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Chapter 5: Q1E (page 210)

A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let \({\rm{X}}\)denote the number of hoses being used on the self-service island at a particular time, and let\({\rm{Y}}\)denote the number of hoses on the full-service island in use at that time. The joint \({\rm{pmf}}\) of \({\rm{X}}\)and \({\rm{Y}}\) appears in the accompanying tabulation.

a. What is\({\rm{P(X = 1 and Y = 1)}}\)?

b. Compute P(X拢1}and{Y拢1)

c. Give a word description of the event , and compute the probability of this event.

d. Compute the marginal \({\rm{pmf}}\) of \({\rm{X}}\)and of \({\rm{Y}}\). Using \({{\rm{p}}_{\rm{X}}}{\rm{(x)}}\)what is P(X拢1)?

e. Are \({\rm{X}}\)and\({\rm{Y}}\)independent \({\rm{rv's}}\)? Explain

Short Answer

Expert verified

a) \({\rm{P(X = 1 and Y = 1) = 0}}{\rm{.2;}}\)

b) P(X拢1}and{Y拢1) = 0.42;

c) P(X0}and{Y0) = 0.7;

d)

\begin{aligned}{p_X}(0) &= 0.16,{p_X}(1) = 0.34 \hfill \\{p_X}(2) &= 0.5, \hfill \\{p_X}(x) &= 0,\;\;\;x"I \{ 0,1,2\} ; \hfill \\\begin{array}{}{{p_Y}(0) = 0.24,} \\{{p_Y}(1) = 0.38,} \\{{p_Y}(2) = 0.38,} \\{{p_Y}(y) = 0,\;\;\;y"I \{ 0,1,2\} } \\ \begin{gathered}P(X拢1) = 0.5; \hfill \\\; \hfill \\\end{gathered} \end{array} \hfill \\ \end{aligned}

e) dependent

Step by step solution

01

Definition of probability

The proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

02

Determining the \({\rm{P(X = 1 and Y = 1)}}\)

Consider the discrete random variables \({\rm{X and Y}}\)

\({\rm{p(x,y)}}\)is the joint probability mass function

\({\rm{p(x,y) = P(X = x,Y = y)}}\)

and \(\mathop {\rm{{aa}}}\limits_{\rm{x}} {\rm{n}}\mathop {\rm{{aa}}}\limits_{\rm{y}} {\rm{np(x,y) = 1}}\)for every pair \({\rm{(x,y)}}\)

(a) The following is true, as we can see from the definition above.

\({\rm{P(X = 1\;and\;Y = 1) = p(1,1) = 0}}{\rm{.2}}\)

where the value can be found in the supplied table

03

Determining \({\rm{P(X£ 1 and Y£ 1)}}\)

(b) The following vacations

\(\begin{aligned}{}{P(X拢1\;and\;Y拢1)\mathop = \limits^{(1)} p(0,0) + p(0,1) + p(1,0) + p(1,1)} \\{= 0.1 + 0.04 + 0.08 + 0.2} \\{ = 0.42.}\end{aligned}\)

(1): the occurrence

\(\{ X拢1\;and\;Y拢1\} \)

can be written as the sum of four separate events

\({\rm{\{ X = 0,Y = 0\} ,\{ X = 0,Y = 1\} ,\{ X = 1,Y = 0\} ,\{ X = 1,Y = 1\} }}{\rm{.}}\)

04

Determining the probability of this event. 

(c) Because neither random variable is equal to zero, at least one of the hoses is used on both islands.

The likelihood of the given event is as follows:

\(\begin{aligned}{P(X0\;and\;Y0)\mathop= \limits^{(2)} p(1,1) + p(1,2) + p(2,1) + p(2,2)} \\{ = 0.2 + 0.06 + 0.14 + 0.3}\\{ = 0.7.}\end{aligned}\)

05

Determining the marginal \({\rm{pmf}}\) of \({\rm{X}}\) and of \({\rm{Y}}\)

(d) The mass function of marginal probability

random variable that is discrete \({\rm{X}}\) is

\({{\rm{p}}_{\rm{X}}}{\rm{(x) = }}\mathop {\rm{{aa}}}\limits_{\rm{y}} {\rm{np(x,y),}}\;\;\;{\rm{\;for every\;x}}\)

The mass function of marginal probability

random variable that is discrete \({\rm{Y}}\)is

\({{\rm{p}}_{\rm{Y}}}{\rm{(y) = }}\mathop {\rm{{aa}}}\limits_{\rm{x}} {\rm{np(x,y),}}\;\;\;{\rm{\;for every\;y}}{\rm{.}}\)

The following is true for \({\rm{x^I \{ 0,1,2\} }}\). according to the definition.

\({{\rm{p}}_{\rm{X}}}{\rm{(x) = }}\mathop {\rm{{aa}}}\limits_{{\rm{y\^I \{ 0,1,2\} }}} {\rm{np(x,y) = p(x,0) + p(x,1) + p(x,2)}}{\rm{.}}\)

The marginal probability is the sum of each row for every \({\rm{x^I \{ 0,1,2\} }}\). As a result, we've

\(\begin{array}{*{20}{c}}{{{\rm{p}}_{\rm{X}}}{\rm{(0) = 0}}{\rm{.1 + 0}}{\rm{.04 + 0}}{\rm{.02 = 0}}{\rm{.16,}}}\\{{{\rm{p}}_{\rm{X}}}{\rm{(1) = 0}}{\rm{.08 + 0}}{\rm{.2 + 0}}{\rm{.06 = 0}}{\rm{.34,}}}\\{{{\rm{p}}_{\rm{X}}}{\rm{(2) = 0}}{\rm{.06 + 0}}{\rm{.14 + 0}}{\rm{.3 = 0}}{\rm{.5,}}}\end{array}\)

which calculates \({\rm{X }}\)marginal \({\rm{pmf}}\)

Similarly, for \({\rm{Y}}\)marginal \({\rm{pmf}}\), we get

\(\begin{array}{*{20}{c}}{{{\rm{p}}_{\rm{Y}}}{\rm{(0) = 0}}{\rm{.1 + 0}}{\rm{.08 + 0}}{\rm{.06 = 0}}{\rm{.24}}}\\{{{\rm{p}}_{\rm{Y}}}{\rm{(1) = 0}}{\rm{.04 + 0}}{\rm{.2 + 0}}{\rm{.14 = 0}}{\rm{.38}}}\end{array}\)

\({{\rm{p}}_{\rm{Y}}}{\rm{(2) = 0}}{\rm{.02 + 0}}{\rm{.06 + 0}}{\rm{.3 = 0}}{\rm{.38}}\)

where we add up the probabilities in each column

To summaries, we have \({\rm{X and Y}}\)marginal \({\rm{pmf}}\), respectively, supplied with

\(\begin{aligned}{}{{{\rm{p}}_{\rm{X}}}{\rm{(0) = 0}}{\rm{.16}}}\\{{{\rm{p}}_{\rm{X}}}{\rm{(1) = 0}}{\rm{.34}}}\\{{{\rm{p}}_{\rm{X}}}{\rm{(2) = 0}}{\rm{.5,}}}\\{{{\rm{p}}_{\rm{X}}}{\rm{(x) = 0,}}\;\;\;{\rm{x"I \{ 0,1,2\} }}{\rm{.}}}\\{{{\rm{p}}_{\rm{Y}}}{\rm{(1) = 0}}{\rm{.38,}}}\\{{{\rm{p}}_{\rm{Y}}}{\rm{(2) = 0}}{\rm{.38,}}}\\{{{\rm{p}}_{\rm{Y}}}{\rm{(y) = 0,}}\;\;\;{\rm{y"I \{ 0,1,2\} }}{\rm{.}}}\end{aligned}\)

Using \({\rm{X }}\)minimal \({\rm{pmf}}\), we get

\(P(X拢1) = {p_X}(0) + {p_X}(1) = 0.16 + 0.34 = 0.5.\)

06

Determining \({\rm{X}}\) and\({\rm{Y}}\)independent \({\rm{rv's}}\)

(e) If and only if, two random variables \({\rm{X and Y}}\)are independent.

1.\({\rm{p(x,y) = }}{{\rm{p}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{p}}_{\rm{Y}}}{\rm{(y)}}\)

where \({\rm{X and Y}}\)are discrete \({\rm{rv's}}\), for every \({\rm{(x,y)}}\)

2. \({\rm{f(x,y) = }}{{\rm{f}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{f}}_{\rm{Y}}}{\rm{(y)}}\)

where \({\rm{X and Y}}\)are discrete \({\rm{rv's}}\), for every \({\rm{(x,y)}}\)

Otherwise, they are reliant on the following factors:

\(\begin{array}{*{20}{c}}{{{\rm{p}}_{\rm{X}}}{\rm{(2) \times }}{{\rm{p}}_{\rm{Y}}}{\rm{(2) = 0}}{\rm{.5 \times 0}}{\rm{.38 = 0}}{\rm{.19}}}\\{{\rm{p(2,2) = 0}}{\rm{.3}}}\end{array}\)

It leads us to the conclusion that the following holds true for the pair \({\rm{(2,2)}}\)

\(\begin{array}{*{20}{c}}{p(2,2) = 0.30.19 = {p_X}(2) \times {p_Y}(2)} \\{p(2,2){p_X}(2) \times {p_Y}(2)}\end{array}\)

The random variables are so dependent.

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Most popular questions from this chapter

A health-food store stocks two different brands of a certain type of grain. Let \(X = \)the amount (lb) of brand A on hand and \(Y = \)the amount of brand B on hand. Suppose the joint pdf of X and Y is

\(f(x,y) = \left\{ {\begin{array}{*{20}{c}}{kxy\;\;\;\;\;\;\;x \ge 0,\;y \ge 0,\;20 \le x + y \le 30}\\{0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise}\end{array}} \right\}\)

a. Draw the region of positive density and determine the value of k.

b. Are X and Y independent? Answer by first deriving the marginal pdf of each variable.

c. Compute \(P\left( {X + Y \le 25} \right)\).

d. What is the expected total amount of this grain on hand?

e. Compute \(Cov\left( {X, Y} \right)\)and\(Corr\left( {X, Y} \right)\).

f. What is the variance of the total amount of grain on hand?

A particular brand of dishwasher soap is sold in three sizes: \({\rm{25oz,40oz}}\), and\({\rm{65oz}}\). Twenty percent of all purchasers select a\({\rm{25 - 0z}}\)box,\({\rm{50\% }}\)select a\({\rm{40 - 0z}}\)box, and the remaining\({\rm{30\% }}\)choose a\({\rm{65}}\)-oz box. Let\({{\rm{X}}_{\rm{1}}}\)and\({{\rm{X}}_{\rm{2}}}\)denote the package sizes selected by two independently selected purchasers.

a. Determine the sampling distribution of\({\rm{\bar X}}\), calculate\({\rm{E(\bar X)}}\), and compare to\({\rm{\mu }}\).

b. Determine the sampling distribution of the sample variance\({{\rm{S}}^{\rm{2}}}\), calculate\({\rm{E}}\left( {{{\rm{S}}^{\rm{2}}}} \right)\), and compare to\({{\rm{\sigma }}^{\rm{2}}}\).

The joint pdf of pressures for right and left front tires.

a. Determine the conditional pdf of \({\rm{Y}}\) given that \({\rm{X = x}}\) and the conditional pdf of \({\rm{X}}\) given that \({\rm{Y = y}}\).

b. If the pressure in the right tire is found to be \({\rm{22}}\) psi, what is the probability that the left tire has a pressure of at least \({\rm{25}}\) psi? Compare this to.

c. If the pressure in the right tire is found to be \({\rm{22}}\) psi, what is the expected pressure in the left tire, and what is the standard deviation of pressure in this tire?

Suppose that when the pH of a certain chemical compound is\(5.00\), the pH measured by a randomly selected beginning chemistry student is a random variable with a mean of\(5.00\)and a standard deviation .2. A large batch of the compound is subdivided and a sample is given to each student in a morning lab and each student in an afternoon lab. Let\(X = \)the average pH as determined by the morning students and\(Y = \)the average pH as determined by the afternoon students.

a. If pH is a normal variable and there are\(25\)students in each lab, compute\(P\left( { - .1 \le X - Y \le - .1} \right)\)

b. If there are\(36\)students in each lab, but pH determinations are not assumed normal, calculate (approximately)\(P\left( { - .1 \le X - Y \le - .1} \right)\).

A stockroom currently has \({\rm{30}}\) components of a certain type, of which \({\rm{8}}\) were provided by supplier \({\rm{1,10}}\)by supplier \({\rm{2}}\) , and \({\rm{12}}\) by supplier \({\rm{3}}\). Six of these are to be randomly selected for a particular assembly. Let \({\rm{X = }}\) the number of supplier l's components selected, \({\rm{Y = }}\) the number of supplier \({\rm{2}}\) 's components selected, and \({\rm{p(x,y)}}\) denote the joint pmf of \({\rm{X}}\) and\({\rm{Y}}\).

a. What is \({\rm{p(3,2)}}\) ? (Hint: Each sample of size \({\rm{6}}\) is equally likely to be selected. Therefore, \({\rm{p(3,2) = }}\) (number of outcomes with \({\rm{X = 3}}\) and \({\rm{Y = 2)/(}}\) total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.)

b. Using the logic of part (a), obtain \({\rm{p(x,y}}\) ). (This can be thought of as a multivariate hypergeometric distribution-sampling without replacement from a finite population consisting of more than two categories.)

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