# 2.6. Binary (Bernoulli) distribution¶

Systems that have binary outcomes (pass/fail; yes/no) must obey the probability principle that: $$p(\text{pass}) + p(\text{fail}) = 1$$. That is, the sum of the probabilities of the two possible outcomes must add up to exactly one. A Bernoulli distribution only has a single parameter, $$p_1$$, the probability of observing event 1. The probability of the second event is the difference with 1: that is $$p_2 = 1 - p_1$$.

An example: a histogram for a system that produces 70% acceptable product, $$p(\text{pass}) = 0.7$$, could look like: If each observation is independent of the other, then:

• For the above system where $$p(\text{pass}) = 0.7$$, what is probability of seeing the following sequential outcomes: pass, pass, pass (3 times in a row)?

$$(0.7)(0.7)(0.7) = 0.343$$, about one third

• What is the probability of seeing the sequence: pass, fail, pass, fail, pass, fail?

$$(0.7)(0.3)(0.7)(0.3)(0.7)(0.3) = 0.0093$$, less than 1%

Another example: you work in a company that produces tablets. The machine creates acceptable, unbroken tablets 97% of the time, so $$p_\text{acceptable} = 0.97$$, so $$p_\text{defective} = 0.03$$.

• In a future batch of 850,000 tablets, how many tablets are expected to be defective? (Most companies will call this quantity “the cost of waste”.)

$$850000 \times (1-0.97) = 25500$$ tablets per batch will be defective

• You take a random sample of $$n$$ tablets from a large population of $$N$$ tablets. What is the chance that all $$n$$ tablets are acceptable if $$p$$ is the Bernoulli population parameter of finding acceptable tablets:

Sample size

$$p$$ = 95%

$$p$$ = 97%

$$n=10$$

$$n=50$$

$$n=100$$

• Are you surprised by the large reduction in the number of defective tablets for only a small increase in $$p$$? It is for this reason that a well-performing process producing accetable product does not need to have inspection of every product produced.