Bernoulli Trials

One trial, two outcomes

Roll a dice once and ask one question: is it a six? There are only two answers. A trial with exactly two outcomes is called a Bernoulli trial. The outcome being counted is called a success and the other a failure, whether or not it is good news: a faulty part on a production line can be the success.

The probability of a success is written p, so the probability of a failure is 1 − p. For a six on a fair dice, $p=\frac{1}{6}$p = 1/6 and $1-p=\frac{5}{6}$1 − p = 5/6. For a player who scores 70% of free throws, one throw is a Bernoulli trial with p = 0.7.

One Bernoulli trial: a success with probability p, or a failure with probability 1 − p. The two branches add up to 1.

One trial as a random variable

Let X be 1 for a success and 0 for a failure. Its mean is E(X) = 1 × p + 0 × (1 − p) = p.

Because $1^2=1$1² = 1 and $0^2=0$0² = 0, $E\left(X^2\right)=p$E(X²) = p as well. So Var$\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2=p-p^2=p\left(1-p\right)$(X) = E(X²) − [E(X)]² = p − p² = p(1 − p). For a six on a dice, the mean is $\frac{1}{6}$1/6 and the variance is $\frac{1}{6}\times \frac{5}{6}=\frac{5}{36}$1/6 × 5/6 = 5/36.

Repeating the trial

Roll the dice again. A run of Bernoulli trials needs two more things. The probability of success must be the same p on every trial: the second roll gives a six with probability $\frac{1}{6}$1/6, just as the first did. And the trials must be independent: the result of the first roll does not change the chances on the second.

With both in place, the probability of a run of results is a product. Two rolls give four outcomes: a six then a six with probability $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$1/6 × 1/6 = 1/36, a six then not a six with $\frac{1}{6}\times \frac{5}{6}=\frac{5}{36}$1/6 × 5/6 = 5/36, not a six then a six with $\frac{5}{36}$5/36, and two failures with $\frac{5}{6}\times \frac{5}{6}=\frac{25}{36}$5/6 × 5/6 = 25/36. Check: 1 + 5 + 5 + 25 = 36, so the four add up to $\frac{36}{36}=1$36/36 = 1.

Two trials, S for success and F for failure. The second pair of branches is p and 1 − p after a success and after a failure alike: that is independence with a fixed p. Each outcome is the product along its path.

The conditions for a binomial count

Counting the successes in a run of trials leads to the binomial distribution. It rests on four conditions: a fixed number of trials n, two outcomes on each trial, the same probability p on every trial, and trials that are independent of each other.

All four must hold. Tossing the same coin 20 times meets them: 20 trials, heads or tails, $p=\frac{1}{2}$p = 1/2 each time, and no toss affects another. Rolling a dice 10 times and counting sixes meets them too, with $p=\frac{1}{6}$p = 1/6.

The number of trials does not decide it. A small n can meet every condition, and a large n can still fail one.

When a condition fails

Draw two cards from a pack of 52 without putting the first one back, and count the hearts. The first card is a heart with probability $\frac{13}{52}=\frac{1}{4}$13/52 = 1/4. But the second card depends on the first. After a heart, 12 of the 51 cards left are hearts, so the chance is $\frac{12}{51}$12/51. After another suit, 13 of the 51 are hearts, so it is $\frac{13}{51}$13/51.

The probability changes with what happened before, so the draws are not independent, and they do not form a run of Bernoulli trials. If the first card is put back and the pack shuffled, every draw is a heart with probability $\frac{13}{52}=\frac{1}{4}$13/52 = 1/4 again, and the conditions hold.

Two cards drawn without replacement. The second pair of branches is different after a heart $\left(\frac{12}{51}\right)$(12/51) and after another suit $\left(\frac{13}{51}\right)$(13/51), so the second draw depends on the first.

An application

In the application below, the conditions are checked twice: once for a student who guesses every answer on a quiz, where they hold, and once for question cards drawn from a box without replacement, where they fail.