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A discrete random variable is one that can only take on a finite number of different values.

A, B, and C are the three occurrences that we are provided. We have their probability as well.

Obviously, X has a Binomial Distribution, with \({\rm{n = 100}}\) (clients) and \({\rm{p = 0}}{\rm{.2}}\) as parameters (probability of customer using a debit card).

So, success is when a consumer uses his debit card, and failure is when he does not (so the failure probability is \({\rm{0}}{\rm{.5 + 0}}{\rm{.3 = 0}}{\rm{.8}}\)).

So, here's what we've come up with:

\({\rm{X}} \sim {\rm{Bin(100,0}}{\rm{.2)}}\)

Expected value and variance must be determined. We'll utilise the following proposition to do this.

The following is true for a binomial random variable X with parameters n, p, and \({\rm{q = 1 - p}}\).

\(\begin{aligned}{\rm{E(X) = np;}}\\ V(X) &= np(1 - p) \\ &= npq; \\&^{\rm{\sigma }}{\rm{x = }}\sqrt {{\rm{npq}}} {\rm{.}}\end{aligned}\)

The following is true:

\(\begin{aligned} E(X) &= np \\ &= 100 \times 0{\rm{.2}}\\ & = 20 \end{aligned}\)

Then, the difference is:

\(\begin{aligned} V(X) & = npq \\ &= 20 \times 0 {\rm{.8}}\\ &= 16 \end{aligned}\)

Therefore, the value is: \(\begin{array}{c}{\rm{E(X) = 20}}\\{\rm{V(X) = 16}}\end{array}\).

The random variable X has a Binomial Distribution with \({\rm{n = 100}}\) and \(\begin{array}{c}{\rm{p = 0}}{\rm{.5 + 0}}{\rm{.2 }}\\{\rm{ = 0}}{\rm{.7}}\end{array}\) (the likelihood that the consumer would use a credit or debit card). So, success here is defined by the fact that the consumer does not pay in cash.

The following is true:

\(\begin{aligned}E(X) &= np \\ &= 100 \times 0 {\rm{.7}}\\ &= 70 \end{aligned}\)

The variance is then obtained as:

\(\begin{aligned} V(X) &= npq \\ &= 20 \times 0 {\rm{.7 \times 0}}{\rm{.3}}\\ & = 21 \end{aligned\)

Therefore, the value is: \(\begin{array}{c}{\rm{E(X) = 70}}\\{\rm{V(X) = 21}}\end{array}\).