Solution manual for statistics 2nd edition by lock

We find these two probabilities and add them up, using the total probability rule.. We find these two probabilities and add them up, using the total probability rule.. The tree diagram is

P.5 We have

P (B if A) = P (A and B)

0.1 0.4 = 0.25.

P.6 No! If A and B were disjoint, that means they cannot both happen at once, which means P (A and B) =

0 Since we are told that P (A and B) = 0.1, not zero, the events are not disjoint.

P.7 We need to check whether P (A and B) = P (A) · P (B) Since P (A and B) = 0.1 and P (A) · P (B) =

0.4 · 0.3 = 0.12, the events are not independent We can also check from an earlier exercise that P (B if A) = 0.25 = P (B) = 0.3.

P.8 We have P (not A) = 1 − P (A) = 1 − 0.8 = 0.2.

P.12 We have

P (B if A) = P (A and B)

0.25 0.8 = 0.3125.

P.13 No! If A and B were disjoint, that means they cannot both happen at once, which means P (A and B) =

0 Since we are told that P (A and B) = 0.25, not zero, the events are not disjoint.

P.14 We need to check whether P (A and B) = P (A) · P (B) Since P (A and B) = 0.25 and P (A) · P (B) =

0.8 · 0.4 = 0.32, the events are not independent We can also check from an earlier exercise that P (B if A) = 0.3125 = P (B) = 0.4.

P.15 Since A and B are independent, knowing that B occurs gives us no additional information about A,

P.18 Since A and B are independent, we have P (A and B) = P (A)·P (B) = 0.7·0.6 = 0.42 By the additive

rule, we have P (A or B) = P (A) + P (B) − P (A and B) = 0.7 + 0.6 − 0.42 = 0.88.

P.19 There are two cells that are included as part of event A, so P (A) = 0.2 + 0.1 = 0.3.

P.20 There are two cells that are included as part of event not B, so P (not B) = 0.1 + 0.3 = 0.4.

P.21 There is one cell that is in both event A and event B, so we have P (A and B) = 0.2.

P.22 There are three cells that represent being in event A or event B or both, so we have P (A or B) =

0.2 + 0.4 + 0.1 = 0.7 We could also use the additive rule P (A or B) = P (A) + P (B) − P (A and B) = 0.3 + 0.6 − 0.2 = 0.7.

P.23 We have

P (A if B) = P (A and B)

0.2 0.6 = 0.333.

P.24 We have

P (B if A) = P (A and B)

0.2 0.3 = 0.667.

P.25 No! If A and B were disjoint, that means they cannot both happen at once, which means P (A and B) =

0 We see in the table that P (A and B) = 0.2, not zero, so the events are not disjoint.

P.26 We need to check whether P (A and B) = P (A) · P (B) Since P (A and B) = 0.2 and P (A) · P (B) =

0.3 · 0.6 = 0.18, the events are not independent We can also check from an earlier exercise that P (B if A) = 0.667 = P (B) = 0.6.

P.27 The two events are disjoint, since if at least one skittle is red then all three can’t be green However,

they are not independent or complements

P.28 The two events are disjoint because both teams cannot win They are also complements because if

Australia does not win, then South Africa wins However, they are not independent

P.29 The two events are independent, as Australia winning their rugby match will not change the probability

that Poland wins their chess match However, they are not disjoint or complements

P.30 The two events are not disjoint or complements, as it is possible to have the rolls be {3,5} where the

first die is a 3 and the sum is 8 To check independence we need to find

P (A) = 1/6 and P (B) = P ({2, 6} or {3, 5} or {4, 4} or {5, 3} or {6, 3}) = 5/36 There is only one possibility for the intersection so P (A and B) = P ( {3, 5}) = 1/36 We then check that

P (A if B) = P (A and B)

1/36 1/6 = 1/6 = P (B) = 5/36

so A and B are not independent We can also verify that

P (A and B) = 1/36 = P (A) · P (B) = (1/6) · (5/36) = 5/216

P.31 (a) It will not necessarily be the case that EXACTLY 1 in 10 adults are left-handed for every sample.

We can only conclude that approximately 10% will be left-handed in the “long run” (for very largesamples)

(b) The three outcomes each have probability 13 only if they are equally likely This may not be the case

for the results of baseball pitches

(c) To find the probability of two consecutive 1’s on independent dice rolls we should multiply the abilities instead of adding them Using the multiplicative rule, the probability that two consecutiverolls land with a 1 is 16×1

prob-6 = 361.

(d) A probability that is not between 0 and 1 does not make sense

P.32 (a) We are told that P (C) = 0.26, P (W ) = 0.13, and P (C and W ) = 0.03.

(b) We use the additive rule for find the probability of one eventor another.

P (C or W ) = P (C) + P (W ) − P (C and W ) = 0.26 + 0.13 − 0.03 = 0.36

The probability that a movie is either a comedy or produced by Warner Bros is 0.36

(c) We are finding the probability that a movie is a comedyif it is produced by Warner Bros, so we have

P (C if W ) = P (C and W )

P (W ) =

0.03 0.13 = 0.23.

About 23% of movies produced by Warner Bros are comedies

(d) We are finding the probability that a movie is produced by Warner Brosif it is a comedy, so we have

P (W if C) = P (C and W )

0.03 0.26 = 0.115.

About 11.5% of comedies are produced by Warner Bros

(e) To find the probability a movie is not a comedy we have P (not C) = 1 − P (C) = 1 − 0.26 = 0.74 (f) Saying C and W are disjoint would mean that Warner Bros never makes any comedies This is not true since we see that 3% of all the movies are comedies from Warner Bros It is clear that C and W

are not disjoint

(g) Saying C and W are independent would mean that knowing a movie comes from Warner Bros would

give us no information about whether it is a comedy and knowing it is a comedy would give us no

information about whether it came from Warner Bros This is almost true: P (C if W ) = 0.23 which

is close to P (C) = 0.26 and P (W if C) = 0.115 which is close to P (W ) = 0.13, but neither result is

exactly the same so the two events are not independent (at least for these approximated probabilityvalues.)

P.33 (a) We are finding P (MP ) There are a total of 303 inductees and 206 of them are performers, so

we have

P (MP ) =206

303 = 0.680The probability that an inductee selected at random will be a performer is 0.680

(b) We are finding P (not F ) There are a total of 303 inductees and 256 of them do not have any female

members, so we have

P (not F ) = 256

303 = 0.845The probability that an inductee selected at random will not have any female members is 0.845

(c) In this case, we are interested only in inductees who are performers, and we want to know the probability

they have female members, P (F if MP ). There are 206 performers and 38 of those have femalemembers, so we have

P (F if MP ) = 38

206 = 0.184.

(d) In this case, we are interested only in inductees that do not have any female members, and we want

to know the probability of not being a performer, P (not MP if not F ) There are 256 inductees with

no female members and 88 of them are not performers, so we have

P (not MP if not F ) = 88

256 = 0.344.

(e) We are finding P (MP and not F ) Of the 303 inductees, there are 168 that are performers with no

female members, so we have

P (MP and not F ) = 168

303 = 0.554

(f) We are finding P (not MP or F ) Of the 303 inductees, 38 + 9 + 88 = 135 are either not performers or

have female members (or both), so we have

P (not MP or F ) = 135

303 = 0.446Notice that this is the complement of the event found in part (e)

P.34 (a) We are finding P (C) There are a total of 268 inductees and 233 of them were born in Canada,

so we have

P (C) =233

268 = 0.869The probability is 0.869 that an inductee selected at random is Canadian It is remarkable how muchCanada has dominated the sport!

(b) We are finding P (not D) There are a total of 268 inductees and 85 of them played defense while the

other 183 do not, so we have

P (not D) = 183

268 = 0.683The probability that an inductee selected at random will not be a defenseman is 0.683

(c) We are finding P (C and D) Of the 268 inductees, there are 74 that are defensemen born in Canada,

so we have

P (Cand D) = 74

268 = 0.276

(d) We are finding P (C or D) Of the 268 inductees, 233 were born in Canada and 85 are defensemen and

74 are both, so we have

that a Canadian inductee plays defense, P (D if C) There are 233 Canadians and 74 of them play

defense, so we have

P (D if C) = 74

233 = 0.318

(f) In this case, we are interested only in defensemen, and we want to know the probability that a

defense-man inductee is also Canadian, P (C if D) There are 85 defensemen and 74 of them are Canadian, so

(b) The probability that itis blue is 20/80 = 0.25 so the probability that it is not blue is 1 − 0.25 = 0.75.

(c) The single piece can be red or orange, but not both, so these are disjoint events The probability the

randomly selected candy is red or orange is 11/80 + 12/80 = 23/80 = 0.2875.

(d) The probability that the first one is blue is 20/80 = 0.25 When we put it back and mix them up, the

probability that the next one is blue is also 0.25 By the multiplication rule, since the two selections

are independent, the probability both selections are blue is 0.25 · 0.25 = 0.0625.

(e) The probability that the first one is red is 11/80 Once that one is taken (since we don’t put it back

and we eat it instead), there are only 79 pieces left and 11 of those are green By the multiplication

rule, the probability of a red then a green is (11/80) · (11/79) = 0.191.

P.36 (a) There are 18 yellow ones out of a total of 80, so the probability that we pick a yellow one is

18/80 = 0.225.

(b) The probability that itis brown is 8/80 = 0.10, so the probability that it is not brown is 1−0.10 = 0.90.

(c) The single piece can be blue or green, but not both, so these are disjoint events The probability the

randomly selected candy is blue or green is 20/80 + 11/80 = 31/80 = 0.3875.

(d) The probability that the first one is red is 11/80 = 0.1375 When we put it back and mix them up, the

probability that the next one is red is also 0.1375 By the multiplication rule, since the two selections

are independent, the probability both selections are red is 0.1375 · 0.1375 = 0.0189.

(e) The probability that the first one is yellow is 18/80 Once that one is taken (since we don’t put it back

and we eat it instead), there are only 79 pieces left and 20 of those are blue By the multiplication

rule, the probability is (18/80) · (20/79) = 0.0570.

P.37 Let S denote successfully making a free throw and F denote missing it.

(a) As free throws are independent, we can multiply the probabilities

P (Makes two) = P (S1 and S2) = P (S1 · P (S2) = 0.908 × 0.908 = 0.824 (b) The probability of missing one free throw is P (F ) = 1 − 0.908 = 0.092 So,

P (Misses two) = P (F1 and F2) = P (F1 · P (F2) = 0.092 × 0.092 = 0.008

(c) He can either miss the first and make the second shot, or make the first and miss the second So,

P (Makes exactly one) = P (S1 and F2) + P (F1and S2) = 0.908 × 0.092 + 0.092 × 0.908 = 0.167

P.38 Let CBM and CBW denote the events that a man or a woman is colorblind, respectively.

(a) As 7% of men are colorblind, P (CBM ) = 0.07.

(b) As 0.4% of women are colorblind, P (not CBW ) = 1 − P (CBW ) = 1 − 0.004 = 0.996.

(c) The probability the woman is not colorblind is 0.996, and the probability that the man is not

color-blind is 1− 0.07 = 0.93 As the man and woman are selected independently, we can multiply their

probabilities:

P (Neither is Colorblind) = P (not CBM) · P (not CBW ) = 0.93 × 0.996 = 0.926.

(d) The event that “At least one is colorblind” is the complement of part (d) that “Neither is Colorblind”

so we have

P (At least one is Colorblind) = 1 − P (Neither is Colorblind) = 1 − 0.926 = 0.074

We could also do this part as

P (CBM or CBW ) = P (CBM)+P (CBW )−P (CBM and CBW ) = 0.07+0.004−(0.07)(0.004) = 0.074

P.39 (a) The probability that a women is not color-blind is 1 − 0.04 = 0.996, and the probability that

a man is not colorblind is 1− 0.07 = 0.93 As all events are independent, we can multiply their

probabilities:

P (Nobody is Colorblind) = 0.99625× 0.9315= 0.305

(b) The event “At least one is Colorblind” is the complement of the event “Nobody is Colorblind” whichhas its probability computed in part (a), so

P (At least one is Colorblind) = 1 − P (Nobody is Colorblind) = 1 − 0.305 = 0.695

(c) The probability that the randomly selected student is a man is 15/40 = 0.375 and the probability that

it is a women is 25/40 = 0.625 Using the additive rule for disjoint events,

P (Colorblind) = P (Colorblind Man OR Colorblind Woman)

= P (Colorblind and Man) + P (Colorblind and Woman).

By the multiplicative rule,

P (Colorblind and Man) = P (Man) · P (Colorblind if Man) = 0.375 × 0.07 = 0.02625

(d) Use conditional probability:

P (dies at 90 if lives till 90) = P (dies at 90 and lives till 90)

P (lives till 90) =

P (dies at 90)

P (lives till 90)=

0.0294 17429/100000 = 0.169.

(e) Use conditional probability:

P (dies at 90 if lives till 80) = P (dies at 90 and lives till 80)

P (lives till 80) =

P (dies at 90)

P (lives till 80)=

0.0294 50344/100000 = 0.058.

(f) As 85,995 out of 100000 men live to age 60 and 17,429 out of 100000 live to age 90, the probabilitythat a man dies between the ages of 60 and 89 is 85995−17429100000 = 0.686.

(g) Use conditional probability:

P (lives till 90 if lives till 60) = P (lives till 90 and lives till 60)

P (lives till 60) =

P (lives till 90)

P (lives till 60) =

17429/100000 85995/100000 = 0.203.

P.41 (a) The probability that the S&P 500 increased on a randomly selected day is 423/756 = 0.5595.

(b) Assuming independence, the probability that the S&P 500 increases for two consecutive days is 0.5595× 0.5595 = 0.3130 (using the multiplicative rule) The probability that the S&P 500 increases on a day given that it increased the day before remains 0.5595 if the events are independent.

(c) The probability that the S&P 500 increases for two consecutive days is 234/755 = 0.3099. Theprobability that the S&P 500 increases on a day, given that it increased on the previous day is

P (Increase both days)

P (Increase first day) =

0.3099 0.5595 = 0.5539

(d) The difference between the results in part (b) and part (c) is very small and insignificant, so we havelittle evidence that daily changes are not independent (However, since the question does not ask for

a formal hypothesis test, other answers are acceptable.)

P.42 If you are served one of the pancakes at random, let A be the event that the side facing you is burned

and B be the event that the other side is burned We want to find P (B if A) As only one of three pancakes

is burned on both sides, P (A and B) = 1/3 As 3 out of 6 total sides are burned, P (A) = 3/6 = 1/2 So,

P (B if A) = P (A and B)

P (A) =

1/3 1/2 =

23

Section P.2 Solutions

P.43 Here is the tree with the missing probabilities filled in.

Note that the probabilities of all branches arising from a common point must sum to one, so

P (I) + P (II) + P (III) = 1 =⇒ P (I) = 1 − 0.43 − 0.31 = 0.26

P (A if II) + P (B if II) = 1 =⇒ P (A if II) = 1 − 0.24 = 0.76

We obtain the probabilities at the end of each pair of branches with the multiplicative rule, so

P (II and B) = P (II) · P (B if II) = 0.43(0.24) = 0.1032

P (III and A) = P (III) · P (A if III) = 0.31(0.80) = 0.248

P.44 Here is the tree with the missing probabilities filled in.

We use the conditional probability rule to find

P (A if I) = P (I and A)

P (I) =

0.09 0.18 = 0.5

We use the complement rule (sum of a branches from one point equals one) to find

P (B if I) = 1 − P (A if I) = 1 − 0.5 = 0.5

We use the multiplicative rule to find

P (I and B) = P (I) · P (B if I) = 0.18(0.5) = 0.09 From the multiplicative rule for III and A we have

P (III and A) = P (III) · P (A if III) =⇒ 0.268 = P (III) · 0.67 =⇒ P (III) = 0.268

0.67 = 0.4

Using the complement rule several more times

P (II) = 1 − P (I) − P (III)) = 1 − 0.18 − 0.4 = 0.42

P (A if II) = 1 − P (B if II) = 1 − 0.45 = 0.55

P (B if III) = 1 − P (A if III) = 1 − 0.67 = 0.33

Finally, several more multiplicative rules along pairs of branches give

P (II and A) = P (II) · P (A if II) = 0.42(0.55) = 0.231

P (II and B) = P (II) · P (B if II) = 0.42(0.45) = 0.189

P (III and B) = P (III) · P (B if III) = 0.40(0.33) = 0.132

P.45 Here is the tree with the missing probabilities filled in.

First,the sum of all the joint probabilities for all pairs of branches must be one so we have

P (A and I) = 1 − (0.225 + 0.16 + 0.45 + 0.025 + 0.025) = 0.115

Using the total probability rule we have

P (I) = P (I and A) + P (I and B) + P (I and C) = 0.115 + 0.225 + 0.16 = 0.5

P (II) = P (II and A) + P (II and B) + P (II and C) = 0.45 + 0.025 + 0.025 = 0.5

The remaining six probabilities all come from the conditional probability rule For example,

P (A if I) = P (A and I)

P (I) =

0.115 0.5 = 0.23

P.46 Here is the tree with the missing probabilities filled in.

Using the complement rule (sum of all branches from one point must be one), we have

P (II) = 1 − P (I) = 1 − 0.38 = 0.62

P (C if I) = 1 − P (A if I) − P (B if I) = 1 − 0.00 − 0.19 = 0.81

By the conditional probability rule , we have

P (C if II) = P (II and C)

P (II) =

0.00 0.62 = 0.00

One more complement rule gives

P (B if II) = 1 − P (A if II) − P (C if II) = 1 − 0.56 − 0.00 = 0.44

Finally, we apply the multiplicative rule several times to get

P (I and A) = P (I) · P (A if I) = 0.38(0.00) = 0.00

P (I and B) = P (I) · P (B if I) = 0.38(0.19) = 0.0722

P (I and C) = P (I) · P (C if I) = 0.38(0.81) = 0.3078

P (II and A) = P (II) · P (A if II) = 0.62(0.56) = 0.3472

P (II and B) = P (II) · P (B if II) = 0.62(0.44) = 0.2728

P.47 We use the multiplicative rule to see P (B and R) = P (B) · P (R if B) = 0.4 · 0.2 = 0.08.

P.48 We use the multiplicative rule to see P (A and S) = P (A) · P (S if A) = 0.6 · 0.1 = 0.06.

P.49 This conditional probability is shown directly on the tree diagram Since we assume A is true, we

follow the A branch and then find the probability of R, which we see is 0.9

P.50 This conditional probability is shown directly on the tree diagram Since we assume B is true, we

follow the B branch and then find the probability of S, which we see is 0.8

P.51 We see in the tree diagram that there are two ways for R to occur We find these two probabilities and

add them up, using the total probability rule We see that P (A and R) = 0.6 · 0.9 = 0.54 so that top branch can be labeled 0.54 We also see that P (B and R) = 0.4 · 0.2 = 0.08 Since either A or B must occur (since

these are the only two branches in this part of the tree), these are the only two ways that R can occur, so

P (R) = P (A and R) + P (B and R) = 0.54 + 0.08 = 0.62.

P.52 We see in the tree diagram that there are two ways for S to occur We find these two probabilities

and add them up, using the total probability rule We see that P (A and S) = 0.6 · 0.1 = 0.06 so this branch can be labeled 0.06 We also see that P (B and S) = 0.4 · 0.8 = 0.32 Since either A or B must occur (since

these are the only two branches in this part of the tree), these are the only two ways that S can occur, so

We see that P (A if S) = 0.158.

We see that P (B if R) = 0.129.

P.55 We first create the tree diagram using the information given, and use the multiplication rule to fill in

the probabilities at the ends of the branches For example, for the top branch, the probability of having 1

occupant in an owner-occupied housing unit is 0.65 · 0.217 = 0.141.

(a) We see at the end of the branch with rented and 2 occupants that the probability is 0.091

(b) There are two branches that include having 3 or more occupants and we use the addition rule to see

that the probability of 3 or more occupants is 0.273 + 0.132 = 0.405.

(c) This is a conditional probability (or Bayes’ rule) We have:

P (rent if 1) = P (rent and 1 person)

P (1 person) =

0.127 0.141 + 0.127 =

0.127 0.268 = 0.474

If a housing unit has only 1 occupant, the probability that it is rented is 0.474

P.56 We use F to denote the event of having fibromyalgia and R to denote the event of having restless leg

syndrome The tree diagram is shown, and we use the multiplication rule to find the probabilities at the end

of the branches For example, for the top branch, we have P (F and R) = 0.02 · 0.33 = 0.0066.

We are finding P (F if R) By adding the probabilities of the R branches, we see that P (R) = 0.0066 + 0.0294 = 0.036 The conditional probability is

P (F if R) = P (F and R)

P (R) =

0.0066 0.036 = 0.183.

If a person has restless leg syndrome, the probability that the person will also have fibromyalgia is about18%

P.57 We are given

P (Positive if no Cancer) = 86.6/1000 = 0.0866,

P (Positive if Cancer) = 1 − 1.1/1000 = 0.9989, and

P (Cancer) = 1/38 = 0.0263

Applying Bayes’ rule we have

P (Cancer if Positive) = P (Cancer)P (Positive if Cancer)

P (no Cancer)P (Positive if no Cancer) + P (Cancer)P (Positive if Cancer)

(1− 0.0263) · 0.0866 + 0.0263 · 0.9989

= 0.2375

P.58 Let F , C, and S denote the three types of pitches (fastball, curveball, spitball) and K denote the pitch

is a strike We are given the probability for each type of pitch

P (F ) = 0.60 P (C) = 0.25 P (S) = 1 − P (F ) − P (C) = 1 − 0.60 − 0.25 = 0.15

and the conditional probability for a strike with each pitch

P (K if F ) = 0.70 P (K if C) = 0.50 P (K if S) = 0.30

We need to find the probability of a certain type of pitch (curveball) given that we know the outcome (strike).This calls for an application of Bayes’ rule:

If we see Slippery throw a strike, there’s a 21.2% chance it was with a curveball

P.59 (a) Using the formula for conditional probability,

P (Free if Spam) = P (Free and Spam)

P (Spam) =

0.0357 0.134 = 0.266

(b) Using the formula for conditional probability,

P (Spam if Free) = P (Free and Spam)

P (Free) =

0.0357 0.0475 = 0.752

P.60 Using the Bayes’ rule,

P (Spam if Text) = P (Spam)P (Text if Spam)

0.134 · 0.3855 0.0701 = 0.737

P.61 Using Bayes’ rule,

P (Spam if Text and Free) = P (Spam)P (Text and Free if Spam)

P (Spam)P (Text and Free if Spam) + P (not Spam)P (Text and Free if not Spam)

= 0.134 · 0.17 0.134 · 0.17 + 0.866 · 0.0006

= 0.978.

P.62 Applying the total probability rule,

P (Free and Text) = P (Spam)P (Text and Free if Spam) + P (not Spam)P (Text and Free if not Spam)

= 0.134 · 0.17 + 0.866 · 0.0006

= 0.0233.

Again applying the total probability rule,

P (Free and not Text) = P (Free) − P (Free and Text) = 0.0475 − 0.0233 = 0.0242

and

P (Free and not Text if Spam) = P (Free if Spam) − P (Free and Text if Spam) = 0.2664 − 0.1700 = 0.0964.

So, applying Bayes’ rule,

P (Spam if Free and not Text) = P (Spam)P (Free and not Text if Spam)

P (Free and not Text) =

0.134 · 0.0964 0.0242 = 0.534.

Section P.3 Solutions

P.63 A discrete random variable, as it can only take the values {0, 1, 2, , 10}.

P.64 A discrete random variable, as it can only take the values {0, 0.1, 0.2, , 1}).

P.65 A discrete random variable, as an ace must appear somewhere between the 1st and 48th card dealt.

P.66 Not a random variable, as it does not have a numerical outcome.

P.67 A continuous random variable, as it can take any value greater than or equal to 0 lbs.

P.68 There are two conditions for a probability function The first is that all the probabilities are between 0

and 1 and that is true here The second is that the probabilities add up to 1.0, and that is true here because

0.4 + 0.3 + 0.2 + 0.1 = 1.0.

P.69 We see that P (X = 3) = 0.2 and P (X = 4) = 0.1 so P (X = 3 or X = 4) = 0.2 + 0.1 = 0.3.

P.70 We have P (X > 1) = P (X = 2 or X = 3 or X = 4) = 0.3 + 0.2 + 0.1 = 0.6.

P.71 We have P (X < 3) = P (X = 1 or X = 2) = 0.4 + 0.3 = 0.7.

P.72 We have P (X is an odd number) = P (X = 1 or X = 3) = 0.4 + 0.2 = 0.6.

P.73 We have P (X is an even number) = P (X = 2 or X = 4) = 0.3 + 0.1 = 0.4.

P.74 The probability values have to add up to 1.0 so we have ? = 1.0 − (0.1 + 0.1 + 0.2) = 0.6.

P.75 The probability values have to add up to 1.0 so we have ? = 1.0 − (0.2 + 0.2 + 0.2) = 0.4.

P.76 The probability values have to add up to 1.0 but the sum of the two values there is already greater

than 1 (we see 0.5 + 0.6 = 1.1) Since a probability cannot be a negative number, this cannot be a probability

function

P.77 The probability values have to add up to 1.0 but the sum of the values there is already greater than

1 (we see 0.3 + 0.3 + 0.3 + 0.3 = 1.2) Since a probability cannot be a negative number, this cannot be a

probability function

P.78 (a) We multiply the values of the random variable (in this case, 1, 2, and 3) by the corresponding

probability and add up the results We have

μ = 1(0.2) + 2(0.3) + 3(0.5) = 2.3

The mean of this random variable is 2.3

(b) To find the standard deviation, we subtract the mean of 2.3 from each value, square the difference,multiply by the probability, and add up the results to find the variance; then take a square root to findthe standard deviation

σ2 = (1− 2.3)2· 0.2 + (2 − 2.3)2· 0.3 + (3 − 2.3)2· 0.5 = 0.61

=⇒ σ = √ 0.61 = 0.781

P.79 (a) We multiply the values of the random variable (in this case, 10, 20, and 30) by the corresponding

probability and add up the results We have

μ = 10(0.7) + 20(0.2) + 30(0.1) = 14

The mean of this random variable is 14

(b) To find the standard deviation, we subtract the mean of 14 from each value, square the difference,multiply by the probability, and add up the results to find the variance; then take a square root to findthe standard deviation

σ2 = (10− 14)2· 0.7 + (20 − 14)2· 0.2 + (30 − 14)2· 0.1 = 44

=⇒ σ = √ 44 = 6.63

P.80 (a) We multiply the values of the random variable (in this case, 20, 30, 40, and 50) by the

corre-sponding probability and add up the results We have

μ = 20(0.6) + 30(0.2) + 40(0.1) + 50(0.1) = 27

The mean of this random variable is 27

(b) To find the standard deviation, we subtract the mean of 27 from each value, square the difference,multiply by the probability, and add up the results to find the variance; then take a square root to findthe standard deviation

σ2 = (20− 27)2· 0.6 + (30 − 27)2· 0.2 + (40 − 27)2· 0.1 + (50 − 27)2· 0.1 = 101

=⇒ σ = σ = √ 101 = 10.05

P.81 (a) We multiply the values of the random variable (in this case, 10, 12, 14, and 16) by the

corre-sponding probability and add up the results We have

σ2 = (10− 13)2· 0.25 + (12 − 13)2· 0.25 + (14 − 13)2· 0.25 + (16 − 13)2· 0.25 = 5

=⇒ σ = √ 5 = 2.236

P.82 (a) We see that 0.217 + 0.363 + 0.165 + 0.145 + 0.067 + 0.026 + 0.018 = 1.001 This is different from

1 just by round-off error on the individual probabilities

(b) We have p(1) + p(2) = 0.217 + 0.363 = 0.580.

(c) We have p(5) + p(6) + p(7) = 0.067 + 0.026 + 0.018 = 0.111.

(d) It is easiest to find this probability using the complement rule, since more than 1 occupant is thecomplement of 1 occupant for this random variable The answer is 1− p(1) = 1 − 0.217 = 0.783.

P.83 (a) We see that 0.362 + 0.261 + 0.153 + 0.114 + 0.061 + 0.027 + 0.022 = 1, as expected.

The average household size for an owner-occupied housing unit in the US is 2.635 people

(b) To find the standard deviation, we subtract the mean of 2.635 from each value, square the difference,multiply by the probability, and add up the results to find the variance; then take a square root to findthe standard deviation

The average household size for a renter-occupied housing unit in the US is 2.42 people

(b) To find the standard deviation, we subtract the mean of 2.42 from each value, square the difference,multiply by the probability, and add up the results to find the variance; then take a square root to findthe standard deviation

σ2 = (1− 2.42)2· 0.362 + (2 − 2.42)2· 0.261 + · · · + (7 − 2.42)2· 0.022

= 2.3256

=⇒ σ = √ 2.3256 = 1.525

P.86 Let the random variable X measure fruit fly lifetimes (in months).

(a) The probabilities must add to 1, so the proportion of dying in the second month is

P.87 (a) As the probabilities must sum to 1, so

P (X = 4) = 1 − (0.29 + 0.3 + 0.2 + 0.17) = 1 − 0.96 = 0.04 (b) P (X < 2) = P (X = 0) + P (X = 1) = 0.29 + 0.3 = 0.59

(c) To find the mean we use

P.88 (a) This is a conditional probability, P (A if B), where the event A is X = 1 and the conditioning

event B is {X = 1 or X = 2} From the probability function we see that P (A) = 0.30 and P (B) = 0.30 + 0.20 = 0.50 Note also that P (A and B) = P (A) since X = 1 is the only outcome they have in

P (D) = 0.20 + 0.15 + 0.10 + 0.05 = 0.50 Note also that P (C and D) = P (C) since X = 5 and X = 6

are the only outcomes they have in common We have

A fruit fly that makes it safely through it’s first two months will have a 30% chance of living to five orsix months

P.89 Let X1, X2, and X3 represent the sales on each of three consecutive days Since daily sales areindependent, we can multiple their probabilities So, the probability that the no cars are sold in threeconsecutive days is

P (X1= 0 and X2= 0 and X3= 0) = P (X1= 0)· P (X2= 0)· P (X3= 0)

= 0.29 · 0.29 · 0.29

= 0.293

= 0.0244

P.90 (a) There are three possible values for the random variable, {0, 1, 2} Let S denote he successfully

makes a shot and F that he fails to make it.

P (X = 2) = P (S1 and S2) = P (S1 · P (S2) = 0.908 · 0.908 = 0.8245

P (X = 0) = P (F1and F2) = P (F1 · P (F2) = 0.092 · 0.092 = 0.0085

P (X = 1) = P (F1and S2) + P (S1 and F2) = 0.092 · 0.908 + 0.908 · 0.092 = 0.1671

So, the probability distribution of X is

p(x) 0.0085 0.1671 0.8245

(b) The mean number of free throws made in two attempts is

μ = 0(0.0085) + 1(0.1671) + 2(0.8245) = 1.816

P.91 (a) If the woman dies during the first year the organization loses $100000 − c, if the woman dies

during the second year the organization loses $100000−2c, and so on If the woman does not die during the five year contract the organization earns 5c dollars, and the probability of this is 1 − (0.00648 + 0.00700 + 0.00760 + 0.00829 + 0.00908) = 0.96155 The probability distribution of the profit (as a function of the yearly fee c) is given below.

P.92 (a) We know that P (X = $29.95) = 2 · P (X = $39.95) and P (X = $23.95) = 3 · P (X = $39.95) It

(b) To find P (X > 2) we use the complement rule to find

 56

2

= 0.116

(b) The event “more than three turns to finish” or X > 3 includes X = 4, 5, 6, , an infinite number of

possible outcomes! Fortunately we can use the complement rule

P (X > 3) = 1 − (p(1) + p(2) + p(3))

= 1−



16

 56

0+

16

 56

1+

16

 56

2

= 1− [0.1667 + 0.1389 + 0.1157]

= 1− 0.4213 = 0.5787

Section P.4 Solutions

P.95 This is a binomial random variable, with n = 10 and p = 1/6.

P.96 Not binomial, since the number of rolls is not fixed.

P.97 Not binomial, since it is not clear what is counted as a success.

P.98 This is a binomial random variable, with n = 75 and p = 0.3.

P.99 This is a binomial random variable, with n = 100 and p = 0.51.



(0.32)(0.74) = 15(0.32)(0.74) = 0.324.

P.109 We first calculate that

87



(0.97)(0.11) = 8(0.97)(0.11) = 0.383.

P.110 We first calculate that

103



(0.43)(0.67) = 120(0.43)(0.67) = 0.215.

P.111 We first calculate that

128

P.116 A probability function gives the probability for each possible value of the random variable This is a

binomial random variable with n = 3 and p = 0.49 (since we are counting the number of girls not boys).

The probability of 0 girls is:

P (X = 0) =

30