Applications of Bernoulli Random Variables and Probabilities

In other chapters, the mathematical form of the binomial distribution is explained together with the connection with Bernoulli trials. Other chapters that should be read in conjunction with this one are Problems with Binomial Distributions, Bernoulli Mean and Variance and Features of Binomial Distributions.

The title of this chapter refers to Bernoulli random variables but essentially, this is the same as considering problems to do with binomial random variables. We can justify this by the fact that a binomial random variable $X$X  in an experiment with $n$n independent trials, is the sum of $n$n identical Bernoulli random variables, say, $X=B_1+B_2+...+B_n$X=B1​+B2​+...+Bn​.

As we have seen, this way of looking at a binomial random variable makes it easy to develop expressions for the mean and variance of the relevant binomial distribution.

$\mu_X=E\left(X\right)=np$μX​=E(X)=np

where $p$p is the probability of success on a Bernoulli trial, and

$Var\left[X\right]=np\left(1-p\right)$Var[X]=np(1−p)

We also developed the formula for the probabilities associated with each possible number of successes in a sequence of $n$n Bernoulli trials.

$Pr\left(X=r\right)=\binom{n}{r}p^r\left(1-p\right)^{n-r}$Pr(X=r)=(nr)pr(1−p)n−r

When considering situations that might be modelled as binomial experiments, it is essential to identify the features of the situation that will constitute a trial, a success and the sequence of trials. 

Example 1

A six-sided die is to be tossed $21$21 times and the number of times neither a $3$3 nor a $4$4 occurs is to be recorded. If this is to be seen as a binomial experiment, what should we mean by a trial, a success and the sequence of trials? What is the probability of observing the expected value of the random variable representing the number of successes?

In this situation, a trial is a single throw of the die. A success is the non-occurrence of both a $3$3 and a $4$4. The experiment is the sequence of $21$21 tosses.

By a counting argument, we see that the probability of a success on a single trial is $\frac{2}{3}$23​. So, the expected or mean number of successes over the whole experiment is $21\times\frac{2}{3}=14$21×23​=14. 

Then, the probability of observing this many successes is $\binom{21}{14}\times\left(\frac{2}{3}\right)^{14}\times\left(1-\frac{2}{3}\right)^{\left(21-14\right)}\approx0.182.$(2114)×(23​)14×(1−23​)(21−14)≈0.182.

 

Example 2

A student with a flair for working with spreadsheets makes a program in which a row of the spreadsheet has several adjacent cells that are individually coloured either red or green at random whenever the sheet is re-calculated. The formulas in the cells determine that each cell will be green with probability $0.25$0.25 and red otherwise.

The fourth row has four such adjacent cells. The fifth row has five. The sixth has six, and so on. The student is interested in the probability that there will be exactly four green cells in any given row when the sheet is re-calculated.

What are the probabilities of observing four green cells for rows four, five and six? What is the highest probability attainable of seeing exactly four green cells as further rows are added?

You should play with the following applet, which does the same thing as the student's spreadsheet.  You could use this simulation to calculate relative frequencies over many performances of the recalculation, and compare your experimental results with the theoretical calculations provided. 

Think about the event 'no cells are green'. In which row do you think this event will occur most often? Check your intuition with the help of the applet.

Now, think about the event 'exactly one cell is green'. In which row should this event occur most often?

You might continue in this way, considering the event 'exactly two green cells'.

The question below is about the event 'exactly four green cells'. (The applet will not answer the question for you in this case. So, it will be necessary to do some calculations.) Can you explain why the first three rows need not be considered in this $4$4-cell question?

_{(please be patient with the 50 trial button - we are working on making this faster!)}

Here, each row of the spreadsheet is a separate experiment. The coloured cells in a row are the Bernoulli trials for that experiment and a success occurs when a cell is coloured green. Letting the random variable $X_i$Xi​ be the number of successes in row $i$i, we calculate as follows:

$Pr\left(X_4=4\right)=\binom{4}{4}\times0.25^4\times0.75^0\approx0.0039$Pr(X4​=4)=(44)×0.254×0.750≈0.0039

$Pr\left(X_5=4\right)=\binom{5}{4}\times0.25^4\times0.75^1\approx0.0146$Pr(X5​=4)=(54)×0.254×0.751≈0.0146

$Pr\left(X_6=4\right)=\binom{6}{4}\times0.25^4\times0.75^2\approx0.0329$Pr(X6​=4)=(64)×0.254×0.752≈0.0329

In this sequence of experiments, we see that the random variable $X_{16}$X16​ has expected value $16\times0.25=4$16×0.25=4 and we might suspect that the probability of observing exactly four green cells increases up to $n=16$n=16 and decreases thereafter. Calculating as before, we have

$Pr\left(X_{14}=4\right)=\binom{14}{4}\times0.25^4\times0.75^{10}\approx0.2202$Pr(X14​=4)=(144)×0.254×0.7510≈0.2202

$Pr\left(X_{15}=4\right)=\binom{15}{4}\times0.25^4\times0.75^{11}\approx0.2252$Pr(X15​=4)=(154)×0.254×0.7511≈0.2252

$Pr\left(X_{16}=4\right)=\binom{16}{4}\times0.25^4\times0.75^{12}\approx0.2252$Pr(X16​=4)=(164)×0.254×0.7512≈0.2252

$Pr\left(X_{17}=4\right)=\binom{17}{4}\times0.25^4\times0.75^{13}\approx0.2209$Pr(X17​=4)=(174)×0.254×0.7513≈0.2209

$Pr\left(X_{18}=4\right)=\binom{18}{4}\times0.25^4\times0.75^{14}\approx0.2130$Pr(X18​=4)=(184)×0.254×0.7514≈0.2130

Thus, the evidence is strong that the highest attainable probability of seeing exactly $4$4 green cells is approximately $0.2252$0.2252 and this occurs when $n=16$n=16.

 

Worked Examples

Question 1

 

Question 2

 

Question 3