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Chapter 5: Q3E (page 210)

A certain market has both an express checkout line and a superexpress checkout line. Let \({{\rm{X}}_{\rm{1}}}\) denote the number of customers in line at the express checkout at a particular time of day, and let \({{\rm{X}}_{\rm{2}}}\) denote the number of customers in line at the superexpress checkout at the same time. Suppose the joint pmf of \({{\rm{X}}_{\rm{1}}}\)and\({{\rm{X}}_{\rm{2}}}\) is as given in the accompanying table

a. What is \({\rm{P(}}{{\rm{X}}_{\rm{1}}}{\rm{ = 1}}{{\rm{X}}_{\rm{2}}} = 1)\), that is, the probability that there is exactly one customer in each line?

b. What is \({\rm{P(}}{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{X}}_{\rm{2}}})\),that is, the probability that the numbers of customers in the two lines are identical? c. Let A denote the event that there are at least two more customers in one line than in the other line. Express A in terms of \({{\rm{X}}_{\rm{1}}}\) and \({{\rm{X}}_{\rm{2}}}\), and calculate the probability of this event.

d. What is the probability that the total number of customers in the two lines is exactly four? At least four?

Short Answer

Expert verified
  1. The probability is \({\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 1,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 1}}} \right){\rm{ = 0}}{\rm{.15;}}\)
  2. The probability is \({\rm{ P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{X}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.4;}}\)
  3. The probability is \({\rm{ P(A) = 0}}{\rm{.22;}}\)
  4. The probability is \({\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}} \ge {\rm{4}}} \right){\rm{ = 0}}{\rm{.46}}{\rm{.}}\)

Step by step solution

01

Definition of Probability

Probability is a metric for determining the possibility of an event occurring. Many things are impossible to forecast with\({\rm{100\% }}\)accuracy. Using it, we can only anticipate the probability of an event occurring, or how probable it is to occur. Probability can range from\({\rm{0}}\)to\({\rm{1}}\), with\({\rm{0}}\)indicating an improbable event and 1 indicating a certain event. Possibility of...

02

Find the probability that there is exactly one customer in each line?

(a):

As an entry, the following probability can be read from the given table.\((1,1)\)-first column/first row

\(\begin{aligned}{\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = 1,}}{{\rm{X}}_{\rm{2}}}{\rm{ = 1}}} \right)\\{\rm{ = p(1,1)}}\\{\rm{ = 0}}{\rm{.15}}{\rm{.}}\end{aligned}\)

Therefore , the probability is \( = 0.15\).

03

 Find the probability that the numbers of customers?

(b):

The following holds

\(\begin{aligned}{\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{X}}_{\rm{2}}}} \right){\rm{ = p(0,0) + p(1,1) + p(2,2) + p(3,3)}}\\{\rm{ = 0}}{\rm{.08 + 0}}{\rm{.15 + 0}}{\rm{.1 + 0}}{\rm{.7}}\\{\rm{ = 0}}{\rm{.4}}\end{aligned}\)

Therefore , the probability is \( = 0.4\).

04

Calculate the probability of this event?

(c):

The following two events can be combined to form the provided event.

\(\left\{ {{{\rm{X}}_{\rm{1}}} \ge {\rm{2 + }}{{\rm{X}}_{\rm{2}}}} \right\}{\rm{ and }}\left\{ {{{\rm{X}}_{\rm{2}}} \ge 2 + {{\rm{X}}_{\rm{1}}}} \right\}.\)

Therefore, we have

\({\rm{A = }}\left\{ {\left\{ {{{\rm{X}}_{\rm{1}}} \ge 2 + {{\rm{X}}_{\rm{2}}}} \right\} \cup \left\{ {{{\rm{X}}_{\rm{2}}} \ge 2 + {{\rm{X}}_{\rm{1}}}} \right\}} \right\}\)

The event \({\rm{A}}\)can be represented as the sum of nine discrete occurrences in form.

\(\left\{ {{{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}} \right\}{\rm{,}}\left\{ {{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}} \right\}\)

where \({{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}\) satisfy the event \({\rm{A}}\)'s specified property. As a result, the chance can be computed as follows:

\(\begin{aligned}{\rm{P(A) = P}}\left( {\left\{ {{X_1} \ge 2 + {X_2}} \right\} \cup \left\{ {{X_2} \ge 2 + {X_1}} \right\}} \right)\\{\rm{ = p(2,0) + p(3,0) + p(4,0) + p(3,1) + p(4,1) + p(4,2) + p(0,2) + p(0,3) + p(1,3)}}\\{\rm{ = 0}}{\rm{.05 + 0}}{\rm{.00 + 0}}{\rm{.00 + 0}}{\rm{.03 + 0}}{\rm{.01 + 0}}{\rm{.05 + 0}}{\rm{.04 + 0}}{\rm{.00 + 0}}{\rm{.04}}\\{\rm{ = 0}}{\rm{.22}}{\rm{.}}\end{aligned}\)

Therefore , the probability is \({\rm{ = }}{\rm{.22}}\).

05

Find the probability that the total number of customers?

(d):

The following holds

\(\begin{aligned}{\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}{\rm{ = 4}}} \right){\rm{ = p(1,3) + p(2,2) + p(3,1) + p(4,0)}}\\{\rm{ = 0}}{\rm{.04 + 0}}{\rm{.1 + 0}}{\rm{.03 + 0}}{\rm{.00}}\\{\rm{ = 0}}{\rm{.17}}{\rm{.}}\end{aligned}\)

And for the second part the following is true

\(\begin{aligned}{l}{\rm{P}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}} \ge {\rm{4}}} \right){\rm{ = p(1,3) + p(2,2) + p(3,1) + p(4,0) + p(4,1) + p(4,2) + p(4,3) + p(3,2) + p(3,3) + p(2,3)}}\\{\rm{ = 0}}{\rm{.46}}\end{aligned}\)

All values can be found at the given table (see \({\rm{(a)}}\)).

Therefore , the probability is \({\rm{ = 0}}{\rm{.46}}\).

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Most popular questions from this chapter

A more accurate approximation to \({\rm{E}}\left( {{\rm{h}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}} \right)} \right)\) in Exercise 95 is

\(h\left( {{\mu _1}, \ldots ,{\mu _n}} \right) + \frac{1}{2}\sigma _1^2\left( {\frac{{{\partial ^2}h}}{{\partial x_1^2}}} \right) + \cdots + \frac{1}{2}\sigma _n^2\left( {\frac{{{\partial ^2}h}}{{\partial x_n^2}}} \right)\)

Compute this for \({\rm{Y = h}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}{\rm{,}}{{\rm{X}}_{\rm{4}}}} \right)\)given in Exercise 93 , and compare it to the leading term \({\rm{h}}\left( {{{\rm{\mu }}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{\mu }}_{\rm{n}}}} \right)\).

The manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal distribution, and the three times are independent of one another. The mean values are\(15,\;30,\;20\)min, respectively, and the standard deviations are\(1,\;2,\;1.5\)min, respectively. What is the probability that it takes at most\(1\)hour of machining time to produce a randomly selected component?

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A binary communication channel transmits a sequence of 鈥渂its鈥� (\({\bf{0s}}{\rm{ }}{\bf{and}}{\rm{ }}{\bf{1s}}\)). Suppose that for any particular bit transmitted, there is a\({\bf{10}}\% \)chance of a transmission error (a zero becoming a one or a one becoming a zero). Assume that bit errors occur independently of one another.

a. Consider transmitting\(1000\)bits. What is the approximate probability that at most\(125\)transmission errors occur?

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