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Need some help.

I'm a first year teacher, moving into a probability unit.

I was looking through my given materials, and think I've found two errors. It's been a long time since I've done probability and I don't want to ask staff for fear of sounding stupid.

I created a Word document that I took a screen shot of, in order to hopefully get some confirmation.

Anything inside the orange boxes are from my course materials, anything outside is work I did myself in Word. The red box on the right is just highlighting where I think the error occurs.

Any help would be greatly appreciated.

enter image description here

Noticed a small error on my part. The absolute last thing I wrote should say 3/16. As the book defines P(A and B) as such, not as 11/16, as I suggested.

SamForte87
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    What reference book is this? Seems worth avoiding. – lulu May 04 '18 at 00:15
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    Note: for the second one, they try to compute the probability that the number is either odd or prime or both (which is not what the problem asks). Of course, they do even that wrong...the winning outcomes in that case would be ${3,5,7,9}$ so the answer to that question would be $\frac 48$. Their error is to treat the two events as independent which is silly. Prime implies odd in this case. – lulu May 04 '18 at 00:18
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    Holy smokes, those provided answers are bad! And you are correct in both cases. – Bram28 May 04 '18 at 00:23
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    +1 for thinking things through. I think you should write Pearson, and the book's authors. Pearson is a big publisher. Be polite, but make it clear, politely, that you expect be some kind of public retraction/correction, An implicit suggestion that this will become very public might me in order - you can point them to this discussion.. Note that the people answering here are all low reputation - many beginners who agree with you that this makes no sense. This is not rocket science. Please let me know if you follow up (my email is on my SE profile). – Ethan Bolker May 04 '18 at 00:27
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    Regarding reputation on this site, high reputation may correlate with competence, but it also depends very much on the length of membership and level of activity on this specific site. In any case, we're unanimous so far. Your explanations of what the answers should be are excellent, which may help. – David K May 04 '18 at 00:44
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    Thank you all for your helpful responses. Teaching this soon. Glad I don't have to question myself. – SamForte87 May 04 '18 at 00:47
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    @EthanBolker: I second that! Dear SamForte87: by contacting Pearson you will do a great service to so many kids far outside of your own classroom! Pearson is a huge publisher, and you probably can't change the fact that their product is adopted for your school. But you can -- and you should! -- bring this to their attention. Believe it or not, in my experience these huge publisher tend to respond well to such criticism, and they do follow up and make corrections. – zipirovich May 04 '18 at 04:41
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    Although this question fits here perfectly, I would like to kindly hint OP at the existence of matheducators.SE. – Dubu May 04 '18 at 13:59
  • I'd wonder if the book has errata posted somewhere on the web. Most probably on the publishers website. – MaxW May 04 '18 at 16:27
  • Please transcribe at least the bare minimal content to the post, rather than putting it in an image. You can format it using MathJax syntax. – jpmc26 May 04 '18 at 23:42
  • As an aside, plain English text in formulae should never be in italics. In Microsoft Word, use double quotation marks to escape such text: v_"input" and v_"output", for example. See https://english.rejbrand.se/rejbrand/msword.asp#mathematics for more information. – Andreas Rejbrand May 05 '18 at 11:15
  • Both of your calculations are correct (though I see others have already confirmed this). – WillO May 05 '18 at 13:52

6 Answers6

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Not only you're right, but the second question is quite weird: the question asks for $P(A\cap B)$ when in the end the final result is supposedly $P(A\cup B)$. Moreover, $A$ contains $B$, which is never pointed out; the term "overlapping" is very misleading in this case. I would throw this material away or be very suspicious about every statement it makes if I were you.

Arnaud Mortier
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I'm pretty sure its an error on the course materials. The ideas of a number being odd and a number being prime are dependent on each other (an even number cannot be prime unless it is $2$), so the rule $P(A)*P(B)=P(A\cap B)$ is not valid here (it's only valid for independent events). So: trust your instincts, you are correct.

Rhys Hughes
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8

What is truly amazing to me is how $11/16$ could be thought to be an answer in the second question. The provided solution at one point even explicitly lists out the $3$ odd primes in the set $\{3, 4, 5, 6, 7, 8, 9, 10\}$, and then after getting the correct answer, proceeds to continue on and perform completely irrelevant computations.

It should immediately obvious that, for a single choice from a set of $8$ numbers, the denominator of any resulting probability cannot be $16$.

I cannot even begin to speculate as to how $0.17$ was obtained for the first question, as a decimal answer gives little insight compared to a fraction.

As a teaching tool, I would advise that for the second exercise, you actually show the students how the teacher's guide went through the solution, step by step. Then ask at each step whether the reasoning was correct. If not, why not? Where does the solution break down? What would a correct solution look like? Approach it like a critique: students will be very interested in seeing that even a publisher of a textbook can make mistakes from time to time.

heropup
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6

Both your answers are right, and it seems that the course material is wrong.

Green
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I figured the first one a bit different, but got the same answer you did.

When you pull the first card there is a 3 in 7 chance of it being a card you want.

Assuming success, when you pull the second card there is a 2 in 6 chance.

Again assuming success, when you pull the third card there is a 1 in 5 chance.

The odds of all three happening are an AND condition.

$$ P = \frac{3}{7} \cdot \frac{2}{6} \cdot \frac{1}{5} = \frac{1}{35} $$

The second one is clearly self contradictory within the text. The story part says "AND" and the solution part says "OR".

Furthermore, the two conditions are not independent. If the number is a prime in that range, then it is odd, so the being odd condition is irrelevant. Since there are three primes out of eight numbers, your odds are $\frac{3}{8} $

Clearly you should view the rest of the material in this book with a skeptical eye.

Ced

Cedron Dawg
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1

Three playing cards are chosen from a set numbered from $1$ to $7$. What is the probability that the cards chosen are numbered $1$, $2$, and $3$?

Your answer is correct.

An alternate approach:

There are $$\binom{7}{3}$$ ways to select a subset of three of the seven cards. Of these, just one contains the cards numbered $1$, $2$, and $3$. Hence, the probability of selecting the cards numbered $1$, $2$, and $3$ when three cards are drawn from seven cards numbered from $1$ through $7$ is $$\frac{\dbinom{3}{3}}{\dbinom{7}{3}} = \frac{1}{35}$$

As Martin Argerami pointed out in the comments, the answer in the text is six times this, which suggests that the author(s) divided the $3!$ sequences in which the numbers $1$, $2$, and $3$ could be obtained by the $\binom{7}{3}$ ways of selecting three-element subsets of a set with seven elements.

You can use this example to demonstrate that the method of counting favorable outcomes must be consistent with the method of counting the number of elements in the sample space. You can show them two ways of solving the problem correctly:

  1. Divide the number of sequences in which the cards numbered $1$, $2$, and $3$ are obtained by the number of sequences of three cards that can be obtained from the seven available cards.
  2. Divide the number of three-element subsets containing the cards numbered $1$, $2$, and $3$ by the number of subsets of three cards that can be obtained from the seven available cards.

What you cannot do is what the author(s) apparently did, making the mistake of treating the favorable outcomes as if order mattered and the sample space as if order does not matter.

You put $8$ tiles, numbered $3$ through $10$ into a bag. What is the probability of choosing a number that is both an odd number and a prime number?

You correctly determined that $3$ of the $8$ numbers are both odd and prime. The authors inexplicably treated the events of being odd and being prime as independent events even though an integer larger than $2$ can only be prime if it is odd.

You can use this example to demonstrate that probabilities can only be multiplied if events are independent.

Your correct answer to that question also allowed you to correctly answer the question the author(s) meant to ask:

You put $8$ tiles, numbered $3$ through $10$ into a bag. What is the probability of choosing a number that is an odd number or a prime number?

It is disheartening that the text was published with these errors. Please send the publisher a list of errata. Since the errors are present, you can critique them in class so that your students know to avoid making such errors themselves.

N. F. Taussig
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