
  &= \dfrac {3}{8}.
\end{align}</math>

===Rolling dice===
What is probability that when three independent fair six-sided dice are rolled, exactly two yield sixes?

====Solution====
[[File:binomial_trial_dice.svg|lang=egl|thumb|Probabilities of rolling ''k'' sixes from ''n'' independent fair dice, with crossed out dice denoting non-six rolls &ndash; 2 sixes out of 3 dice is circled]]
On one die, the probability of rolling a six, <math>p = \tfrac{1}{6}</math>. Thus, the probability of not rolling a six, <math>q = 1 - p = \tfrac{5}{6}</math>.

As above, the probability of exactly two sixes out of three,
:<math>\begin{align}
P(2)
  &= {3 \choose 2} p^{2} q^{3-2} \\
  &= 3 \times \left(\tfrac{1}{6}\right)^2 \times \left(\tfrac{5}{6}\right)^1 \\
  &= \dfrac {5}{72} \approx 0.069.
\end{align}</math>