Chapter 3: ProbabilityContent check — Statistics (OpenIntro), Chapter 3: Probability

Here are four examples. (i) Whether someone gets sick in the next month or not is an apparently random process with outcomes sick and not. (ii) We can generate a random process by randomly picking a person and measuring that person’s height. The outcome of this process will be a positive number. (iii) Whether the stock market goes up or down next week is a seemingly random process with possible outcomes up, down, and no_change. Alternatively, we could have used the percent change in the stock market as a numerical outcome. (iv) Whether your roommate cleans her dishes tonight probably seems like a random process with possible outcomes cleans_dishes and leaves_dishes.

The random process is a die roll, and at most one of these outcomes can come up. This means they are disjoint outcomes.

$P($1 or 4 or 5⁠$)=P($1⁠$)+P($4⁠$)+P($5⁠$)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$

Yes. Each loan is categorized in only one level of homeownership.

Mortgage: $\frac{4789}{10000}=0.479$⁠. Own: $\frac{1353}{10000}=0.135$⁠.

$P($mortgage or own⁠$)=P($mortgage⁠$)+P($own⁠$)=0.479+0.135=0.614$⁠.

$P(A)=P($1 or 2⁠$)=P($1⁠$)+P($2⁠$)=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}$⁠.

Similarly, $P(B)=1/3$⁠.

Outcomes 2 and 3.

Yes, events $B$ and $D$ are disjoint because they share no outcomes.

The events $A$ and $D$ share an outcome in common, 2, and so are not disjoint.

Since $B$ and $D$ are disjoint events, use the Addition Rule: $P(B$ or $D)=P(B)+P(D)=\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$⁠.

There are 52 cards and 13 diamonds. If the cards are thoroughly shuffled, each card has an equal chance of being drawn, so the probability that a randomly selected card is a diamond is $P(♢)=\frac{13}{52}=0.250$⁠.

Likewise, there are 12 face cards, so $P($⁠face card⁠$)=\frac{12}{52}=\frac{3}{13}=0.231$⁠.

If $A$ and $B$ are disjoint, $A$ and $B$ can never occur simultaneously.

If $A$ and $B$ are disjoint, then the last $P(A\text{ and }B)$ term of in the General Addition Rule formula is 0 (see part (a)) and we are left with the Addition Rule for disjoint events.

Both the counts and corresponding probabilities (e.g. $3839/10000=0.384$⁠) are shown. Notice that the number of loans represented in the left circle corresponds to $3839+950=4789$⁠, and the number represented in the right circle is $950+545=1495$⁠.

The solution is represented by the intersection of the two circles: 0.095.

This is the sum of the three disjoint probabilities shown in the circles: $0.384+0.095+0.055=0.534$ (off by 0.001 due to a rounding error).

The probabilities of

do not sum to 1. The second probability in

is negative. This leaves

, which sure enough satisfies the requirements of a distribution. One of the three was said to be the actual distribution of US household incomes, so it must be (c).

The outcomes are disjoint and each has probability $1/6$⁠, so the total probability is $4/6=2/3$⁠.

We can also see that $P(D)=\frac{1}{6}+\frac{1}{6}=1/3$⁠. Since $D$ and $D^c$ are disjoint, $P(D)+P(D^c)=1$⁠.

Brief solutions:

$A^c=\{$3, 4, 5, 6⁠$\}$ and $B^c=\{$1, 2, 3, 5⁠$\}$⁠.

Noting that each outcome is disjoint, add the individual outcome probabilities to get $P(A^c)=2/3$ and $P(B^c)=2/3$⁠.

$A$⁠ and $A^c$ are disjoint, and the same is true of $B$⁠ and $B^c$⁠. Therefore, $P(A)+P(A^c)=1$ and $P(B)+P(B^c)=1$⁠.

The complement of $A$⁠: when the total is equal to 12.

$P(A^c)=1/36$⁠.

Use the probability of the complement from part (b), $P(A^c)=1/36$⁠, and the equation for the complement: $P($⁠less than 12⁠$)=1-P($12⁠$)=1-1/36=35/36$⁠.

First find $P($6⁠$)=5/36$⁠, then use the complement: $P($⁠not 6⁠$)=1-P($6⁠$)=31/36$⁠.

First find the complement, which requires much less effort: $P($2 or 3⁠$)=1/36+2/36=1/12$⁠. Then calculate $P(B)=$ $1-$ $P(B^c)=$ $1-$ $1/12=$ $11/12$⁠.

As before, finding the complement is the clever way to determine $P(D)$⁠. First find $P(D^c)=P($11 or 12⁠$)=2/36+1/36=1/12$⁠. Then calculate $P(D)=1-P(D^c)=11/12$⁠.

The probability the first person is left-handed is $0.09$⁠, which is the same for the second person. We apply the Multiplication Rule for independent processes to determine the probability that both will be left-handed: $0.09\times 0.09=0.0081$⁠.

It is reasonable to assume the proportion of people who are ambidextrous (both right- and left-handed) is nearly 0, which results in $P($⁠right-handed⁠$)=1-0.09=0.91$⁠. Using the same reasoning as in part (a), the probability that both will be right-handed is $0.91\times 0.91=0.8281$⁠.

The abbreviations RH and LH are used for right-handed and left-handed, respectively. Since each are independent, we apply the Multiplication Rule for independent processes:

$$\begin{array}{cc}\hfill P(\text{all five are RH})& =P(\text{first }\text{= }\text{RH, }\text{second }\text{= }\text{RH, }\text{..., }\text{fifth }\text{= }\text{RH})\hfill \\ \hfill & =P(\text{first = RH})\times P(\text{second = RH})\times \cdots \times P(\text{fifth = RH})\hfill \\ \hfill & =0.91\times 0.91\times 0.91\times 0.91\times 0.91=0.624\hfill \end{array}$$

Using the same reasoning as in (a), $0.09\times 0.09\times 0.09\times 0.09\times 0.09=$ $0.0000059$

Use the complement, $P($⁠all five are RH⁠$)$⁠, to answer this question:

$$\begin{array}{c}\hfill P(\text{not all RH})=1-P(\text{all RH})=1-0.624=0.376\end{array}$$

Brief answers are provided.

This can be written in probability notation as $P($⁠a randomly selected person is male and right-handed⁠$)=0.455$⁠.

0.207.

0.045.

0.0093.

False. These are independent trials.

False. There are red face cards.

True. A card cannot be both a face card and an ace.

10 tosses. Fewer tosses mean more variability in the sample fraction of heads, meaning there’s a better chance of getting at least 60% heads.

100 tosses. More flips means the observed proportion of heads would often be closer to the average, 0.50, and therefore also above 0.40.

100 tosses. With more flips, the observed proportion of heads would often be closer to the average, 0.50.

10 tosses. Fewer flips would increase variability in the fraction of tosses that are heads.

${0.5}^{10}$ = 0.00098.

${0.5}^{10}$ = 0.00098.

$P$⁠(at least one tails) = $1-P$⁠(no tails) = $1-({0.5}^{10})\approx 1-0.001=0.999$⁠.

No, there are voters who are both independent and swing voters.

Each Independent voter is either a swing voter or not. Since 35% of voters are Independents and 11% are both Independent and swing voters, the other 24% must not be swing voters.

0.47.

0.53.

P(Independent) $\times$ P(swing) = $0.35\times 0.23=0.08$⁠, which does not equal P(Independent and swing) = 0.11, so the events are dependent.

If the class is not graded on a curve, they are independent. If graded on a curve, then neither independent nor disjoint – unless the instructor will only give one A, which is a situation we will ignore in parts

and

. (b) They are probably not independent: if you study together, your study habits would be related, which suggests your course performances are also related. (c) No. See the answer to part (a) when the course is not graded on a curve. More generally: if two things are unrelated (independent), then one occurring does not preclude the other from occurring.

$0.16+0.09=0.25$⁠.

$0.17+0.09=0.26$⁠.

Assuming that the education level of the husband and wife are independent: $0.25\times 0.26=0.065$⁠. You might also notice we actually made a second assumption: that the decision to get married is unrelated to education level.

The husband/wife independence assumption is probably not reasonable, because people often marry another person with a comparable level of education. We will leave it to you to think about whether the second assumption noted in part (c) is reasonable.

Each of the four outcome combination are disjoint, all probabilities are indeed non-negative, and the sum of the probabilities is $0.1081+$ $0.0121+$ $0.0615+$ $0.8183=$ $1.00$⁠.

If the photo is about fashion and the ML algorithm prediction was correct, then the ML algorithm my have a value of pred_fashion:

$$\begin{array}{c}\hfill P(\text{𝚖𝚊𝚌𝚑\_𝚕𝚎𝚊𝚛𝚗 }\text{is }\text{pred\_fashion}|\text{𝚝𝚛𝚞𝚝𝚑 }\text{is }\text{fashion})\end{array}$$

The equation for conditional probability indicates we should first find
$P(\text{𝚖𝚊𝚌𝚑\_𝚕𝚎𝚊𝚛𝚗 }$ $\text{is }$ $\text{pred\_fashion }$ $\text{and }$ $\text{𝚝𝚛𝚞𝚝𝚑 }$ $\text{is }$ $\text{fashion})=$ $0.1081$ and $P(\text{𝚝𝚛𝚞𝚝𝚑 }\text{is }\text{fashion})=$ $0.1696$⁠.
Then the ratio represents the conditional probability: $0.1081/0.1696=0.6374$⁠.

This probability is $\frac{P(\text{𝚖𝚊𝚌𝚑\_𝚕𝚎𝚊𝚛𝚗 }\text{is }\text{pred\_not, }\text{𝚝𝚛𝚞𝚝𝚑 }\text{is }\text{fashion})}{P(\text{𝚝𝚛𝚞𝚝𝚑 }\text{is }\text{fashion})}=$ $\frac{0.0615}{0.1696}=$ $0.3626$⁠.

The total equals 1.

Under the condition the photo is about fashion, the ML algorithm must have either predicted it was about fashion or predicted it was not about fashion. The complement still works for conditional probabilities, provided the probabilities are conditioned on the same information.

$P($result = died $|$ inoculated = no⁠$)=$ $\frac{P(\text{𝚛𝚎𝚜𝚞𝚕𝚝 }\text{= }\text{died }\text{and }\text{𝚒𝚗𝚘𝚌𝚞𝚕𝚊𝚝𝚎𝚍 }\text{= }\text{no})}{P(\text{𝚒𝚗𝚘𝚌𝚞𝚕𝚊𝚝𝚎𝚍 }\text{= }\text{no})}=$ $\frac{0.1356}{0.9608}=$ $0.1411$⁠.

$P($result = died $|$ inoculated = yes⁠$)=$ $\frac{P(\text{𝚛𝚎𝚜𝚞𝚕𝚝 }\text{= }\text{died }\text{and }\text{𝚒𝚗𝚘𝚌𝚞𝚕𝚊𝚝𝚎𝚍 }\text{= }\text{yes})}{P(\text{𝚒𝚗𝚘𝚌𝚞𝚕𝚊𝚝𝚎𝚍 }\text{= }\text{yes})}=$ $\frac{0.0010}{0.0392}=$ $0.0255$ (if we avoided rounding errors, we’d get $6/244=0.0246$⁠). The death rate for individuals who were inoculated is only about 1 in 40 while the death rate is about 1 in 7 for those who were not inoculated.

Brief answers:

Observational.

No, we cannot infer causation from this observational study.

Accessibility to the latest and best medical care. There are other valid answers for part (c).

The answer is 0.0382, which can be verified using Figure 3.16.

There were only two possible outcomes: lived or died. This means that 100% - 97.54% = 2.46% of the people who were inoculated died.

The samples are large relative to the difference in death rates for the “inoculated” and “not inoculated” groups, so it seems there is an association between inoculated and outcome. However, as noted in the solution to Guided Practice 3.33, this is an observational study and we cannot be sure if there is a causal connection. (Further research has shown that inoculation is effective at reducing death rates.)

Brief solutions:

$1/6$⁠.

$1/36$⁠.

$\frac{P(Y=\text{ 1 and }X=\text{ 1})}{P(X=\text{ 1})}=$ $\frac{1/36}{1/6}=$ $1/6$⁠.

The probability is the same as in part (c): $P(Y=1)=1/6$⁠. The probability that $Y=1$ was unchanged by knowledge about $X$⁠, which makes sense as $X$ and $Y$ are independent.

He has forgotten that the next roulette spin is independent of the previous spins. Casinos do employ this practice, posting the last several outcomes of many betting games to trick unsuspecting gamblers into believing the odds are in their favor. This is called the gambler’s fallacy.

The tree diagram is shown to the right.

Identify which two joint probabilities represent students who passed, and add them: $P($⁠passed⁠$)=0.7566+0.1254=0.8820$⁠.

$P($⁠construct tree diagram $|$ passed⁠$)=\frac{0.7566}{0.8820}=0.8578$⁠.

The tree diagram, with three primary branches, is shown to the right. Next, we identify two probabilities from the tree diagram. (1) The probability that there is a sporting event and the garage is full: 0.14. (2) The probability the garage is full: $0.0875+0.14+0.0225=0.25$⁠. Then the solution is the ratio of these probabilities: $\frac{0.14}{0.25}=0.56$⁠. If the garage is full, there is a 56% probability that there is a sporting event.

Short answer:

$$\begin{array}{cc}\hfill P(A_2|B)& =\frac{P(B|A_2)P(A_2)}{P(B|A_1)P(A_1)+P(B|A_2)P(A_2)+P(B|A_3)P(A_3)}\hfill \\ \hfill & =\frac{(0.25)(0.35)}{(0.7)(0.2)+(0.25)(0.35)+(0.05)(0.45)}\hfill \\ \hfill & =0.35\hfill \end{array}$$

Each probability is conditioned on the same information that the garage is full, so the complement may be used: $1.00-0.56-0.35=0.09$⁠.

(a) No, but we could if A and B are independent. (b-i) 0.21. (b-ii) 0.79. (b-iii) 0.3. (c) No, because 0.1 $\ne$ 0.21, where 0.21 was the value computed under independence from part (a). (d) 0.143.

No, 0.18 of respondents fall into this combination.

$0.60+0.20-0.18=0.62$⁠.

$0.18/0.20=0.9$⁠.

$0.11/0.33\approx 0.33$⁠.

No, otherwise the answers to (c) and (d) would be the same.

$0.06/0.34\approx 0.18$⁠.

No. There are 6 females who like Five Guys Burgers.

$162/248=0.65$⁠.

$181/252=0.72$⁠.

Under the assumption of a dating choices being independent of hamburger preference, which on the surface seems reasonable: $0.65\times 0.72=0.468$⁠.

$(252+6-1)/500=0.514$⁠.

0.84

0.0714. Even when a patient tests positive for lupus, there is only a 7.14% chance that he actually has lupus. House may be right.

The three probabilities we computed were actually one marginal probability, $P($Q1$=$not_picked⁠$)$⁠, and two conditional probabilities:

$$\begin{array}{cc}\hfill & P(\text{𝚀𝟸}=\text{not\_picked }|\text{ 𝚀𝟷}=\text{not\_picked})\hfill \\ \hfill & P(\text{𝚀𝟹}=\text{not\_picked }|\text{ 𝚀𝟷}=\text{not\_picked, }\text{𝚀𝟸}=\text{not\_picked})\hfill \end{array}$$

Using the General Multiplication Rule, the product of these three probabilities is the probability of not being picked in 3 questions.

$P($⁠being picked to answer all three questions⁠$)={\left(\frac{1}{15}\right)}^3=0.00030$⁠.

First determine the probability of not winning. The tickets are sampled without replacement, which means the probability you do not win on the first draw is $29/30$⁠, $28/29$ for the second, ..., and $23/24$ for the seventh. The probability you win no prize is the product of these separate probabilities: $23/30$⁠. That is, the probability of winning a prize is $1-23/30=7/30=0.233$⁠.

When the tickets are sampled with replacement, there are seven independent draws. Again we first find the probability of not winning a prize: ${(29/30)}^7=0.789$⁠. Thus, the probability of winning (at least) one prize when drawing with replacement is 0.211.

There is about a 10% larger chance of winning a prize when using sampling without replacement. However, at most one prize may be won under this sampling procedure.

0.3.

0.3.

0.3.

$0.3\times 0.3=0.09$⁠.

Yes, the population that is being sampled from is identical in each draw.

$2/9\approx 0.22$⁠.

$3/9\approx 0.33$⁠.

$\frac{3}{10}\times \frac{2}{9}\approx 0.067$⁠.

No, e.g. in this exercise, removing one chip meaningfully changes the probability of what might be drawn next.

$P{(}^1$⁠leggings, ${}^2$⁠jeans, ${}^3$⁠jeans⁠$)=\frac{5}{24}\times \frac{7}{23}\times \frac{6}{22}=0.0173$⁠. However, the person with leggings could have come 2nd or 3rd, and these each have this same probability, so $3\times 0.0173=0.0519$⁠.

If they sell a little more or a little less, this should not be a surprise. Hopefully Chapter 1 helped make clear that there is natural variability in observed data. For example, if we would flip a coin 100 times, it will not usually come up heads exactly half the time, but it will probably be close.

100% - 25% - 60% = 15% of students do not buy any books for the class. Part

is represented by the first two lines in the table below. The expectation for part

is given as the total on the line $y_i\times P(Y=y_i)$⁠. The result of part

is the square-root of the variance listed on in the total on the last line: $\sigma =\sqrt{Var(Y)}=\$69.28$⁠.

She will make $X$ dollars on the TV but spend $Y$ dollars on the toaster oven: $X-Y$⁠.

$E(X-Y)=$ $E(X)-$ $E(Y)=$ $175-$ $23=$ $\$152$⁠. She should expect to make about $152.

No, since there is probably some variability. For example, the traffic will vary from one day to next, and auction prices will vary depending on the quality of the merchandise and the interest of the attendees.

$E(\$6000\times X+\$2000\times Y)=$ $\$6000\times 0.020+$ $\$2000\times 0.002=$ $\$124$⁠.

No. While stocks tend to rise over time, they are often volatile in the short term.

One concern is whether traffic patterns tend to have a weekly cycle (e.g. Fridays may be worse than other days). If that is the case, and John drives, then the assumption is probably not reasonable. However, if John walks to work, then his commute is probably not affected by any weekly traffic cycle.

The equation for Elena can be written as

$$\begin{array}{c}\hfill (1)\times X+(-1)\times Y\end{array}$$

The variances of $X$ and $Y$ are 625 and 64. We square the coefficients and plug in the variances:

$$\begin{array}{c}\hfill {(1)}^2\times Var(X)+{(-1)}^2\times Var(Y)=1\times 625+1\times 64=689\end{array}$$

The variance of the linear combination is 689, and the standard deviation is the square root of 689: about $26.25.

13.

No, these 27 students are not a random sample from the university’s student population. For example, it might be argued that the proportion of smokers among students who go to the gym at 9 am on a Saturday morning would be lower than the proportion of smokers in the university as a whole.

E(X) = 3.59. SD(X) = 9.64.

E(X) = -1.41. SD(X) = 9.64.

No, the expected net profit is negative, so on average you expect to lose money.

5% increase in value.

E = -0.0526. SD = 0.9986.

Brief answers:

$0.1157\times 0.1157\times 0.1157=$ $0.0015$⁠.

${(1-0.1157)}^3=0.692$

This has positive probability. Anyone between 179.5 cm and 180.5 cm will have a measured height of 180 cm. This is probably a more realistic scenario to encounter in practice versus Example 3.74.

Approximate answers are OK.

$(29+32)/144=0.42$⁠.

$21/144=0.15$⁠.

$(26+12+15)/144=0.37$⁠.

Invalid. Sum is greater than 1.

Valid. Probabilities are between 0 and 1, and they sum to 1. In this class, every student gets a C.

Invalid. Sum is less than 1.

Invalid. There is a negative probability.

Valid. Probabilities are between 0 and 1, and they sum to 1.

Invalid. There is a negative probability.

0.8247.

E = $3.90. SD = $0.34.

E = $27.30. SD = $0.89.

$Var\left(\frac{X_1+X_2}{2}\right)$
$=Var(\frac{X_1}{2}+\frac{X_2}{2})$
$=\frac{Var(X_1)}{2^2}+\frac{Var(X_2)}{2^2}$
$=\frac{{\sigma }^2}{4}+\frac{{\sigma }^2}{4}$
$={\sigma }^2/2$

$Var\left(\frac{X_1+X_2+\cdots +X_n}{n}\right)$
$=Var(\frac{X_1}{n}+\frac{X_2}{n}+\cdots +\frac{X_n}{n})$
$=$ $\frac{Var(X_1)}{n^2}+$ $\frac{Var(X_2)}{n^2}+$ $\cdots +$ $\frac{Var(X_n)}{n^2}$
$=\frac{{\sigma }^2}{n^2}+\frac{{\sigma }^2}{n^2}+\cdots +\frac{{\sigma }^2}{n^2}$ (there are $n$ of these terms)
$=n\frac{{\sigma }^2}{n^2}$
$={\sigma }^2/n$