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Chapter 5: Q20E (page 212)

Let\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{,}}{{\rm{X}}_{\rm{4}}}{\rm{,}}{{\rm{X}}_{\rm{5}}}\), and \({{\rm{X}}_{\rm{6}}}\) denote the numbers of blue, brown, green, orange, red, and yellow M\&M candies, respectively, in a sample of size\({\rm{n}}\). Then these \({{\rm{X}}_{\rm{i}}}\) 's have a multinomial distribution. According to the M\&M Web site, the color proportions are\({{\rm{p}}_{\rm{1}}}{\rm{ = }}{\rm{.24,}}{{\rm{p}}_{\rm{2}}}{\rm{ = }}{\rm{.13}}\), \({{\rm{p}}_{\rm{3}}}{\rm{ = }}{\rm{.16,}}{{\rm{p}}_{\rm{4}}}{\rm{ = }}{\rm{.20,}}{{\rm{p}}_{\rm{5}}}{\rm{ = }}{\rm{.13}}\), and\({{\rm{p}}_{\rm{6}}}{\rm{ = }}{\rm{.14}}\).

a. If\({\rm{n = 12}}\), what is the probability that there are exactly two M\&Ms of each color?

b. For\({\rm{n = 20}}\), what is the probability that there are at most five orange candies? (Hint: Think of an orange candy as a success and any other color as a failure.)

c. In a sample of\({\rm{20M \backslash Ms}}\), what is the probability that the number of candies that are blue, green, or orange is at least \({\rm{10}}\) ?

Short Answer

Expert verified

a. The probability is \({\rm{0}}{\rm{.00213}}\).

b. The probability is \({\rm{0}}{\rm{.8042}}\).

c. The probability is \(0.87248.\)

Step by step solution

01

Definition

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.

02

Calculating the probability

(a):

The probability mass function of multinomial distribution is

\(\begin{aligned}{{\rm{f}}_{{\rm{XYZ}}}}{\rm{(x,y,z) = P(X = x,Y = y,Z = z)}}\\{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{x!y!z!}}}}{\rm{p}}_{\rm{1}}^{\rm{x}}{\rm{p}}_{\rm{2}}^{\rm{y}}{\rm{p}}_{\rm{3}}^{\rm{z}}{\rm{,}}\end{aligned}\)

where, analogously, this can be written in the same manner for given \({\rm{6}}\) random variables.

Given the probabilities

\(\begin{aligned}{*{20}{l}}{{{\rm{p}}_{\rm{1}}}{\rm{ = 0}}{\rm{.24,}}}&{{{\rm{p}}_{\rm{2}}}{\rm{ = 0}}{\rm{.13,}}}&{{{\rm{p}}_{\rm{3}}}{\rm{ = 0}}{\rm{.16,}}}\\{{{\rm{p}}_{\rm{4}}}{\rm{ = 0}}{\rm{.2,}}}&{{{\rm{p}}_{\rm{5}}}{\rm{ = 0}}{\rm{.13,}}}&{{{\rm{p}}_{\rm{6}}}{\rm{ = 0}}{\rm{.13,}}}\end{aligned}\)

where they sum to\({\rm{1}}\), the requested probability can be computed as follows

\({{\rm{f}}_{\rm{X}}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{x}}_{\rm{3}}}{\rm{,}}{{\rm{x}}_{\rm{4}}}{\rm{,}}{{\rm{x}}_{\rm{5}}}{\rm{,}}{{\rm{x}}_{\rm{6}}}} \right){\rm{ = P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{ = }}{{\rm{x}}_{\rm{3}}}{\rm{,}}{{\rm{X}}_{\rm{4}}}{\rm{ = }}{{\rm{x}}_{\rm{4}}}{\rm{,}}{{\rm{X}}_{\rm{5}}}{\rm{ = }}{{\rm{x}}_{\rm{5}}}{\rm{,}}{{\rm{X}}_{\rm{6}}}{\rm{ = }}{{\rm{x}}_{\rm{6}}}} \right)\)

for \({{\rm{x}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{ = \ldots = }}{{\rm{x}}_{\rm{6}}}{\rm{ = 2}}\) is

\(\begin{aligned}{{\rm{f}}_{\rm{X}}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{x}}_{\rm{3}}}{\rm{,}}{{\rm{x}}_{\rm{4}}}{\rm{,}}{{\rm{x}}_{\rm{5}}}{\rm{,}}{{\rm{x}}_{\rm{6}}}} \right)\\{\rm{ = }}\frac{{{\rm{12!}}}}{{{\rm{2!2!2!2!2!2!}}}}{\rm{ \times 0}}{\rm{.2}}{{\rm{4}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.1}}{{\rm{3}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.1}}{{\rm{6}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.}}{{\rm{2}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.1}}{{\rm{3}}^{\rm{2}}}{\rm{ \times 0}}{\rm{.1}}{{\rm{3}}^{\rm{2}}}{\rm{ = 0}}{\rm{.00213}}{\rm{.}}\end{aligned}\)

03

Calculating the probability

(b):

Binomial distribution with parameters \({\rm{n}}\) and \({\rm{p(q = 1 - p)}}\) has probability mass function

\({\rm{P(X = x) = }}\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{x}}\end{array}} \right){{\rm{p}}^{\rm{x}}}{{\rm{q}}^{{\rm{n - x}}}}{\rm{,}}\;\;\;{\rm{xI \{ 0,1,2, \ldots n\} }}{\rm{.}}\)

The parameter \({\rm{p}}\) is the probability of success, therefore

\({{\rm{p}}_{\rm{4}}}{\rm{ = 0}}{\rm{.2}}\)

which means that

\({\rm{q = 1 - 0}}{\rm{.2 = 0}}{\rm{.8}}\)

The parameter \({\rm{n}}\) is 20. Hence, the requested probability is

\({\rm{P(X拢5) = }}\sum\limits_{{\rm{x = 0}}}^{\rm{5}} {\left( {\begin{aligned}{*{20}{c}}{{\rm{20}}}\\{\rm{x}}\end{aligned}} \right)} {\rm{0}}{\rm{.}}{{\rm{2}}^{\rm{x}}}{\rm{ \times 0}}{\rm{.}}{{\rm{8}}^{{\rm{20 - x}}}}{\rm{ = 0}}{\rm{.8042}}\)

04

Calculating the probability

(c):

Look at the blue, green, and orange candies as a success, and all other colors as failure. In this case the parameter \({\rm{p}}\) is

\({\rm{p = }}{{\rm{p}}_{\rm{1}}}{\rm{ + }}{{\rm{p}}_{\rm{3}}}{\rm{ + }}{{\rm{p}}_{\rm{4}}}{\rm{ = 0}}{\rm{.24 + 0}}{\rm{.16 + 0}}{\rm{.2 = 0}}{\rm{.6}}\)

which means that

\({\rm{q = 1 - 0}}{\rm{.6 = 0}}{\rm{.4}}\)

The parameter \({\rm{n}}\) is \({\rm{20}}\) . Hence, the requested probability is

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Most popular questions from this chapter

A restaurant serves three fixed-price dinners costing \({\rm{12, 15and 20}}\). For a randomly selected couple dining at this restaurant, let \({\rm{X = }}\)the cost of the man鈥檚 dinner and \({\rm{Y = }}\)the cost of the woman鈥檚 dinner. The joint \({\rm{pmf's}}\) of \({\rm{X and Y }}\)is given in the following table:

a. Compute the marginal \({\rm{pmf's}}\)of \({\rm{X and Y }}\)

b. What is the probability that the man鈥檚 and the woman鈥檚 dinner cost at most \({\rm{15}}\)each?

c. Are \({\rm{X and Y }}\)independent? Justify your answer.

d. What is the expected total cost of the dinner for the two people?

e. Suppose that when a couple opens fortune cookies at the conclusion of the meal, they find the message 鈥淵ou will receive as a refund the difference between the cost of the more expensive and the less expensive meal that you have chosen.鈥� How much would the restaurant expect to refund?

You have two lightbulbs for a particular lamp. Let\({\rm{X = }}\)the lifetime of the first bulb and\({\rm{Y = }}\)the lifetime of the second bulb (both in\({\rm{1000}}\)s of hours). Suppose that\({\rm{X}}\)and\({\rm{Y}}\)are independent and that each has an exponential distribution with parameter\({\rm{\lambda = 1}}\).

a. What is the joint pdf of\({\rm{X}}\)and\({\rm{Y}}\)?

b. What is the probability that each bulb lasts at most\({\rm{1000}}\)hours (i.e.,\({\rm{X拢1}}\)and\({\rm{Y拢1)}}\)?

c. What is the probability that the total lifetime of the two bulbs is at most\({\rm{2}}\)? (Hint: Draw a picture of the region before integrating.)

d. What is the probability that the total lifetime is between\({\rm{1}}\)and\({\rm{2}}\)?

The mean weight of luggage checked by a randomly selected tourist-class passenger flying between two cities on a certain airline is\({\bf{40}}\)lb, and the standard deviation is\({\bf{10}}\)lb. The mean and standard deviation for a business class passenger is\({\bf{30}}\)lb and\({\bf{6}}\)lb, respectively.

a. If there are\({\bf{12}}\)business-class passengers and\({\bf{50}}\)tourist-class passengers on a particular flight, what is the expected value of total luggage weight and the standard deviation of total luggage weight?

b. If individual luggage weights are independent, normally distributed RVs, what is the probability that total luggage weight is at most\({\bf{2500}}\)lb?

Suppose the expected tensile strength of type-A steel is \({\rm{105ksi}}\)and the standard deviation of tensile strength is \({\rm{8ksi}}\). For type-B steel, suppose the expected tensile strength and standard deviation of tensile strength are \({\rm{100ksi}}\)and \({\rm{6ksi}}\), respectively. Let \({\rm{\bar X = }}\)the sample average tensile strength of a random sample of \({\rm{40}}\) type-A specimens, and let \({\rm{\bar Y = }}\)the sample average tensile strength of a random sample of \({\rm{35}}\)type-B specimens.

a. What is the approximate distribution of \({\rm{\bar X ? of \bar Y?}}\)

b. What is the approximate distribution of \({\rm{\bar X - \bar Y}}\)? Justify your answer.

c. Calculate (approximately) \(P( - 1拢\bar X - \bar Y拢1)\)

d. Calculate. If you actually observed , would you doubt that \({{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}{\rm{ = 5?}}\)

Show that if\({\rm{X}}\)and\({\rm{Y}}\)are independent rv's, then\({\rm{E(XY) = E(X) \times E(Y)}}\). Then apply this in Exercise\({\rm{25}}{\rm{.}}\)[A1] (Hint: Consider the continuous case with\({\rm{f(x,y) = }}\)\({{\rm{f}}_{\rm{X}}}{\rm{(x) \times }}{{\rm{f}}_{\rm{Y}}}{\rm{(y)}}\).)

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