NZ Level 8 (NZC) Level 3 (NCEA) [In development]

Bernoulli Trials and Sequences

Lesson

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.

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.

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.

Does drawing a marble without replacement from a bag containing purple, yellow and red marbles, and noting the number of purple marbles describe a Bernoulli sequence?

Yes

ANo

BYes

ANo

B

Census data shows that $30%$30% of the population in a particular country have red hair. Find the probability that more than half of a random sample of $6$6 people have red hair.

Give your answer correct to four decimal places.

Investigate situations that involve elements of chance: A calculating probabilities of independent, combined, and conditional events B calculating and interpreting expected values and standard deviations of discrete random variables C applying distributions such as the Poisson, binomial, and normal

Apply probability distributions in solving problems