Need help with this really tricky math problem

[william1324]

A box contains all 26 letters of the alphabet. What is the probability that the first two letters drawn will be the letters a, b, c, d, e, or f?

 

[Dr.Peterson]

A box contains all 26 letters of the alphabet. What is the probability that the first two letters drawn will be the letters a, b, c, d, e, or f?

I assume it contains exactly one of each.

What is the probability that the first letter drawn will be one of those 6? What about the second?

What part of this do you find tricky?

 

[pka]

A box contains all 26 letters of the alphabet. What is the probability that the first two letters drawn will be the letters a, b, c, d, e, or f?

This question is very poorly stated. We are to assume that each letter is in the box one and only one time time.
And two letters are drawn out of the box without replacement.
What does either of these have to do with the question? [imath]\dbinom{6}{2}=15~\&~\dbinom{26}{2}=325~?[/imath]

 

[william1324]

I assume it contains exactly one of each.

What is the probability that the first letter drawn will be one of those 6? What about the second?

What part of this do you find tricky?

Both. How do I set this up to solve it?

 

[william1324]

This question is very poorly stated. We are to assume that each letter is in the box one and only one time time.
And two letters are drawn out of the box without replacement.
What does either of these have to do with the question? [imath]\dbinom{6}{2}=15~\&~\dbinom{26}{2}=325~?[/imath]

Well, according to the source where I got this problem. The answer key claims that the final result is 3/115

 

[Dr.Peterson]

Both. How do I set this up to solve it?

Please answer my questions! They are the start of the process (using one method).

pka has suggested a different method, using combinations rather than compound probability.

Part of the reason we are hinting at different ways is that you have given us no information about what you know, so we can't tell what method will be most appropriate for you. This is why we ask you to show work, even just an attempt: it shows us where you are in your own thinking.

Please read this:

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[pka]

Well, according to the source where I got this problem. The answer key claims that the final result is 3/115

Are you sure that you posted the exact problem as given to you?
The given answer does not seem to it.
Is it that, first two letters drawn will form a subset of [imath]\{a,b,c,d,e,f\}~?[/imath]

 

[william1324]

Are you sure that you posted the exact problem as given to you?
The given answer does not seem to it.
Is it that, first two letters drawn will form a subset of [imath]\{a,b,c,d,e,f\}~?[/imath]

Yes, I'm sure that it is the exact problem. Yes, the first letters drawn will be from the subset (a,b,c,d,e,f).

 

[Dr.Peterson]

Yes, I'm sure that it is the exact problem. Yes, the first letters drawn will be from the subset (a,b,c,d,e,f).

Then their answer is wrong. The question is, what answer do you get when you try it? Please show us some method, so we can help you with it.

 

[william1324]

Then their answer is wrong. The question is, what answer do you get when you try it? Please show us some method, so we can help you with it.

Well, I had a session with a math tutor of mine, I acquired some time ago, just recently and she told me that the method I should've used, after consulting with a colleague of hers, is (5/26) (5/25) without replacement. The result is 25/650, which can be reduced to 5/130.
The picture below was my attempt of solving this probability, using the help from my tutor. I should warn you though that this method I used may be bit too artificial for usual math procedures, maybe. Let me know if it was. Thank you.

 

[Dr.Peterson]

Well, I had a session with a math tutor of mine, I acquired some time ago, just recently and she told me that the method I should've used, after consulting with a colleague of hers, is (5/26) (5/25) without replacement. The result is 25/650, which can be reduced to 5/130.
The picture below was my attempt of solving this probability, using the help from my tutor. I should warn you though that this method I used may be bit too artificial for usual math procedures, maybe. Let me know if it was. Thank you.

No, that answer is not quite correct (and it's very wrong to say there is only one method you "should have used").

The probability of success on the first selection is not 5/26, but 6/26, because there are 6 acceptable letters out of 26.

The probability of success on the second selection is 5/25, because there are now 5 remaining acceptable letters out of 25 remaining letters.

So the answer is 6/26 * 5/25 = 3/65.

Your work seems to be for a different problem:

You used 24 instead of 26; do you speak a language with only 24 letters in its alphabet? And you decreased the number of acceptable letters by 2 when you removed only one. Can you explain this?

I also don't see why you then multiplied by 3/5. The problem you are working on must be something very different from what you showed us. Or does "artificial" mean "doing whatever I need to do to get the book's answer"?

 

[pka]

This question is very poorly stated. We are to assume that each letter is in the box one and only one time time.
And two letters are drawn out of the box without replacement.
What does either of these have to do with the question? [imath]\dbinom{6}{2}=15~\&~\dbinom{26}{2}=325~?[/imath]

So the answer is 6/26 * 5/25 = 3/65.

You used 24 instead of 26; do you speak a language with only 24 letters in its alphabet? And you decreased the number of acceptable letters by 2 when you removed only one. Can you explain this?

I also don't see why you then multiplied by 3/5. The problem you are working on must be something very different from what you showed us. Or does "artificial" mean "doing whatever I need to do to get the book's answer"?

In the PO a twenty-six alphabet was specified. The subset [imath]\mathcal{S}=\{a,b,,c,d,e,f\}[/imath] contains six letters so there are [imath]15[/imath] two element subsets of [imath]\mathcal{S}[/imath] . Because there are [imath]325[/imath] two element subsets of the alphabet, then the probability that a randomly selected two element set chosen from the alphabet is a subset of [imath]\mathcal{S}[/imath] in [imath]\dfrac{15}{325}=\dfrac{3}{65}[/imath] that agrees with Prof Peterson.

 

[william1324]

No, that answer is not quite correct (and it's very wrong to say there is only one method you "should have used").

The probability of success on the first selection is not 5/26, but 6/26, because there are 6 acceptable letters out of 26.

The probability of success on the second selection is 5/25, because there are now 5 remaining acceptable letters out of 25 remaining letters.

So the answer is 6/26 * 5/25 = 3/65.

Your work seems to be for a different problem:
View attachment 28274
You used 24 instead of 26; do you speak a language with only 24 letters in its alphabet? And you decreased the number of acceptable letters by 2 when you removed only one. Can you explain this?

I also don't see why you then multiplied by 3/5. The problem you are working on must be something very different from what you showed us. Or does "artificial" mean "doing whatever I need to do to get the book's answer"?

Yes, pretty much that whole "doing whatever I need to do to get the book's answer" part. This problem really bothered so much so that I started to make things up as I go along. Is the answer that you provided the correct method and solution to the problem?

 

[Dr.Peterson]

Yes, pretty much that whole "doing whatever I need to do to get the book's answer" part. This problem really bothered so much so that I started to make things up as I go along. Is the answer that you provided the correct method and solution to the problem?

Each of us provided a different, correct, solution. That's why I said, "it's very wrong to say there is only one method you 'should have used'", and why I've tried all along to find out what methods you have learned. It turns out, I think, that my method was what's most familiar to you (or at least to your tutor). Pka's method is better for many more complicated problems; this one was quite simple (apart from the poor wording).

Never, ever try reverse-engineering a solution; if the answer you were given is wrong, it can drive you insane (as in this case), and even if it's right, you usually "learn" a "method" that has no understanding behind it.

But what book is this, that has such a poorly-written problem and a wrong answer for it??

 

[pka]

Yes, pretty much that whole "doing whatever I need to do to get the book's answer" part. This problem really bothered so much so that I started to make things up as I go along. Is the answer that you provided the correct method and solution to the problem?

william, do you know that in most cases of basic textbooks if they have answers at all, that answer key was produced be a graduate student or sometimes an advanced undergraduate. They are paid by the problem solved and the rate is minimal. I can tell you that rarely does the book's author even see the answer key. Therefore, it is not surprising to find mistakes in the back-of-the-book. As to your question, there may be a correct answer but no one correct method.

 

[william1324]

Each of us provided a different, correct, solution. That's why I said, "it's very wrong to say there is only one method you 'should have used'", and why I've tried all along to find out what methods you have learned. It turns out, I think, that my method was what's most familiar to you (or at least to your tutor). Pka's method is better for many more complicated problems; this one was quite simple (apart from the poor wording).

Never, ever try reverse-engineering a solution; if the answer you were given is wrong, it can drive you insane (as in this case), and even if it's right, you usually "learn" a "method" that has no understanding behind it.

But what book is this, that has such a poorly-written problem and a wrong answer for it??

Well, as it turns out, this book that I have happens to be the latest edition of the CliffsNotes CBEST. I completely agree with pka's response, that the people who wrote these problems, who happen to be graduates and undergraduates, that are apparently not paid enough to give a damn about double checking their work just to survive another day at life as a college student. Dr. Peterson, I want to thank you very much for taking the time to investigate this probability question (if you can even call it that) that I posted on this forum. I appreciate the advice you gave me about solving these types of questions and never again attempt to reverse engineer any math problem, just because I knew the answer. Or else, it will drive me mad. I would like to move on from this problem and continue my studying to become a teacher, if you don't mind. You take care and I'll be sure to post anything else I find to be too difficult to solve on my own.

william, do you know that in most cases of basic textbooks if they have answers at all, that answer key was produced be a graduate student or sometimes an advanced undergraduate. They are paid by the problem solved and the rate is minimal. I can tell you that rarely does the book's author even see the answer key. Therefore, it is not surprising to find mistakes in the back-of-the-book. As to your question, there may be a correct answer but no one correct method.

Wow, really? I'm never aware of this fact. Those grads are paid to solve problems at a minimal rate? That's rough! Thank you pka, for sharing this fact. I'll keep that in mind whenever I encounter a book of math problems with answer keys, and expect, at least, some of the answers to be wrong. Thank you for taking the time of solving this poorly-written problem, with possibly some missing information. I like to move from this problem, if you don't mind. I appreciated the knowledge that you shared and learned that mistakes will happen and I can never trust what I see in the answers. I have to question it and ask "is this answer truly the correct one?" You take care and I'll be sure to post any more tricky math problems I could use your help on.