Stats without Tears
Solutions for Chapter 5
Updated 20 Sept 2014
(What’s New?)
Copyright © 2012–2023 by Stan Brown, BrownMath.com
Updated 20 Sept 2014
(What’s New?)
Copyright © 2012–2023 by Stan Brown, BrownMath.com
(b)
S = { | HHH HTH THH TTH | } |
---|---|---|
HHT HTT THT TTT |
(c) Three events out of eight equally likely events: P(2H) = 3/8
Common mistake: Sometimes students write the sample space correctly but miss one of the combinations of 2 heads. I wish I could offer some “magic bullet” for counting correctly, but the only advice I have is just to be really careful.
Service type | Prob. |
---|---|
Landline and cell | 58.2% |
Landline only | 37.4% |
Cell only | 2.8% |
No phone | 1.6% |
Total | 100.0% |
(a) In a probability model, the probabilities must add to 1 (= 100%). The given probabilities add to 62.6%. What is the missing 37.4%? They’ve accounted for cell and landline, cell only, and nothing; the remaining possibility is landline only. The model is shown at right.
(b) P(Landline) = P(Landline only) + P(Landline and cell)
P(Landline) = 37.4% + 58.2% = 95.6%
Remark: “Landline” and “cell” are not disjoint events, because a given household could have both. But “landline only” and “landline and cell” are disjoint, because a given house can’t both have a cell phone with landline and have no cell phone with landline.
(b) That A and B are complementary means that one or the other must happen, but not both. Therefore P(B) = P(not A) → P(B) = 0.3
(c) Since the events are complementary, they can’t both happen: P(A and B) = 0
Common mistake: Many students get (c) wrong, giving an answer of 1. If events are complementary, they can’t both happen at the same time. That means P(A and B) must be 0, the probability of something impossible.
Maybe those students were thinking of P(A or B). If A and B are complementary, then one or the other must happen, so P(A or B) = P(A) + P(B) = 1. But part (c) was about probability and, not probability or.
This is the difference between theoretical and empirical probability. A truly impossible event has a theoretical probability of zero. But the 0 out of 412 figure is an empirical probability (based on past experience). Empirical probabilities are just estimates of the “real” theoretical probability. From the empirical 0/412, you can tell that the theoretical probability is very low, but not necessarily zero. In plain language, an unresolved complaint is unlikely, but just because it hasn’t happened yet doesn’t mean it can’t happen.
Common mistake: Students often try some sort of complicated calculation here. You would have to do that if conditions were stated on all five of those cards, but they weren’t. Think about it: any card has a 1/4 chance of being a spade.
(a) 0.0171 × 0.0171 = 0.0003
(b) The events are not independent. When a married couple are at home together or out together, any attack that involves one of them will involve the other also.
(b) About 10.38% of American adults in 2006 were divorced. If you randomly selected an American adult in 2006, there was a 0.1038 probability that he or she was divorced.
(c) Empirical or experimental
(d)
P(divorcedC) = 1−P(divorced) =
1−22.8/219.7 ≈ 0.8962
About 89.62% of American adults in 2006 were not divorced
(or, had a marital status other than divorced).
(e) P(man and married) = 63.6/219.7 ≈ 0.2895 (You can’t use a formula on this one.)
(f) Add up P(man) and P(not man but married):
P(man or married) = 106.2/219.7 + 64.1/219.7 ≈ 0.7751
Alternative solution: By formula:
P(man or married) = P(man) + P(married) − P(man and married)
P(man or married) = 106.2/219.7 + 127.7/219.7 − 63.6/219.7 = 0.7751
Remember, math “or” means one or the other or both.
(g) What proportion of males were never married? 30.3/106.2 = 28.53%.
(h) P(man | married) uses the sub-subgroup of men within the subgroup of married persons.
P(man | married) = 63.6/127.7 = 0.4980
49.80% of married persons were men.
Remark: You might be surprised that it’s under 50%. Isn’t polygamy illegal in the US? Yes, it is. But the table considers only resident adults. Women tend to marry slightly earlier than men, so fewer grooms than brides are under 18. Also, soldiers deployed abroad are more likely to be male.
(i) P(married | man) used the sub-subgroup of married persons within the subgroup of men.
P(married | men) = 63.6/106.2 = 0.5989
59.89% of men were married.
(a)
3 of 20 M&Ms are yellow, so 17 are not yellow. You want
the probability of three non-yellows in a row:
(17/20)×(16/19)×(15/18) ≈
0.5965
(b) The probability is zero, since there are only two reds to start with.
(a) Since the companies are independent, you can use the simple multiplication rule:
P(A bankrupt and W bankrupt) = P(A bankrupt) × P(W bankrupt)
P(A bankrupt and W bankrupt) = .9 × .8 = 0.72
At this point you could compute (b), but it’s little messy because you need the probability that A fails and W is okay, plus the probability that A is okay and W fails. (c) looks easier, so do that first.
(c) “Neither bankrupt” means both are okay. Again, the events are independent so you can use the simple multiplication rule.
P(neither bankrupt) = P(A okay and W okay)
P(A okay) = 1−.9 = 0.1; P(W okay) = 1−.8 = 0.2
P(neither bankrupt) = .1 × .2 = 0.02
(b) is now a piece of cake.
P(only one bankrupt) = 1 − P(both bankrupt) − P(none bankrupt)
P(only one bankrupt) = 1 − .72 − .02 = 0.26
Remark: If you have time, it’s always good to check your work and work out (b) the long way. You have only independent events (whether A is okay or fails, whether W is okay or fails) and disjoint events (A fails and W okay, A okay and W fails). The “okay” probabilities were computed in part (c).
P(only one bankrupt) = (A bankrupt and W okay) or (A okay and W bankrupt)
P(only one bankrupt) = (.9 × .2) + (.1 × .8) = 0.26
Common mistake: When working this out the long way, students often solve only half the problem. But when you have probability of exactly one out of two, you have to consider both A-and-not-W and W-and-not-A.
You can’t use the “or” formula here, even if you studied it. That computes the probability of one or the other or both, but you need the probability of one or the other but not both.
Remark: If you computed all three probabilities the long way, pause a moment to check your work by adding them to make sure you get 1. Whenever possible, check your work with a second type of computation.
(a) 15(You can assume independence because it’s a small sample from a large population.) P(red1 and red2 and red3) = 0.13×0.13×0.13 = 0.0022
(b)
P(red) = 0.13; P(redC) = 1−0.13 = 0.87.
P(red1C and
red2C and
red3C) = 0.87×0.87×0.87 or 0.87³ =
0.6585
Common mistake: Students sometimes compute 1−.13³. But .13³ is the probability that all three are red, so 1−.13³ is the probability that fewer than three (0, 1, or 2) are red. You need the probability that zero are red, not the probability that 0, 1, or 2 are red. Think carefully about where your “not” condition must be applied!
(c)
The complement is your friend with
“at least” problems. The complement of “at least one is
green” is “none of them is green”, which is the same as
“every one is something other than green.”
P(green) = 0.16, P(non-green) = 1−0.16 = 0.84.
P(≥1 green of 3) = 1 − P(0 green of 3) = 1 −
P(3 non-green of 3) = 1−0.84³ ≈
0.4073
(d)
(Sequences are the most practical way to solve this
one.)
(A) G1 and G2C and G3C;
(B) G1C and G2 and G3C;
(C) G1C and G2C and G3
.16×(1−.16)×(1−.16) +
(1−.16)×.16×(1−.16) +
(1−.16)×(1−.16)×.16 ≈
0.3387
P(all 5 attended) = 0.45^5 = 0.0185
P(at least 1 had not attended) = 1 − 0.0185 = 0.9815
(cherry1 and orange2) or (orange1 and cherry2)
Common mistake: There are two ways to get one of each: cherry followed by orange and orange followed by cherry. You have to consider both probabilities.
There are 11+9 = 20 sourballs in all, and Grace is choosing the sourballs without replacement (one would hope!), so the probabilities are:
(11/20)×(9/19) + (9/20)×(11/19) = 99/190 or about 0.5211
P(win ≥1) = 1−P(win 0) = 1−P(lose 5).
P(lose) = 1−P(win) = 1−(1/500) = 499/500
P(lose 5) = [P(lose)]5 = (499/500)^5 = 0.9900
P(win ≥1) = 1−P(lose 5) = 1−0.9900 = 0.0100 or 1.00%
Common mistake: If you compute 1−(499/500)5 in one step and get 0.00996008, be careful with your rounding! 0.00996... rounds to 0.0100 or 1%, not 0.0010 or 0.1%.
Common mistake: 1/500 + 1/500 + ... is wrong. You can add probabilities only when events are disjoint, and wins in the various years are not disjoint events. It is possible (however unlikely) to win more than once; otherwise it would make no sense for the problem to talk about winning “at least once”.
Common mistake: You can’t multiply 5 by anything. Take an analogy: the probability of heads in one coin flip is 50%. Does that mean that the probability of heads in four flips is 4×50% = 200%? Obviously not! Any process that leads to a probability >1 must be incorrect.
Common mistake: 1−(1/500)5 is wrong. (1/500)5 is the probability of winning five years in a row, so 1−(1/500)5 is the probability of winning 0 to 4 times. What the problem asks is the probability of winning 1 to 5 times.
(a) P(not first and not second) = P(not first) × P(not second) = (1−.7)×(1−.6) = 0.12
(c) P(first and second) = P(first) × P(second) = .7×.6 = 0.42
(b) 1−.12−.42 = 0.46
Alternative: You could compute (b) directly too, using sequences:
P(exactly one copy recorded) =
P(first and not second) + P(second and not first) =
P(first)×(1−P(second)) + P(second)×(1−P(first)) =
.7×(1−.6) + .6×(1−.7) = 0.46
A very common mistake on problems like this is writing down only one of the sequences. When you have exactly one success (or exactly any definite number), almost always there are multiple ways to get to that outcome.
You can’t use the “or” formula here, even if you studied it. That computes the probability of one or the other or both, but you need the probability of one or the other but not both.
(b) The probability of not getting a ticket on a given morning is 1−0.27 = 0.73. The probability of getting no tickets on five mornings in a row is therefore 0.735 ≈ 0.2073 or about 21%.
P(man) = 106.2/219.7 ≈0.4834
P(man|divorced) = 9.7/22.8 ≈ 0.4254
Since P(man|divorced) ≠ P(man), the events are not independent.
Alternative solution: You could equally well show that P(divorced|man) ≠ P(divorced):
P(divorced|man) = 9.7/106.2 ≈ 0.0913
P(divorced) = 22.8/219.7 ≈ 0.1038
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