Bernoulli Trials and Sequences

Further information relating to the ideas in this chapter can be found in the chapters Binomial Distributions, Problems with Binomial Distributions and Bernoulli Mean and Variance.

A sequence of independent Bernoulli trials gives rise to a binomial experiment. It is important to be able to know precisely when we are dealing with a sequence of independent Bernoulli trials because only then can we safely model the situation using a binomial distribution.

Clearly, the trials must all belong to the same experiment. Most importantly, they should be independent. That is, the outcome of a trial should not be affected by the outcome of any other trial. Said another way, the probability of success in a trial should be the same in every trial, no matter what other successes have occurred.

Example 1

Thirteen cards are to be drawn randomly, one-by-one from a standard deck of $52$52 playing cards and not replaced in the deck. (There are $26$26 'red' cards and $26$26 'black' cards in the deck.) The number of black cards drawn in total could be any number from zero to thirteen and a probability is to be assigned to the occurrence of each possible total.

Can this experiment be modelled using a binomial distribution?

At the first trial, the probability of drawing a black card is $0.5$0.5 but as the experiment progresses the probability of drawing a black card can vary depending on what black cards have already been drawn. If the first $12$12 cards were all black, for example, then the probability that the thirteenth card is black is $\frac{14}{40}$1440​ or $0.35$0.35. We see that the trials are not independent and conclude that the binomial $\left(52,0.5\right)$(52,0.5) distribution would give unreliable results. 

 

Example 2

Suppose we are satisfied that an experiment can be modelled using the binomial distribution. We have defined a random variable $X$X and determined the parameters of the distribution, $n$n and $p$p. We have $X\sim B\left(21,0.25\right)$X~B(21,0.25).

What is the probability of observing $15$15 successes, and what is the probability of observing $0$0 successes?

For the $15$15 successes, we evaluate $\binom{21}{15}\times0.25^{15}\times0.75^6$(2115)×0.2515×0.756. This can be done if we know how to evaluate the binomial symbol $\binom{21}{15}$(2115). Recall that this is $\frac{21!}{15!\times6!}=54264$21!15!×6!​=54264, and so, the required probability is about $9\times10^{-6}$9×10−6.

For the case of $0$0 successes, we evaluate $\binom{21}{0}\times0.25^0\times0.75^{21}$(210)×0.250×0.7521. Recall that $0!=1$0!=1 or note that $\binom{21}{0}=1$(210)=1 because there is one way in which zero things can be chosen out of twenty-one. Then, working as before, we find the probability to be about $2.4\times10^{-3}$2.4×10−3.

 

 

 

Worked Examples

QUESTION 1

QUESTION 2