It’s been a while since we’ve looked at probability. Here, we’ll look at three questions that we received last year. In each case, we have to detect an error! They’re good examples of what can go wrong, and what to do when your answer appears to be wrong.
Seats in a taxi: Multiple ways to solve
This is from late October:
I have what appears to be a simple probability question but cannot agree with the given answer.
The question is:
A group of 2 women and 4 men are going to travel in two taxis. Each taxi can hold only three people. What is the probability that the first taxi will take 2 women and 1 man?
What I have done is to say pick one woman from the 6 people, then the second woman from the 5 people remaining. Then there are four ways of selecting the man. As we are just picking a group of people, order does not matter.
So the probability is (2/6 × 1/5) × 4 = 4/15.
The given answer is 1/5. I don’t see how that is arrived at.
I must be overlooking something obvious as the question seems too basic to get wrong.
Any help would be appreciated.
With thanks.
The first thing to do is to be sure of the meaning of the problem. From experience with such problems, we can fill in some gaps: It appears to be assumed that who gets in which taxi is totally random (unlike real life). Also, clearly the second taxi will take the remaining three men, so we can ignore what happens there.
I answered:
Hi, Jonathan.
These are always an interesting challenge, because it can be hard to keep your focus on “order doesn’t matter”. (I hardly ever consider a probability problem too simple to get wrong!)
I’m going to try to correct your answer, and then give a couple other ways.
The key error in your work is that if you say order doesn’t matter, then you can’t talk about the first or second woman (or about picking one, and then another). Or at least, if you do, then you have to compensate.
But also, it isn’t appropriate to multiply this probability by the number of ways to choose the man. You are mixing together two approaches to the problem (combining probabilities, and counting possibilities).
You can multiply two probabilities (to get the probability that both happen), or multiply ways two things can happen (to get the number of ways both can happen), but you have to be careful mixing these ideas (as we will below).
What I would do, if I chose to combine probabilities as you are doing, is to initially let order matter, and then adjust. The probability that the first two chosen (in order) are women and the third is a man is
P(woman first)×P(woman second)×P(man third) = 2/6 × 1/5 × 4/4 = 1/15
Then, there are 3C2 = 3 ways to choose the seats for the women (WWM, WMW, MWW), so we multiply by 3 to get the probability that one of these three orders will happen:
1/15 × 3 = 1/5
Here I multiplied by 4/4, not just by 4, to get the probability of WWM; then multiplied by 3 as the number of ways to choose an order for everyone. But even this would leave me a little unsure, until I carefully thought about how they are being chosen. Do I need to count the arrangements of WWM (3 ways), or of W1W2M (6 ways)? It’s the former, because I am just choosing locations, and then will fill the two women’s places from left to right, and the man’s.
Now, there is enough uncertainty about that reasoning, that I would want to get the same answer by a different method in order to be sure. So here are the two ways I did it before looking at your answer:
- Combinations: We want to divide the number of ways to choose two women and one man by the total number of ways to choose any three people.
The numerator is 2C2 × 4C1 = 4.
The denominator is 6C3 = 20.
So the probability is 4/20 = 1/5.
Here we are consistently ignoring order, so this feels like the most natural method. The four ways to choose the riders, in the numerator, are W1W2M1, W1W2M2, W1W2M3, and W1W2M4.
- Permutations: Even though order doesn’t matter in the problem, we can still use permutations in the work! This time, we want to divide the number of ways to arrange two women and one man by the total number of ways to arrange any three people.
The numerator is the number of ways to arrange WWM, namely 3C2 = 3 as before, times the number of ways to arrange the two women in those two seats, and the one man in the third. This gives 3 × 2P2 × 4P1 = 3 × 2! × 4 = 24.
The denominator is the number of ways to arrange 3 out of 6 people, namely 6P3 = 6×5×4 = 120.
So the probability is 24/120 = 1/5.
The 24 ways to arrange the riders are
W1W2M1, W1W2M2, W1W2M3, W1W2M4, W2W1M1, W2W1M2, W2W1M3, W2W1M4,
W1M1W2, W1M2W2, W1M3W2, W1M4W2, W2M1W1, W2M2W1, W2M3W1, W2M4W1,
M1W1W2, M2W1W2, M3W1W2, M4W1W2, M1W2W1, M2W2W1, M3W2W1, M4W2W1.
(That isn’t really the second way I did it; instead, I did the following, which is a little different.)
- We can also consider all 6 people, looking for the probability that both women are in the first 3 positions in a row of 6. This time, we want to divide the number of ways to arrange them as required, by the total number of ways to arrange all six people.
The numerator is the number of ways to choose two of the first three seats for women, namely 3C2 = 3 as before, times the number of ways to arrange the two women in those seats, and the four men in the remaining 4. This gives 3 × 2P2 × 4P4 = 3 × 2! × 4! = 144.
The denominator is the number of ways to arrange all 6 people, namely 6P6 = 6! = 720.
So the probability is 144/720 = 1/5.
Some similar issues (about whether order matters) arise in Permutation vs Combination: Clarifying Our Terms.
So we can either choose or arrange people in the first taxi, or arrange them in both taxis. There may be other ways as well – in fact, there almost surely are!
Clarifying details
Jonathan replied:
Hi Doctor Peterson. Thank you for your reply. I multiplied by 4 because I assumed that as there are 4 men, there must be 4 ways of selecting the man once the 2 women have been picked. I can see the error now I have seen the various ways you have tackled this.
I didn’t think of permutations because I assumed it doesn’t matter where each passenger is seated: I thought of it more in terms of a committee. But then they have to be picked in some sort of order…
Probability can be confusing and it appears to depend on how you interpret the problem. Thank you once again for your help.
Technically, if we interpret the problem as meaning something else, we will get a different answer. But he likely means “how you see the problem”, which is different. Different perspectives on the same problem will give the same answer.
I responded, first to the second paragraph::
Exactly. When you multiply probabilities, you are assuming order; if you recall the binomial distribution, that amounts to the same thing, multiplying probabilities pxq1-x, but multiplying that by nCx to account for the number of possible orders.
And then to the last paragraph:
Yes, that’s why, as I said, I don’t trust myself completely until I’ve obtained the same result two different ways. I generally try to imagine carrying out a compound process step by step, and make sure that does exactly what is needed by the problem.
Another place I’ve said that is Combinatorics: Multiple Methods, Subtle Wording.
Choosing two cards: Is the book wrong?
This came in at the start of October:
In a “Teach Yourself” math book is this:
A card is picked at random from a pack. It is replaced and a card is again picked at random. Find the probability of . . .
(a) first a heart and second a king.
(b) a heart and a king (in either order).
The book gives the answers, and I got question (a) correct: 13/52 × 4/52 = 52/2704 = 1/52.
It’s (b) I don’t understand. The book gives the answer of 1/26. I see that 26 is half of 52. And I see that in comparison to (a), in (b) you get 2 chances to get that same card combination. I need to see the actual math that gets me to 1/26.
Lastly, question
(c): at least one diamond.
The probability of picking one diamond is 13/52 = 1/4. But we’re picking two cards here. Since the second card picked could be any card, I’m thinking the probability of selecting any card is 52/52. So 1/4 x 52/52 = 1/4. But the book gives the answer of 7/16. I want to see the math of how to get 7/16.
Thanks!
Jim
Parts (a) and (c) are classic problems of selection with replacement; (b) is a little odd, and needs special care.
I answered:
Hi, Jim.
Let’s take a look.
A card is picked at random from a pack. It is replaced and a card is again picked at random. Find the probability of . . .
(a) first a heart and second a king.
Since a standard deck contains 13 hearts, and 4 kings, and we are drawing with replacement,
P(heart, then king) = P(heart)×P(king) = 13/52 × 4/52 = 1/4 × 1/13 = 1/52
as you say.
With replacement, the events (outcomes for the first card and for the second card) are independent, and can be multiplied together for an “and” problem like this.
The next looks like an “and” problem, but is really an “or”:
(b) a heart and a king (in either order).
This is tricky, and I think the book got it wrong. Here is one way to get what I think is the correct answer:
P(heart, then king, or king, then heart) =
P(heart, then king) + P( king, then heart) – P(king of hearts, then king of hearts) =
1/4 × 1/13 + 1/13 × 1/4 – 1/52 × 1/52 = 1/52 + 1/52 – 1/(52×52) = (52+52-1)/(52×52) = 103/2704
This follows the rule that P(A or B) = P(A) + P(B) – P(A and B). The author evidently forgot that it is possible to get both a heart followed by a king, and a king followed by a heart, by both cards being the king of hearts.
But this answer is very close to the claimed answer; 1/26 ≈ 0.03846, while 103/2704 ≈ 0.03809.
The difference is the probability of both cards being the king of hearts, namely \(\left(\frac{1}{52}\right)^2=0.00037\). This wouldn’t be needed if the selection were without replacement, since you couldn’t draw that king of hearts twice! The author probably missed that because it’s usually the “without replacement” case that requires more thinking.
Now, if we selected without replacement, we would have to take into account whether the first card is both a heart and a king, because that would affect the probability for the second card. Using the approach we used above, we have $$P(\text{heart, then king}) = P(\text{king of hearts, then other king})+P(\text{non-king heart, then any king})\\=\frac{1}{52}\cdot\frac{3}{51}+\frac{12}{52}\cdot\frac{4}{51}=\frac{3+48}{52\cdot51}=\frac{1}{52}\\P(\text{king, then heart}) = P(\text{king of hearts, then other heart})+P(\text{non-heart king, then any heart})\\=\frac{1}{52}\cdot\frac{12}{51}+\frac{3}{52}\cdot\frac{13}{51}=\frac{12+39}{52\cdot51}=\frac{1}{52}\\P(\text{king and heart in either order})=\frac{1}{52}+\frac{1}{52}=\frac{1}{26}$$
which is the book’s answer! (We didn’t have to subtract anything this time, because, without replacement, both cards can’t be both king and heart.)
The third question is a classic “at least one” problem, with a standard technique:
c): at least one diamond.
The easiest way to get this is to see that
P(at least one diamond) = P(not no diamonds) =
1 – P(no diamonds) = 1 – (3/4 × 3/4) = 1 – 9/16 = 7/16
This can also be done as an “or” problem:
P(at least one diamond) = P(first is diamond or second is diamond) =
P(first is diamond) + P(second is diamond) – P(both are diamonds) =
1/4 + 1/4 – 1/4 × 1/4 = 1/2 – 1/16 = 7/16
The various rules I’ve used are discussed in
Jim closed:
Thank you for the fast response! I appreciate the time and effort you put into your explanations; they are quite helpful.
The book appears to be Teach Yourself Mathematics, by Trevor Johnson and Hugh Neill (or one of several related editions). Here is the actual problem and solution, as I found it in the one book I had access to:
Most, if not all, textbooks have occasional errors; I have no reason to think this one has more than usual.
No two days in a row: Is the website wrong?
The next question is from a teacher, in July:
My problem:
Vanessa travels to school each weekday by bus or by car. The car option is only available if her cousin is available to drive which only happens on 30% of the days.
What is the probability that next week her mode of travel to school will not be the same on any two consecutive days?
I am not sure how to do this one.
Since no attempt was shown, a hint was appropriate, rather than a full solution. I answered:
Hi, Carla.
Interesting problem! I think the hardest part is to figure out exactly what it means. Once we manage that, you’ll probably have no trouble.
I’ll look at the last line first. What does it mean to not have the same mode on any two consecutive days? I take it to mean that during those 5 days (assuming “weekday” means the same in your culture as in mine), she must take either BCBCB or CBCBC. Do you agree?
Since every day will be either Bus or Car, the only way to never have BB or CC is to alternate like this. This rephrasing greatly simplifies the problem!
Then, I think we have to suppose that she takes the car whenever she can. It’s conceivable that she could randomly decide, when the cousin is available, whether she wants to take advantage of that, but that would mean we’d need more information (namely, how likely she is to choose the bus when it isn’t the only choice). Another possibility is that she wants to take a different mode each day, so that she will take the car only if she took the bus the day before, even if it is available.
This sort of thinking (deciding the meaning by looking for the interpretation that makes it a reasonable problem) should not be necessary, but it often is!
There’s actually another issue in the wording of the problem: “30% of the days” almost sounds as if it could mean that the car will be available on exactly 30% of these particular 5 days. But that would be 1.5 days, which wouldn’t make sense. So I take it to mean “30% of all days”.
So, what is the probability that she will take the bus on day 1? What is the probability that she will take the car on day 2? And so on.
Let me know what you think, both about the interpretation of the problem, and the next steps.
Since we didn’t get an answer, I’ll solve it now.
On any given day, \(P(\text{ Car available})=0.30\). We’re taking this to mean \(P(\text{ Takes car})=0.30\).
Therefore, \(P(\text{ Takes bus})=1-0.30=0.70\).
Then \(P(\text{BCBCB})=0.70\times0.30\times0.70\times0.30\times0.70=0.70^3\times0.30^2=0.03087\), and \(P(\text{CBCBC})=0.30\times0.70\times0.30\times0.70\times0.30=0.70^2\times0.30^3=0.01323\).
So our answer is \(P(\text{BCBCB or CBCBC})=0.03087+0.01323=0.0441\).
Having said that, I found the problem at
https://www.transum.org/Maths/Activity/Tree_Diagrams/Problems.asp
and find that none of the answers I try are judged correct, so either I am interpreting it differently than they intend, or I am making some silly error. I also haven’t been able to get the first problem “correct”, which I also consider somewhat ambiguous. So at this point I seem to be stuck. I’ll come back to look at it again later.
I put “.0441” in as the answer, and it was called wrong. Was I interpreting the problem differently than the site’s authors?
Getting the site to accept my answers
Carla never answered. But in editing this post, I went back to the site, and was able to get it to say all my answers were correct (sort of). To do that, I had to give what I consider a wrong answer to question 1a; but the rest of my answers were called right once I included a leading zero in decimal answers! This is just bad programming.
Here are the first and last problems on the page, with my answers as I first entered them:
That’s what left me confused. Here’s what it says when I add the leading zeros:
So my answer to #6 does agree with their interpretation; but I still have 1a “wrong”. I interpreted “not more than an hour late on both days” as “it is not true that it will be late on both days”, so that the answer is \(1-0.2^2=1-0.04=0.96\).
When I instead take it as “on time on both days”, which a careless reader might think, then I get \((1-0.2)^2=0.8^2=0.64\):
So it turns out that it is problem 1a that is truly ambiguous; problem 6 is no problem once you think through it.
This experience reminds me how frustrated students can be when computer feedback isn’t clear!