91Ó°ÊÓ

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

Given: \({\rm{X = }}\)the number of suits represented in the five-card hand.

A standard deck of cards contains \({\rm{52}}\) cards, of which \({\rm{26}}\) are red and \({\rm{26}}\) are black, \({\rm{13}}\)are of each suit (hearts, diamonds, spades, clubs) and of which 4 are of each denomination (\({\rm{A,2}}\)to \({\rm{10,J,Q,K}}\) ). The face cards are the jacks J, queens \({\rm{Q}}\) and kings K.

(a) The probability is the number of favorable outcomes divided by the number of possible outcomes:

\({\rm{P( spade ) = }}\frac{{{\rm{\# of favorable outcomes }}}}{{{\rm{\# of possible outcomes }}}}{\rm{ = }}\frac{{{\rm{13}}}}{{{\rm{52}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{4}}}\)

Note that once one spade has been drawn, there are \({\rm{51}}\) cards left in the deck of cards, of which there are only \({\rm{12}}\)spades.

Multiplication rule for independent events:

\({\rm{P(A and B) = P(A) \times P(B)}}\)

There are four suits, thus we can obtain all hearts, all clubs, all spades or all diamonds. Each of these options will have the same probability:

\(\begin{array}{c}{\rm{p(1) = 4P( all are spades )}}\\{\rm{ = 4 \times }}\frac{{{\rm{13}}}}{{{\rm{52}}}}{\rm{ \times }}\frac{{{\rm{12}}}}{{{\rm{51}}}}{\rm{ \times }}\frac{{{\rm{11}}}}{{{\rm{50}}}}{\rm{ \times }}\frac{{{\rm{10}}}}{{{\rm{49}}}}{\rm{ \times }}\frac{{\rm{4}}}{{{\rm{48}}}}\\{\rm{\gg 0}}{\rm{.002}}\end{array}\)

There are \({\rm{6}}\) different combinations of two suits. In the five-card hand with exactly \({\rm{2}}\) suits, we either have 1 card of one suit and \({\rm{4}}\) cards of the other suits (\({\rm{5}}\)ways to order and \({\rm{2}}\) suits), or \({\rm{2}}\) cards of one suit and \({\rm{3}}\) cards of the other suit (\({\rm{10}}\) ways to order and \({\rm{2}}\) suits)

\({\rm{P(}}\)only spades and hearts with at least one of each suit)

\(\begin{array}{l}{\rm{ = 2 \times 5 \times }}\frac{{{\rm{13}}}}{{{\rm{52}}}}{\rm{ \times }}\frac{{{\rm{12}}}}{{{\rm{51}}}}{\rm{ \times }}\frac{{{\rm{11}}}}{{{\rm{50}}}}{\rm{ \times }}\frac{{{\rm{10}}}}{{{\rm{49}}}}{\rm{ \times }}\frac{{{\rm{13}}}}{{{\rm{48}}}}{\rm{ + 2 \times 10 \times }}\frac{{{\rm{13}}}}{{{\rm{52}}}}{\rm{ \times }}\frac{{{\rm{12}}}}{{{\rm{51}}}}{\rm{ \times }}\frac{{{\rm{11}}}}{{{\rm{50}}}}{\rm{ \times }}\frac{{{\rm{13}}}}{{{\rm{49}}}}{\rm{ \times }}\frac{{{\rm{12}}}}{{{\rm{48}}}}\\{\rm{\gg 0}}{\rm{.0243}}\end{array}\)

\({\rm{p(2) = 6P(}}\)only spades and hearts with at least one of each suit \({\rm{) = 0}}{\rm{.146}}\)

If we have \({\rm{4}}\) suits in the five-card hand, then one of the suits has two cards in the hand and there are \({\rm{4}}\) possible suits to have the two cards (\({\rm{10}}\)possible ways to order the two spades in five cards and \({\rm{6}}\) possible ways to order the remaining three suits).

\({\rm{P}}\)(\({\rm{2}}\)spades and one of each of the other suits)

\({\rm{ = 6 \times 10 \times }}\frac{{{\rm{13}}}}{{{\rm{52}}}}{\rm{ \times }}\frac{{{\rm{12}}}}{{{\rm{51}}}}{\rm{ \times }}\frac{{{\rm{13}}}}{{{\rm{50}}}}{\rm{ \times }}\frac{{{\rm{13}}}}{{{\rm{49}}}}{\rm{ \times }}\frac{{{\rm{13}}}}{{{\rm{48}}}}{\rm{\gg 0}}{\rm{.066}}\)

\({\rm{p(4) = 4P(2}}\)spades and one of each of the other suits \({\rm{) = 4(0}}{\rm{.066) = 0}}{\rm{.243}}\)

Complement rule:

\({\rm{P(notA) = 1 - P(A)}}\)

The remaining probability can then be determined using the complement rule:

\(\begin{aligned}p(3) &= 1 - p(1) - p(2) - p(4) \\ &= 1 - 0 {\rm{.002 - 0}}{\rm{.146 - 0}}{\rm{.264}}\\ &= 0 {\rm{.588}}\end{aligned}\)

(b)

The expected value \({\rm{\mu }}\) is the sum of the product of each possibility \({\rm{x}}\) with its probability\({\rm{P(x)}}\):

\(\begin{aligned}mu &= \sum {\rm{x}} {\rm{P(x)}}\\ &= 1 \times 0 {\rm{.002 + 2 \times 0}}{\rm{.146 + 3 \times 0}}{\rm{.588 + 4 \times 0}}{\rm{.264}}\\ &= 3 {\rm{.114}}\end{aligned}\)

The variance is the expected value of the squared deviation from the mean:

\(\begin{aligned}{{\rm{\sigma }}^{\rm{2}}} &= \sum {{{{\rm{(x - \mu )}}}^{\rm{2}}}} {\rm{P(x)}}\\ &= (1 - 3 {\rm{.114}}{{\rm{)}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.002 + (2 - 3}}{\rm{.114}}{{\rm{)}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.146 + (3 - 3}}{\rm{.114}}{{\rm{)}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.588 + (4 - 3}}{\rm{.114}}{{\rm{)}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.264}}\\ &= 0{\rm{.405004}}\end{aligned}\)

The standard deviation is the square root of the variance:

\({\rm{\sigma = }}\sqrt {{{\rm{\sigma }}^{\rm{2}}}} {\rm{ = }}\sqrt {{\rm{0}}{\rm{.405004}}} {\rm{\gg 0}}{\rm{.6364}}\)