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Probability is a metric for determining the possibility of an event occurring. Many things are impossible to forecast with\({\rm{100\% }}\)accuracy. Using it, we can only anticipate the probability of an event occurring, or how probable it is to occur. Probability can range from\({\rm{0}}\)to\({\rm{1}}\), with\({\rm{0}}\)indicating an improbable event and 1 indicating a certain event. Possibility of...

We are given random variable \({\rm{X}}\)

This random variable represents the number of customers who are currently waiting. We're also given odds that a randomly chosen customer will have \({\rm{1,2}}\),or \({\rm{3}}\) packages.

(a):

The following holds

\({\rm{P(X = 3,Y = 3) = P(3}}\)customers waiting, \({\rm{3}}\) packages to be wrapped \({\rm{)}}\)

\(\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P(3}}\)customers waiting, \(1\) package each \()\)

\(\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{P(1}}\)package each \(\mid 3\)customers waiting \({\rm{) \times P(3}}\)customers waiting)

\(\mathop = \limits^{(3)} 0.6 \cdot 0.6 \cdot 0.6 \cdot 0.25\)

\( = 0.054,\)

(1): Because a client who is waiting cannot have 0 gifts for wrapping, the only way for three customers to have three packages is if each has just one gift to wrap.

(2): we apply the multiplication rule as follows:

(3): the exercise provides the probabilities. We use the facts that there are three clients waiting and that the quantity of packages submitted is independent of one another for the conditional probability.

The Rule of Multiplication

\({\rm{P(}} \cap {\rm{B) = P(A}}\mid {\rm{B) \times P(B)}}\)

\({\rm{p(3,3) = 0}}{\rm{.054;}}\)

(b):

The following is true

\({\rm{p(4,11) = P(4}}\)customers waiting, \({\rm{11}}\) packages to be wrapped \()\)

\(\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{P(11}}\)packages to be wrapped \(\mid 4\)customers waiting \({\rm{) \times P(3}}\)customers waiting \()\)

\(\mathop {\rm{ = }}\limits^{{\rm{(5)}}} {\rm{4 \times 0}}{\rm{.}}{{\rm{1}}^{\rm{3}}}{\rm{ \times 0}}{\rm{.3 \times 0}}{\rm{.15}}\)

\( = {\bf{0}}.{\bf{00018}}\),

(5): There are four customers in line for the gift-wrapping service, each with one, two, or three parcels. To get a total of \(11\) packages, three clients must each have three packages, and the fourth must have two packages, for a total of \(11\) packages.\(3 + 3 + 3 + 2 = 11\)

Assume that two of the customers have two packages to be packed; this means that the total number of packages to be wrapped can only be ten, implying that three of the customers must have three packages.

Because there are three customers with three packages waiting, the probability of a randomly selected client having three packages is \(0.1\). One customer is waiting with two packages; the probability of such a scenario is also included in the experiment, and it is high.

\({\bf{0}}.{\bf{3}}\)

The likelihood of three clients waiting is

\({\bf{0}}.{\bf{15}}\)

Because there are four different methods to have a total of \(11\) packages, there are a total of four disjoint events of this type:

\(3 + 3 + 3 + 2,3 + 3 + 2 + 3,3 + 2 + 3 + 3,2 + 3 + 3 + 3\)

\({\rm{p(4,11) = 0}}{\rm{.00018}}\)