Bernoulli Trial

Bernoulli Trial

Many real experiments are dichotomous - they end in exactly one of two outcomes. A tossed coin lands heads or tails; a student passes or fails; a manufactured item turns out defective or non-defective; a response to a survey question is yes or no; an egg hatches or does not hatch. Whenever an experiment behaves this way, it is customary to label one outcome a "success" and the other "failure" (or "not success"). The labelling is a matter of convention set by the problem, not a property built into the experiment: in a coin toss, if we decide heads is a success, then tails automatically becomes the failure.

Every time we repeat such an experiment once - toss the coin once, roll the die once, draw one item - we call that repetition a trial. If a coin is tossed 4 times, that is 4 trials, and each trial has exactly two possible outcomes, success or failure. Crucially, the outcome of any one trial does not influence the outcome of any other trial (the trials are independent), and the probability of success stays the same number from trial to trial.

Definition. Trials of a random experiment are called Bernoulli trials if they satisfy both of the following conditions:

1. Each trial has exactly two outcomes: success or failure.
2. The probability of success remains the same in every trial. $p$ denotes the probability of success in a single trial and $q = 1-p$ denotes the probability of failure, so that $p+q=1$ always. Illustration. Throwing a fair die 50 times, with "getting an even number" as success, is a case of 50 Bernoulli trials: each throw independently results in success (an even face) or failure (an odd face), and because the die is fair with six equally likely faces, $p=\dfrac{3}{6}=\dfrac12$ and $q=1-p=\dfrac12$ for every one of the 50 throws. The value of $p$ depends entirely on how "success" is defined for that experiment, not on the experiment alone. If a die is thrown 20 times and success is "an even number", then $p=\dfrac12$ (3 favourable faces out of 6). If, in the very same die, success is instead defined as "a multiple of 3", then $p=\dfrac13$ (only 2 favourable faces - 3 and 6 - out of 6). Both re-definitions still produce Bernoulli trials, because in each case the two conditions above hold: two outcomes, and a probability of success that does not change from throw to throw. Worked example. Six balls are drawn successively from an urn containing 7 red and 9 black balls; is drawing balls a set of Bernoulli trials (i) with replacement (ii) without replacement, taking "red" as success?

(i) With replacement: the number of trials is finite (6), and because the ball drawn is put back before the next draw, the urn composition is reset every time, so the probability of drawing red stays $p=\dfrac{7}{16}$ on all six draws. Both conditions hold, so these are Bernoulli trials.

(ii) Without replacement: the probability of red on the first draw is $\dfrac{7}{16}$; on the second draw it becomes $\dfrac{6}{15}$ if the first ball drawn was red, or $\dfrac{7}{15}$ if the first ball drawn was black - and so on, the exact value depending on what has already been removed. Since the probability of success is clearly not the same for every trial (condition (ii) fails, and the draws are not independent either), these are not Bernoulli trials.