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:

\({\rm{Y}}\)given \({\rm{X}}\) (the number of purchasers \({\rm{X}}\) that buy an extended warranty) has a binomial distribution with\({\rm{p = 60\% = 0}}{\rm{.60}}\).

(a) Definition binomial probability:

\({\rm{P(Y = k)}}{{\rm{ = }}_{\rm{n}}}{{\rm{C}}_{\rm{k}}}{\rm{ \times }}{{\rm{p}}^{\rm{k}}}{\rm{ \times (1 - p}}{{\rm{)}}^{{\rm{n - k}}}}{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}{\rm{ \times }}{{\rm{p}}^{\rm{k}}}{\rm{ \times (1 - p}}{{\rm{)}}^{{\rm{n - k}}}}\)

Using the definition of binomial probability with \({\rm{X = n = 4}}\) and\({\rm{Y = k = 2}}\):

\(\begin{aligned}{\rm{P(Y = 2}}\mid {\rm{X = 4)}}\\{\rm{ = }}\frac{{{\rm{4!}}}}{{{\rm{2!(4 - 2)!}}}}{\rm{ \times 0}}{\rm{.6}}{{\rm{0}}^{\rm{2}}}{\rm{ \times (1 - 0}}{\rm{.60}}{{\rm{)}}^{{\rm{4 - 2}}}}\\{\rm{\gg 0}}{\rm{.3456}}\end{aligned}\)

General multiplication rule:

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

Using the general multiplication rule we then obtain:

\(\begin{aligned}{\rm{P(X = 4,Y = 2)}}\\{\rm{ = P(X = 4) \times P(Y = 2}}\mid {\rm{X = 4)}}\\{\rm{ = 0}}{\rm{.15 \times 0}}{\rm{.3456}}\\{\rm{ = 0}}{\rm{.05184}}\\{\rm{ = 5}}{\rm{.184\% }}\end{aligned}\)

(b) (a) Definition binomial probability:

\({\rm{P(Y = k)}}{{\rm{ = }}_{\rm{n}}}{{\rm{C}}_{\rm{k}}}{\rm{ \times }}{{\rm{p}}^{\rm{k}}}{\rm{ \times (1 - p}}{{\rm{)}}^{{\rm{n - k}}}}{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{k!(n - k)!}}}}{\rm{ \times }}{{\rm{p}}^{\rm{k}}}{\rm{ \times (1 - p}}{{\rm{)}}^{{\rm{n - k}}}}\)

Using the definition of binomial probability with \({\rm{X = n}}\) and\({\rm{Y = k = n}}\):

\(\begin{aligned}{\rm{P(Y = 0}}\mid {\rm{X = 0)}}\\{\rm{ = }}\frac{{{\rm{0!}}}}{{{\rm{0!(0 - 0)!}}}}{\rm{ \times 0}}{\rm{.6}}{{\rm{0}}^{\rm{0}}}{\rm{ \times (1 - 0}}{\rm{.60}}{{\rm{)}}^{{\rm{0 - 0}}}}{\rm{ = 1}}\\{\rm{P(Y = 1}}\mid {\rm{X = 1)}}\\{\rm{ = }}\frac{{{\rm{1!}}}}{{{\rm{1!(1 - 1)!}}}}{\rm{ \times 0}}{\rm{.6}}{{\rm{0}}^{\rm{1}}}{\rm{ \times (1 - 0}}{\rm{.60}}{{\rm{)}}^{{\rm{1 - 1}}}}{\rm{ = 0}}{\rm{.6}}\\{\rm{P(Y = 2}}\mid {\rm{X = 2)}}\\{\rm{ = }}\frac{{{\rm{2!}}}}{{{\rm{2!(2 - 2)!}}}}{\rm{ \times 0}}{\rm{.6}}{{\rm{0}}^{\rm{2}}}{\rm{ \times (1 - 0}}{\rm{.60}}{{\rm{)}}^{{\rm{2 - 2}}}}{\rm{ = 0}}{\rm{.36}}\\{\rm{P(Y = 3}}\mid {\rm{X = 3)}}\\{\rm{ = }}\frac{{{\rm{3!}}}}{{{\rm{3!(3 - 3)!}}}}{\rm{ \times 0}}{\rm{.6}}{{\rm{0}}^{\rm{3}}}{\rm{ \times (1 - 0}}{\rm{.60}}{{\rm{)}}^{{\rm{3 - 3}}}}{\rm{ = 0}}{\rm{.216}}\\{\rm{P(Y = 4}}\mid {\rm{X = 4)}}\\{\rm{ = }}\frac{{{\rm{4!}}}}{{{\rm{4!(4 - 4)!}}}}{\rm{ \times 0}}{\rm{.6}}{{\rm{0}}^{\rm{4}}}{\rm{ \times (1 - 0}}{\rm{.60}}{{\rm{)}}^{{\rm{4 - 4}}}}{\rm{ = 0}}{\rm{.1296}}\end{aligned}\)

General multiplication rule:

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

Using the general multiplication rule we then obtain:

\(\begin{aligned}{\rm{P(X = 0,Y = 0)}}\\{\rm{ = P(X = 0) \times P(Y = 0}}\mid {\rm{X = 0)}}\\{\rm{ = 0}}{\rm{.1 \times 0 = 0}}{\rm{.1}}\\{\rm{P(X = 1,Y = 1) = P(X = 1) \times P(Y = 1}}\mid {\rm{X = 1)}}\\{\rm{ = 0}}{\rm{.2 \times 0}}{\rm{.6 = 0}}{\rm{.12}}\\{\rm{P(X = 2,Y = 2)}}\\{\rm{ = P(X = 2) \times P(Y = 2}}\mid {\rm{X = 2)}}\\{\rm{ = 0}}{\rm{.3 \times 0}}{\rm{.36 = 0}}{\rm{.108}}\\{\rm{P(X = 3,Y = 3)}}\\{\rm{ = P(X = 3) \times P(Y = 3}}\mid {\rm{X = 3)}}\\{\rm{ = 0}}{\rm{.25 \times 0}}{\rm{.216 = 0}}{\rm{.054}}\\{\rm{P(X = 4,Y = 4)}}\\{\rm{ = P(X = 4) \times P(Y = 4}}\mid {\rm{X = 4)}}\\{\rm{ = 0}}{\rm{.15 \times 0}}{\rm{.1296 = 0}}{\rm{.01944}}\end{aligned}\)

Addition rule for disjoint or mutually exclusive events:

\({\rm{P(A or B) = P(A) + P(B)}}\)

Add the corresponding probabilities:

\(\begin{aligned}{\rm{P(X = Y)}}\\{\rm{ = P(X = 0,Y = 0) + P(X = 1,Y = 1) + P(X = 2,Y = 2) + P(X = 3,Y = 3) + P(X = 4,Y = 4)}}\\{\rm{ = 0}}{\rm{.1 + 0}}{\rm{.12 + 0}}{\rm{.108 + 0}}{\rm{.054 + 0}}{\rm{.01944}}\\{\rm{ = 0}}{\rm{.40144}}\\{\rm{ = 40}}{\rm{.144\% }}\end{aligned}\)

(c) General multiplication rule:

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

The joint pmf of \({\rm{X}}\) and \({\rm{Y}}\) is the product of the conditional distribution \({\rm{Y}}\mid {\rm{X}}\) (which is the binomial distribution \({\rm{p = 0}}{\rm{.6}}\) and \({\rm{n = X)}}\) and the probability distribution of\({\rm{X}}\).

\({\rm{p(x,y) = P(X = x) \times P(Y = y}}\mid {\rm{X = x) = P(X = x) \times }}\frac{{{\rm{x!}}}}{{{\rm{y!(x - y)!}}}}{\rm{ \times 0}}{\rm{.6}}{{\rm{0}}^{\rm{y}}}{\rm{ \times (1 - 0}}{\rm{.60}}{{\rm{)}}^{{\rm{x - y}}}}\)

Evaluated for all possible \({\rm{x}}\) and \({\rm{y}}\) values (for which the joint pmf is nonzero).

Note: The columns totals are identical to the probability distribution of\({\rm{X}}\).

The marginal pmf of \({\rm{Y}}\) at \({\rm{Y = y}}\) is the sum of the probabilities of all possible values of \({\rm{x}}\) of the joint pmf with\({\rm{Y = y}}\). Thus, the row totals of the table will be the marginal pmf of \({\rm{Y}}\).

\(\begin{aligned}{{\rm{p}}_{\rm{Y}}}{\rm{(0) = p(0,0) + p(1,0) + p(2,0) + p(3,0) + p(4,0)}}\\{\rm{ = 0}}{\rm{.2478}}{{\rm{p}}_{\rm{Y}}}{\rm{(1)}}\\{\rm{ = p(1,1) + p(2,1) + p(3,1) + p(4,1)}}\\{\rm{ = 0}}{\rm{.3590}}{{\rm{p}}_{\rm{Y}}}{\rm{(2)}}\\{\rm{ = p(2,2) + p(3,2) + p(4,2)}}\\{\rm{ = 0}}{\rm{.2678}}{{\rm{p}}_{\rm{Y}}}{\rm{(3)}}\\{\rm{ = p(3,3) + p(4,4)}}\\{\rm{ = 0}}{\rm{.1058}}{{\rm{p}}_{\rm{Y}}}{\rm{(4)}}\\{\rm{ = p(4,4)}}\\{\rm{ = 0}}{\rm{.0194}}\end{aligned}\)