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Chapter 2: Q57E (page 84)

If \({\rm{P(B}}\mid {\rm{A) > P(B)}}\), show that \({\rm{P}}\left( {{{\rm{B}}^{\rm{'}}}\mid {\rm{A}}} \right){\rm{ < P}}\left( {{{\rm{B}}^{\rm{'}}}} \right)\). (Hint: Add \({\rm{P}}\left( {{{\rm{B}}'}\mid {\rm{A}}} \right)\) to both sides of the given inequality.)

Short Answer

Expert verified

The solution is

\(\begin{array}{l}P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{10}}} \right) = 0.401\\P\left( {{A_1} \cup {A_2} \cup \ldots \cup {A_{25}}} \right) = 0.723\end{array}\)

Step by step solution

01

Definition

In mathematics, an inequality is a relation that compares two numbers or other mathematical expressions in a non-equal way. It's most commonly used to compare two numbers on a number line based on their size.

02

Determine

\({\rm{P(A}}\mid {\rm{B) ,P}}\left( {{{\rm{A}}'}\mid {\rm{B}}} \right)\)

Add \({\rm{P}}\left( {{{\rm{B}}'}{\rm{/A}}} \right)\) to both sides as the book suggests.

\({\rm{P}}\left( {{{\rm{B}}'}{\rm{/A}}} \right){\rm{ + P(B/A) > P(B) + P}}\left( {{{\rm{B}}'}{\rm{/A}}} \right)\)

We determined this in with

\({\rm{P}}\left( {{{\rm{B}}'}{\rm{/A}}} \right){\rm{ + P(B/A) = 1}}\)

By substituting result into our inequality, we come to this.

\({\rm{1 > P(B) + P}}\left( {{{\rm{B}}'}{\rm{/A}}} \right)\)

Moving from one side to the other.

\({\rm{1 - P(B) > P}}\left( {{{\rm{B}}'}{\rm{/A}}} \right)\)

03

Proofing that \({\rm{P}}\left( {{{\rm{B}}^{\rm{'}}}\mid {\rm{A}}} \right){\rm{ < P}}\left( {{{\rm{B}}^{\rm{'}}}} \right)\)

\({\rm{1 - P(B) = P}}\left( {{{\rm{B}}'}} \right)\): One of the laws of probability.

\({\rm{P}}\left( {{{\rm{B}}'}} \right){\rm{ > P}}\left( {{{\rm{B}}'}{\rm{/A}}} \right)\)

Therefore, solution is\({\rm{P}}\left( {{{\rm{B}}'}} \right){\rm{ > P}}\left( {{{\rm{B}}'}{\rm{/A}}} \right)\)

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Most popular questions from this chapter

At a certain gas station, \({\rm{40\% }}\)of the customers use regular gas \(\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{,35\% }}\) use plus gas\(\left( {{{\rm{A}}_{\rm{2}}}} \right)\), and \({\rm{25\% }}\) use premium\(\left( {{{\rm{A}}_{\rm{3}}}} \right)\). Of those customers using regular gas, only \({\rm{30\% }}\) fill their tanks (event \({\rm{B}}\) ). Of those customers using plus, \({\rm{60\% }}\)fill their tanks, whereas of those using premium, \({\rm{50\% }}\)fill their tanks.

a. What is the probability that the next customer will request plus gas and fill the tank\(\left( {{{\rm{A}}_{\rm{2}}} \cap {\rm{B}}} \right)\)?

b. What is the probability that the next customer fills the tank?

c. If the next customer fills the tank, what is the probability that regular gas is requested? Plus? Premium?

Consider randomly selecting a single individual and having that person test drive \({\rm{3}}\) different vehicles. Define events \({{\rm{A}}_{\rm{1}}}\), \({{\rm{A}}_{\rm{2}}}\), and \({{\rm{A}}_{\rm{3}}}\) by

\({{\rm{A}}_{\rm{1}}}\)=likes vehicle #\({\rm{1}}\)\({{\rm{A}}_{\rm{2}}}\)= likes vehicle #\({\rm{2}}\)\({{\rm{A}}_{\rm{3}}}\)=likes vehicle #\({\rm{3}}\)Suppose that\({\rm{ = }}{\rm{.65,}}\)\({\rm{P(}}{{\rm{A}}_3}{\rm{)}}\)\({\rm{ = }}{\rm{.70,}}\)\({\rm{P(}}{{\rm{A}}_{\rm{1}}} \cup {{\rm{A}}_{\rm{2}}}{\rm{) = }}{\rm{.80,P(}}{{\rm{A}}_{\rm{2}}} \cap {{\rm{A}}_{\rm{3}}}{\rm{) = 40,}}\)and\({\rm{P(}}{{\rm{A}}_{\rm{1}}} \cup {{\rm{A}}_{\rm{2}}} \cup {{\rm{A}}_{\rm{3}}}{\rm{) = }}{\rm{.88}}{\rm{.}}\)

a. What is the probability that the individual likes both vehicle #\({\rm{1}}\)and vehicle #\({\rm{2}}\)?

b. Determine and interpret\({\rm{p}}\)(\({{\rm{A}}_{\rm{2}}}\)|\({{\rm{A}}_{\rm{3}}}\)).

c. Are \({{\rm{A}}_{\rm{2}}}\)and \({{\rm{A}}_{\rm{3}}}\)independent events? Answer in two different ways.

d. If you learn that the individual did not like vehicle #\({\rm{1}}\), what now is the probability that he/she liked at least one of the other two vehicles?

Let Adenote the event that the next request for assistance from a statistical software consultant relates to the SPSS package, and let Bbe the event that the next request is for help with SAS. Suppose that P(A)=.30and P(B)=.50.

a. Why is it not the case that P(A)+P(B)=1?

b. Calculate P(A鈥�).

c. Calculate P(A\( \cup \)B).

d. Calculate P(A鈥橽( \cap \)B鈥�).

Three molecules of type A, three of type B, three of type C, and three of type \({\rm{D}}\) are to be linked together to form a chain molecule. One such chain molecule is ABCDABCDABCD, and another is BCDDAAABDBCC.

a. How many such chain molecules are there? (Hint: If the three were distinguishable from one another鈥� \({\rm{A1,\;A2, A3}}\)鈥攁nd the were also, how many molecules would there be? How is this number reduced when the subscripts are removed from the ?)

b. Suppose a chain molecule of the type described is randomly selected. What is the probability that all three molecules of each type end up next to one another (such as in BBBAAADDDCCC)?

A friend of mine is giving a dinner party. His current wine supply includes 8 bottles of zinfandel, \({\rm{10}}\) of merlot, and \({\rm{12}}\) of cabernet (he only drinks red wine), all from different wineries.

a. If he wants to serve \({\rm{3}}\) bottles of zinfandel and serving order is important, how many ways are there to do this?

b. If \({\rm{6}}\) bottles of wine are to be randomly selected from the \({\rm{30}}\) for serving, how many ways are there to do this?

c. If \({\rm{6}}\) bottles are randomly selected, how many ways are there to obtain two bottles of each variety?

d. If \({\rm{6}}\) bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen?

e. If \({\rm{6}}\) bottles are randomly selected, what is the probability that all of them are the same variety?
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