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Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes鈥攈ow likely they are鈥攚hen we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

We are given \({\rm{7}}\) cards

and we select \({\rm{3}}\) out of the \({\rm{7}}\) given. By summing the \({\rm{3}}\) selected cards, we can get any integer from 6 to 18 , where we get six if the \({\rm{3}}\) selected cards are \({\rm{123}}\)

and we get 18 if the three selected cards are

\({\rm{5}}\;\;\;{\rm{6}}\;\;\;{\rm{7}}{\rm{.}}\)

This indicated that the random variable \({\rm{W}}\) can take values from\({\rm{6 to 18}}\).

There are \({\rm{35}}\) different outcomes in total (because outcome \({\rm{(1,4,2)}}\) is the same as outcome \({\rm{(1,4,2)}}\) ). The outcomes are given as follows

\({\rm{(1,2,3),(1,2,4), \ldots ,(1,2,7),(2,3,4),(2,3,5), \ldots ,(2,3,7) \ldots ,(4,6,7),(5,6,7)}}\)

Therefore, the number of ways that we can select \({\rm{3}}\) cards out of \({\rm{7}}\) is

\(\left( {\begin{array}{*{20}{l}}{\rm{7}}\\{\rm{3}}\end{array}} \right){\rm{ = 35,}}\)

which is the total number of outcomes.

n order to find pmf of\({\rm{W}}\), we need to find

\({\rm{p(i) = P(W = i),}}\;\;\;{\rm{i = 6,7, \ldots ,18}}{\rm{.}}\)

The only way for the \({\rm{3}}\) selected cards to sum to \({\rm{6}}\) is if cards

\({\rm{1}}\;\;\;{\rm{2}}\;\;\;{\rm{3}}\)

are selected. Since

\({\rm{P(A) = }}\frac{{{\rm{\# of favorable outcomes in A}}}}{{{\rm{\# of outcomes in the sample space }}}}{\rm{,}}\)

we have that

\({\rm{p(6) = P(W = 6) = }}\frac{{\rm{1}}}{{{\rm{35}}}}{\rm{. }}\)

Similarly, there is only one favorable outcome for values

\({\rm{i = 7,17,18, }}\)

therefore

\({\rm{p(i) = }}\frac{{\rm{1}}}{{{\rm{35}}}}{\rm{,}}\;\;\;{\rm{i = 7,17,18}}{\rm{.}}\)

For example, for\({\rm{i = 15}}\), we have that the number of favorable outcomes is \({\rm{3}}\), and the favorable outcomes are

\({\rm{(7,7,1),(7,6,2),(7,5,3)}}\)

We should note that, for example, outcome

\({\rm{(7,4,4)}}\)

is not possible because there is only one card with number\({\rm{4}}\). Therefore,

\({\rm{p(15) = }}\frac{{\rm{3}}}{{{\rm{35}}}}\)

Use the same method to obtain other probabilities. The pmf of random variable \({\rm{W}}\) is

The Expected Value (mean value) of a discrete random variable \({\rm{X}}\) with set of possible values \({\rm{S}}\) and \({\rm{pmfp(x)}}\) is

\({\rm{E(X) = }}{{\rm{\mu }}_{\rm{X}}}{\rm{ = }}\sum\limits_{{\rm{x^I S}}} {\rm{x}} {\rm{ \times p(x)}}\)

Using this for our random variable\({\rm{W}}\), the following is true

\(\begin{array}{c}{\rm{\mu = 6 \times }}\frac{{\rm{1}}}{{{\rm{35}}}}{\rm{ + 7 \times }}\frac{{\rm{1}}}{{{\rm{35}}}}{\rm{ + 8 \times }}\frac{{\rm{2}}}{{{\rm{35}}}}{\rm{ + \ldots + 18 \times }}\frac{{\rm{1}}}{{{\rm{35}}}}\\{\rm{ = 12}}{\rm{.}}\end{array}\)

The Variance of\({\rm{X}}\), where \({\rm{X}}\) is a discrete random variable \({\rm{X}}\) with set of possible values \({\rm{S}}\) and pmf\({\rm{p(x)}}\), denoted by \({\rm{V(X)}}\) (\({\rm{\sigma }}_{\rm{X}}^{\rm{2}}\)or\({{\rm{\sigma }}^{\rm{2}}}\)) is\({\rm{V(X) = }}\sum\limits_{{\rm{x\hat I S}}} {{{{\rm{(x - \mu )}}}^{\rm{2}}}} {\rm{ \times p(x) = E}}\left( {{{{\rm{(X - \mu )}}} ^ {\rm{2}}}} \right)\)

Therefore, for the random variable \({\rm{W}}\) the following holds\(\begin{aligned}{{\rm{\sigma }}^{\rm{2}}} &= \sum\limits_{{\rm{i = 6}}}^{{\rm{18}}} {{{{\rm{(x - 12)}}}^{\rm{2}}}} {\rm{ \times p(i)}}\\&{\rm{ = (6 - 12}}{{\rm{)}}^{\rm{2}}}{\rm{ \times p(6) + (7 - 12}}{{\rm{)}}^{\rm{2}}}{\rm{ \times p(7) + (8 - 12}}{{\rm{)}}^{\rm{2}}}{\rm{ \times p(8) + \ldots + (18 - 12}}{{\rm{)}}^{\rm{2}}}{\rm{ \times p(18)}}\\&{\rm{ = }}{{\rm{6}}^{\rm{2}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{35}}}}{\rm{ + }}{{\rm{5}}^{\rm{2}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{35}}}}{\rm{ + }}{{\rm{4}}^{\rm{2}}}{\rm{ \times }}\frac{{\rm{2}}}{{{\rm{35}}}}{\rm{ + \ldots + }}{{\rm{6}}^{\rm{2}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{35}}}}\\&{\rm{ = 8}}{\rm{.}}\end{aligned}\)