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The binomial distribution determines the probability of looking at a specific quantity of a hit results in a specific quantity of trials.

We are given,

More than $96\%$of the very largest colleges and universities have some online offerings and we have randomly picked $13$such institutions.

Random variable in simple terms generally refers to variables whose values are unknown, therefore, in this case the random variable X is the number of colleges and universities that offer online offerings.

Make the list of values that you want to use X may take on.

As we can see there is an upper bound for the situation at hand, $13$then X is given by,

$X=0,1,2,....,13.$

The random variable is distributed by the data provided X will keep the track of online offerings.

According to the given information, $96\%$colleges and universities have some online offerings.

The probability distribution of binomial distribution has two parameters $n=numberoftrials$and $p=probabilityofsuccess$. The binomial distribution is of the form:

$X~B(n,p)$

Therefore, according to given information, role="math" localid="1649077939499" $n=13$is the number of colleges that offer online offerings and $$\begin{gathered} p=\frac{96}{100} \\ p=0.96 \end{gathered}$$

Hence,$X~B(13,0.96)$

The expected binomial distribution is calculated as:

$渭=np$where, $渭$is the average number of schools that would offer such courses, $n=13$is the number of colleges offer online offering and $p=0.96$is the probability of success.

$$\begin{gathered} 渭=13脳0.96 \\ 渭=12.48 \end{gathered}$$

The probability that at most ten offer such courses is $$\begin{gathered} {=}_{13}{C}_{10}{(0.96)}^{10}{(0.04)}^3 \\ =0.0135 \end{gathered}$$

Using $TI-83$calculator we have found out that the probability of $12$offering such course and the probability $13$of offering such courses are as follow:

$$\begin{gathered} P(X=12)=0.3186 \\ P(X=13)=0.5882 \end{gathered}$$

Therefore, we can conclude from here that we are more likely to get$13.$