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If $P(A)=0.8, P(B)=0.5$, and $P(A \cup B)=0.9$. What is $P(A \cap B)$?
|
0.9
|
0.9
|
Problem 1.4.5
|
stat
|
math
|
|||
Suppose that the alleles for eye color for a certain male fruit fly are $(R, W)$ and the alleles for eye color for the mating female fruit fly are $(R, W)$, where $R$ and $W$ represent red and white, respectively. Their offspring receive one allele for eye color from each parent. Assume that each of the four possible outcomes has equal probability. If an offspring ends up with either two white alleles or one red and one white allele for eye color, its eyes will look white. Given that an offspring's eyes look white, what is the conditional probability that it has two white alleles for eye color?
|
$\frac{1}{3}$
|
0.33333333
|
Problem 1.3.5
|
stat
|
math
|
|||
Consider the trial on which a 3 is first observed in successive rolls of a six-sided die. Let $A$ be the event that 3 is observed on the first trial. Let $B$ be the event that at least two trials are required to observe a 3 . Assuming that each side has probability $1 / 6$, find $P(A)$.
|
$\frac{1}{6}$
|
0.166666666
|
Problem 1.1.5
|
stat
|
math
|
|||
An urn contains four balls numbered 1 through 4 . The balls are selected one at a time without replacement. A match occurs if the ball numbered $m$ is the $m$ th ball selected. Let the event $A_i$ denote a match on the $i$ th draw, $i=1,2,3,4$. Extend this exercise so that there are $n$ balls in the urn. What is the limit of this probability as $n$ increases without bound?
|
$1 - \frac{1}{e}$
|
0.6321205588
|
Problem 1.3.9
|
stat
|
math
|
|||
Of a group of patients having injuries, $28 \%$ visit both a physical therapist and a chiropractor and $8 \%$ visit neither. Say that the probability of visiting a physical therapist exceeds the probability of visiting a chiropractor by $16 \%$. What is the probability of a randomly selected person from this group visiting a physical therapist?
|
0.68
|
0.68
|
Problem 1.1.1
|
stat
|
math
|
|||
A doctor is concerned about the relationship between blood pressure and irregular heartbeats. Among her patients, she classifies blood pressures as high, normal, or low and heartbeats as regular or irregular and finds that 16\% have high blood pressure; (b) 19\% have low blood pressure; (c) $17 \%$ have an irregular heartbeat; (d) of those with an irregular heartbeat, $35 \%$ have high blood pressure; and (e) of those with normal blood pressure, $11 \%$ have an irregular heartbeat. What percentage of her patients have a regular heartbeat and low blood pressure?
|
15.1
|
15.1
|
1.5.3
|
%
|
stat
|
math
|
||
Roll a fair six-sided die three times. Let $A_1=$ $\{1$ or 2 on the first roll $\}, A_2=\{3$ or 4 on the second roll $\}$, and $A_3=\{5$ or 6 on the third roll $\}$. It is given that $P\left(A_i\right)=1 / 3, i=1,2,3 ; P\left(A_i \cap A_j\right)=(1 / 3)^2, i \neq j$; and $P\left(A_1 \cap A_2 \cap A_3\right)=(1 / 3)^3$. Use Theorem 1.1-6 to find $P\left(A_1 \cup A_2 \cup A_3\right)$.
|
$3(\frac{1}{3})-3(\frac{1}{3})^2+(\frac{1}{3})^3$
|
0.6296296296
|
Problem 1.1.9
|
stat
|
math
|
|||
Let $A$ and $B$ be independent events with $P(A)=$ $1 / 4$ and $P(B)=2 / 3$. Compute $P(A \cap B)$
|
$\frac{1}{6}$
|
0.166666666
|
Problem 1.4.3
|
stat
|
math
|
|||
How many four-letter code words are possible using the letters in IOWA if the letters may not be repeated?
|
24
|
24
|
Problem 1.2.5
|
stat
|
math
|
|||
A boy found a bicycle lock for which the combination was unknown. The correct combination is a four-digit number, $d_1 d_2 d_3 d_4$, where $d_i, i=1,2,3,4$, is selected from $1,2,3,4,5,6,7$, and 8 . How many different lock combinations are possible with such a lock?
|
4096
|
4096
|
Problem 1.2.1
|
stat
|
math
|
|||
An urn contains eight red and seven blue balls. A second urn contains an unknown number of red balls and nine blue balls. A ball is drawn from each urn at random, and the probability of getting two balls of the same color is $151 / 300$. How many red balls are in the second urn?
|
11
|
11
|
Problem 1.3.15
|
stat
|
math
|
|||
A typical roulette wheel used in a casino has 38 slots that are numbered $1,2,3, \ldots, 36,0,00$, respectively. The 0 and 00 slots are colored green. Half of the remaining slots are red and half are black. Also, half of the integers between 1 and 36 inclusive are odd, half are even, and 0 and 00 are defined to be neither odd nor even. A ball is rolled around the wheel and ends up in one of the slots; we assume that each slot has equal probability of $1 / 38$, and we are interested in the number of the slot into which the ball falls. Let $A=\{0,00\}$. Give the value of $P(A)$.
|
$\frac{2}{38}$
|
0.0526315789
|
Problem 1.1.1
|
stat
|
math
|
|||
In the gambling game "craps," a pair of dice is rolled and the outcome of the experiment is the sum of the points on the up sides of the six-sided dice. The bettor wins on the first roll if the sum is 7 or 11. The bettor loses on the first roll if the sum is 2,3 , or 12 . If the sum is $4,5,6$, 8,9 , or 10 , that number is called the bettor's "point." Once the point is established, the rule is as follows: If the bettor rolls a 7 before the point, the bettor loses; but if the point is rolled before a 7 , the bettor wins. Find the probability that the bettor wins on the first roll. That is, find the probability of rolling a 7 or 11 , $P(7$ or 11$)$.
|
$\frac{8}{36}$
|
0.22222222
|
Problem 1.3.13
|
stat
|
math
|
|||
Given that $P(A \cup B)=0.76$ and $P\left(A \cup B^{\prime}\right)=0.87$, find $P(A)$.
|
0.63
|
0.63
|
Problem 1.1.7
|
stat
|
math
|
|||
How many different license plates are possible if a state uses two letters followed by a four-digit integer (leading zeros are permissible and the letters and digits can be repeated)?
|
6760000
|
6760000
|
Problem 1.2.3
|
stat
|
math
|
|||
Let $A$ and $B$ be independent events with $P(A)=$ 0.7 and $P(B)=0.2$. Compute $P(A \cap B)$.
|
0.14
|
0.14
|
Problem 1.4.1
|
stat
|
math
|
|||
Suppose that $A, B$, and $C$ are mutually independent events and that $P(A)=0.5, P(B)=0.8$, and $P(C)=$ 0.9 . Find the probabilities that all three events occur?
|
0.36
|
0.36
|
Problem 1.4.9
|
stat
|
math
|
|||
A poker hand is defined as drawing 5 cards at random without replacement from a deck of 52 playing cards. Find the probability of four of a kind (four cards of equal face value and one card of a different value).
|
0.00024
|
0.00024
|
Problem 1.2.17
|
stat
|
math
|
|||
Three students $(S)$ and six faculty members $(F)$ are on a panel discussing a new college policy. In how many different ways can the nine participants be lined up at a table in the front of the auditorium?
|
362880
|
362880
|
Problem 1.2.11
|
stat
|
math
|
|||
Each of the 12 students in a class is given a fair 12 -sided die. In addition, each student is numbered from 1 to 12 . If the students roll their dice, what is the probability that there is at least one "match" (e.g., student 4 rolls a 4)?
|
$1-(11 / 12)^{12}$
|
0.648004372
|
Problem 1.4.17
|
stat
|
math
|
|||
The World Series in baseball continues until either the American League team or the National League team wins four games. How many different orders are possible (e.g., ANNAAA means the American League team wins in six games) if the series goes four games?
|
2
|
2
|
Problem 1.2.9
|
stat
|
math
|
|||
Draw one card at random from a standard deck of cards. The sample space $S$ is the collection of the 52 cards. Assume that the probability set function assigns $1 / 52$ to each of the 52 outcomes. Let
$$
\begin{aligned}
A & =\{x: x \text { is a jack, queen, or king }\}, \\
B & =\{x: x \text { is a } 9,10, \text { or jack and } x \text { is red }\}, \\
C & =\{x: x \text { is a club }\}, \\
D & =\{x: x \text { is a diamond, a heart, or a spade }\} .
\end{aligned}
$$
Find $P(A)$
|
$\frac{12}{52}$
|
0.2307692308
|
Problem 1.1.3
|
stat
|
math
|
|||
An urn contains four colored balls: two orange and two blue. Two balls are selected at random without replacement, and you are told that at least one of them is orange. What is the probability that the other ball is also orange?
|
$\frac{1}{5}$
|
0.2
|
Problem 1.3.7
|
stat
|
math
|
|||
Bowl $B_1$ contains two white chips, bowl $B_2$ contains two red chips, bowl $B_3$ contains two white and two red chips, and bowl $B_4$ contains three white chips and one red chip. The probabilities of selecting bowl $B_1, B_2, B_3$, or $B_4$ are $1 / 2,1 / 4,1 / 8$, and $1 / 8$, respectively. A bowl is selected using these probabilities and a chip is then drawn at random. Find $P(W)$, the probability of drawing a white chip.
|
$\frac{21}{32}$
|
0.65625
|
Problem 1.5.1
|
stat
|
math
|
|||
Divide a line segment into two parts by selecting a point at random. Use your intuition to assign a probability to the event that the longer segment is at least two times longer than the shorter segment.
|
$\frac{2}{3}$
|
0.66666666666
|
Problem 1.1.13
|
stat
|
math
|
|||
In a state lottery, four digits are drawn at random one at a time with replacement from 0 to 9. Suppose that you win if any permutation of your selected integers is drawn. Give the probability of winning if you select $6,7,8,9$.
|
0.0024
|
0.0024
|
Problem 1.2.7
|
stat
|
math
|
|||
Suppose that a fair $n$-sided die is rolled $n$ independent times. A match occurs if side $i$ is observed on the $i$ th trial, $i=1,2, \ldots, n$. Find the limit of this probability as $n$ increases without bound.
|
$ 1-1 / e$
|
0.6321205588
|
Problem 1.4.19
|
stat
|
math
|
|||
The desired probability is
$$
\begin{aligned}
P\left(B \mid A_1 \cup A_2 \cup A_3\right) & =\frac{P\left(A_1 \cap B\right)+P\left(A_2 \cap B\right)+P\left(A_3 \cap B\right)}{P\left(A_1\right)+P\left(A_2\right)+P\left(A_3\right)} \\
& =\frac{(0.3)(0.6)+(0.2)(0.7)+(0.2)(0.8)}{0.3+0.2+0.2} \\
& =\frac{0.48}{0.70}=0.686 .
\end{aligned}
$$
|
An insurance company sells several types of insurance policies, including auto policies and homeowner policies. Let $A_1$ be those people with an auto policy only, $A_2$ those people with a homeowner policy only, and $A_3$ those people with both an auto and homeowner policy (but no other policies). For a person randomly selected from the company's policy holders, suppose that $P\left(A_1\right)=0.3, P\left(A_2\right)=0.2$, and $P\left(A_3\right)=0.2$. Further, let $B$ be the event that the person will renew at least one of these policies. Say from past experience that we assign the conditional probabilities $P\left(B \mid A_1\right)=0.6, P\left(B \mid A_2\right)=0.7$, and $P\left(B \mid A_3\right)=0.8$. Given that the person selected at random has an auto or homeowner policy, what is the conditional probability that the person will renew at least one of those policies?
|
0.686
|
0.686
|
Example 1.3.11
|
stat
|
math
|
||
The number of possible 13-card hands (in bridge) that can be selected from a deck of 52 playing cards is
$$
{ }_{52} C_{13}=\left(\begin{array}{l}
52 \\
13
\end{array}\right)=\frac{52 !}{13 ! 39 !}=635,013,559,600 .
$$
|
What is the number of possible 13-card hands (in bridge) that can be selected from a deck of 52 playing cards?
|
635013559600
|
635013559600
|
Example 1.2.10
|
stat
|
math
|
||
The number of ways of selecting a president, a vice president, a secretary, and a treasurer in a club consisting of 10 persons is
$$
{ }_{10} P_4=10 \cdot 9 \cdot 8 \cdot 7=\frac{10 !}{6 !}=5040 .
$$
|
What is the number of ways of selecting a president, a vice president, a secretary, and a treasurer in a club consisting of 10 persons?
|
5040
|
5040
|
Example 1.2.5
|
stat
|
math
|
||
Of the 24 remaining balloons, 9 are yellow, so a natural value to assign to this conditional probability is $9 / 24$.
|
At a county fair carnival game there are 25 balloons on a board, of which 10 balloons 1.3-5 are yellow, 8 are red, and 7 are green. A player throws darts at the balloons to win a prize and randomly hits one of them. Given that the first balloon hit is yellow, what is the probability that the next balloon hit is also yellow?
|
$\frac{9}{24}$
|
0.375
|
Example 1.3.5
|
stat
|
math
|
||
The number of ordered samples of 5 cards that can be drawn without replacement from a standard deck of 52 playing cards is
$$
(52)(51)(50)(49)(48)=\frac{52 !}{47 !}=311,875,200 .
$$
|
What is the number of ordered samples of 5 cards that can be drawn without replacement from a standard deck of 52 playing cards?
|
311875200
|
311875200
|
Example 1.2.8
|
stat
|
math
|
||
It is reasonable to assign the following probabilities:
$$
P(A)=\frac{3}{10} \text { and } P(B \mid A)=\frac{7}{9} \text {. }
$$
The probability of obtaining red on the first draw and blue on the second draw is
$$
P(A \cap B)=\frac{3}{10} \cdot \frac{7}{9}=\frac{7}{30}
$$
|
A bowl contains seven blue chips and three red chips. Two chips are to be drawn successively at random and without replacement. We want to compute the probability that the first draw results in a red chip $(A)$ and the second draw results in a blue chip $(B)$.
|
$\frac{7}{30}$
|
0.23333333333
|
Example 1.3.6
|
stat
|
math
|
||
Let $A$ be the event of two spades in the first five cards drawn, and let $B$ be the event of a spade on the sixth draw. Thus, the probability that we wish to compute is $P(A \cap B)$. It is reasonable to take
$$
P(A)=\frac{\left(\begin{array}{c}
13 \\
2
\end{array}\right)\left(\begin{array}{c}
39 \\
3
\end{array}\right)}{\left(\begin{array}{c}
52 \\
5
\end{array}\right)}=0.274 \quad \text { and } \quad P(B \mid A)=\frac{11}{47}=0.234
$$
The desired probability, $P(A \cap B)$, is the product of those numbers:
$$
P(A \cap B)=(0.274)(0.234)=0.064
$$
|
From an ordinary deck of playing cards, cards are to be drawn successively at random and without replacement. What is the probability that the third spade appears on the sixth draw?
|
0.064
|
0.064
|
Example 1.3.7
|
stat
|
math
|
||
Assume that each of the $\left(\begin{array}{c}52 \\ 5\end{array}\right)=2,598,960$ five-card hands drawn from a deck of 52 playing cards has the same probability of being selected.
Suppose now that the event $B$ is the set of outcomes in which exactly three cards are kings and exactly two cards are queens. We can select the three kings in any one of $\left(\begin{array}{l}4 \\ 3\end{array}\right)$ ways and the two queens in any one of $\left(\begin{array}{l}4 \\ 2\end{array}\right)$ ways. By the multiplication principle, the number of outcomes in $B$ is
$$
N(B)=\left(\begin{array}{l}
4 \\
3
\end{array}\right)\left(\begin{array}{l}
4 \\
2
\end{array}\right)\left(\begin{array}{c}
44 \\
0
\end{array}\right)
$$
where $\left(\begin{array}{c}44 \\ 0\end{array}\right)$ gives the number of ways in which 0 cards are selected out of the nonkings and nonqueens and of course is equal to 1 . Thus,
$$
P(B)=\frac{N(B)}{N(S)}=\frac{\left(\begin{array}{l}
4 \\
3
\end{array}\right)\left(\begin{array}{c}
4 \\
2
\end{array}\right)\left(\begin{array}{c}
44 \\
0
\end{array}\right)}{\left(\begin{array}{c}
52 \\
5
\end{array}\right)}=\frac{24}{2,598,960}=0.0000092 .
$$
|
What is the probability of drawing three kings and two queens when drawing a five-card hand from a deck of 52 playing cards?
|
0.0000092
|
0.0000092
|
Example 1.2.11
|
stat
|
math
|
||
Considering only the color of the orchids, we see that the number of lineups of the orchids is
$$
\left(\begin{array}{l}
7 \\
4
\end{array}\right)=\frac{7 !}{4 ! 3 !}=35 \text {. }
$$
|
In an orchid show, seven orchids are to be placed along one side of the greenhouse. There are four lavender orchids and three white orchids. How many ways are there to lineup these orchids?
|
35
|
35
|
Example 1.2.13
|
stat
|
math
|
||
$P(B \mid A)=P(A \cap B) / P(A)=0.3 / 0.4=0.75$.
|
If $P(A)=0.4, P(B)=0.5$, and $P(A \cap B)=0.3$, find $P(B \mid A)$.
|
0.75
|
0.75
|
Example 1.3.2
|
stat
|
math
|
||
The number of possible 5-card hands (in 5-card poker) drawn from a deck of 52 playing cards is
$$
{ }_{52} C_5=\left(\begin{array}{c}
52 \\
5
\end{array}\right)=\frac{52 !}{5 ! 47 !}=2,598,960
$$
|
What is the number of possible 5-card hands (in 5-card poker) drawn from a deck of 52 playing cards?
|
2598960
|
2598960
|
Example 1.2.9
|
stat
|
math
|
||
By the multiplication principle, there are
$(2)(2)(3)(2)(4)(7)(4)=2688$
different combinations.
|
A certain food service gives the following choices for dinner: $E_1$, soup or tomato 1.2-2 juice; $E_2$, steak or shrimp; $E_3$, French fried potatoes, mashed potatoes, or a baked potato; $E_4$, corn or peas; $E_5$, jello, tossed salad, cottage cheese, or coleslaw; $E_6$, cake, cookies, pudding, brownie, vanilla ice cream, chocolate ice cream, or orange sherbet; $E_7$, coffee, tea, milk, or punch. How many different dinner selections are possible if one of the listed choices is made for each of $E_1, E_2, \ldots$, and $E_7$ ?
|
2688
|
2688
|
Example 1.2.2
|
stat
|
math
|
||
Because the system fails if $K_1$ fails and $K_2$ fails and $K_3$ fails, the probability that the system does not fail is given by
$$
\begin{aligned}
P\left[\left(A_1 \cap A_2 \cap A_3\right)^{\prime}\right] & =1-P\left(A_1 \cap A_2 \cap A_3\right) \\
& =1-P\left(A_1\right) P\left(A_2\right) P\left(A_3\right) \\
& =1-(0.15)^3 \\
& =0.9966 .
\end{aligned}
$$
|
A rocket has a built-in redundant system. In this system, if component $K_1$ fails, it is bypassed and component $K_2$ is used. If component $K_2$ fails, it is bypassed and component $K_3$ is used. (An example of a system with these kinds of components is three computer systems.) Suppose that the probability of failure of any one component is 0.15 , and assume that the failures of these components are mutually independent events. Let $A_i$ denote the event that component $K_i$ fails for $i=1,2,3$. What is the probability that the system fails?
|
0.9966
|
0.9966
|
Example 1.4.5
|
stat
|
math
|
||
We are effectively restricting the sample space to $B$; of the probability $P(B)=0.3,0.2$ corresponds to $P(A \cap B)$ and hence to $A$. That is, $0.2 / 0.3=2 / 3$ of the probability of $B$ corresponds to $A$. Of course, by the formal definition, we also obtain
$$
P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{0.2}{0.3}=\frac{2}{3}
$$
|
Suppose that $P(A)=0.7, P(B)=0.3$, and $P(A \cap B)=0.2$. Given that the outcome of the experiment belongs to $B$, what then is the probability of $A$ ?
|
\frac{2}{3}
|
0.66666666666
|
Example 1.3.3
|
stat
|
math
|
||
A coin is flipped 10 times and the sequence of heads and tails is observed. The number of possible 10-tuplets that result in four heads and six tails is
$$
\left(\begin{array}{c}
10 \\
4
\end{array}\right)=\frac{10 !}{4 ! 6 !}=\frac{10 !}{6 ! 4 !}=\left(\begin{array}{c}
10 \\
6
\end{array}\right)=210 .
$$
|
A coin is flipped 10 times and the sequence of heads and tails is observed. What is the number of possible 10-tuplets that result in four heads and six tails?
|
210
|
210
|
Example 1.2.12
|
stat
|
math
|
||
The number of different color displays is
$$
\left(\begin{array}{c}
9 \\
3,4,2
\end{array}\right)=\frac{9 !}{3 ! 4 ! 2 !}=1260
$$
|
Among nine orchids for a line of orchids along one wall, three are white, four lavender, and two yellow. How many color displays are there?
|
1260
|
1260
|
Example 1.2.14
|
stat
|
math
|
||
$$
\begin{aligned}
P(A \cup B \cup C)= & P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C) \\
& -P(B \cap C)+P(A \cap B \cap C) \\
= & 0.43+0.40+0.32-0.29-0.22-0.20+0.15 \\
= & 0.59
\end{aligned}
$$
|
A survey was taken of a group's viewing habits of sporting events on TV during I.I-5 the last year. Let $A=\{$ watched football $\}, B=\{$ watched basketball $\}, C=\{$ watched baseball $\}$. The results indicate that if a person is selected at random from the surveyed group, then $P(A)=0.43, P(B)=0.40, P(C)=0.32, P(A \cap B)=0.29$, $P(A \cap C)=0.22, P(B \cap C)=0.20$, and $P(A \cap B \cap C)=0.15$. Find $P(A \cup B \cup C)$.
|
0.59
|
0.59
|
Example 1.1.5
|
stat
|
math
|
||
For notation, let $B L, B R$, and $W L$ denote drawing blue from left pocket, blue from right pocket, and white from left pocket, respectively. Then
$$
\begin{aligned}
P(B R) & =P(B L \cap B R)+P(W L \cap B R) \\
& =P(B L) P(B R \mid B L)+P(W L) P(B R \mid W L) \\
& =\frac{5}{9} \cdot \frac{5}{10}+\frac{4}{9} \cdot \frac{4}{10}=\frac{41}{90}
\end{aligned}
$$
is the desired probability.
|
A grade school boy has five blue and four white marbles in his left pocket and four blue and five white marbles in his right pocket. If he transfers one marble at random from his left to his right pocket, what is the probability of his then drawing a blue marble from his right pocket?
|
$\frac{41}{90}$
|
0.444444444444444
|
Example 1.3.10
|
stat
|
math
|
||
$$P(A \cup B) =P(A)+P(B)-P(A \cap B)=0.93+0.89-0.87=0.95$$
|
A faculty leader was meeting two students in Paris, one arriving by train from Amsterdam and the other arriving by train from Brussels at approximately the same time. Let $A$ and $B$ be the events that the respective trains are on time. Suppose we know from past experience that $P(A)=0.93, P(B)=0.89$, and $P(A \cap B)=0.87$. Find $P(A \cup B)$.
|
0.95
|
0.95
|
Example 1.1.4
|
stat
|
math
|
||
The number of possible four-letter code words, selecting from the 26 letters in the alphabet, in which all four letters are different is
$$
{ }_{26} P_4=(26)(25)(24)(23)=\frac{26 !}{22 !}=358,800 .
$$
|
What is the number of possible four-letter code words, selecting from the 26 letters in the alphabet?
|
358800
|
358800
|
Example 1.2.4
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stat
|
math
|
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