Confirm my math answers

[Vegitto]

Hi, you guys. I just got home from a test. It went very well, but I didn't quite understand one question. Or rather, I did understand, but I didn't know the correct answer. Here's the question, translated into English of course, and my answer 🙂.

A janitor, standing in front of a door, has a keychain with 8 (eight) keys on it. He tries to open the door, but it is locked. Assume that only one of 8 keys fits.
A:What is the chance that the 8th and last key fits, and the other ones don't? Assume that a value of 1 equals a 100% chance.
B: What is the chance that the 4th key fits? Assume that a value of 1 equals a chance of 100%.

My answers were:
A: (1/8)x(1/7)x(1/6)x(1/5)x(1/4)x(1/3)x(1/2)x(1/1)
B: (1/8)x(1/7)x(1/6)x(1/5)

Are these the correct answers? If not, which answers would be correct?

 

[bmacd]

which math course is this?

-=bmacd=-

 

[Vegitto]

Originally posted by: bmacd
which math course is this?

-=bmacd=-

I don't know what this is in America, but here it's Mathemetics B2, the highest course available for people my age (14-16).

 

[dullard]

Assuming all keys can be mistaken for each other:

A: 1/8
B: 1/8

Your odds are way, way too low.

 

[tfinch2]

Originally posted by: dullard
Assuming all keys can be mistaken for each other:

A: 1/8
B: 1/8

Your odds are way, way too low.

Wouldn't it be more since he tried the first key and it doesn't fit, so the odds are a little higher now that it could be the fourth or last key?

 

[Vegitto]

Originally posted by: dullard
Assuming all keys can be mistaken for each other:

A: 1/8
B: 1/8

Your odds are way, way too low.

Assuming that once a key is tried, it is thrown away. Otherwise you'd get skewed results, and this test wasn't trying to be realistic, it was trying to test whether or not you understood the chapter. 😉

 

[Soccer55]

Originally posted by: Vegitto
Hi, you guys. I just got home from a test. It went very well, but I didn't quite understand one question. Or rather, I did understand, but I didn't know the correct answer. Here's the question, translated into English of course, and my answer 🙂.

A janitor, standing in front of a door, has a keychain with 8 (eight) keys on it. He tries to open the door, but it is locked. Assume that only one of 8 keys fits.
A:What is the chance that the 8th and last key fits, and the other ones don't? Assume that a value of 1 equals a 100% chance.
B: What is the chance that the 4th key fits? Assume that a value of 1 equals a chance of 100%.

My answers were:
A: (1/8)x(1/7)x(1/6)x(1/5)x(1/4)x(1/3)x(1/2)x(1/1)
B: (1/8)x(1/7)x(1/6)x(1/5)

Are these the correct answers? If not, which answers would be correct?

Well, this depends on what the question asking. If you are supposed to start with the 1st key and try every key until you get the right one, then you're on the right track (but with the wrong answer). If you are supposed to randomly pick a key and want to know the probability that it's the right one, then dullard is correct.

-Tom

 

[dullard]

Originally posted by: tfinch2
Wouldn't it be more since he tried the first key and it doesn't fit, so the odds are a little higher now that it could be the fourth or last key?

BEFORE he starts trying the keys, if we know at least one of the 8 keys fits, then his chance that the 8th key fits is 1/8.

If he uses 7 keys, and they all fail, then the chance of the last key fitting is 100%. Of course, that is a silly question, and obviously this case is not what was being asked about.

 

[Vegitto]

Originally posted by: dullard
Assuming all keys can be mistaken for each other:

A: 1/8
B: 1/8

Your odds are way, way too low.

It wasn't about the realistic possibiliy that it'd be true, it was just about the number. I think the first answer was something like 2.54x10^-5 or something..

 

[Soccer55]

Originally posted by: Vegitto

Originally posted by: dullard
Assuming all keys can be mistaken for each other:

A: 1/8
B: 1/8

Your odds are way, way too low.

Assuming that once a key is tried, it is thrown away. Otherwise you'd get skewed results, and this test wasn't trying to be realistic, it was trying to test whether or not you understood the chapter. 😉

ok, then you're on the right track. Here's what you have to do though. For part a, you have to get the probability that the 8th key is right. That means that the previous 7 keys are wrong. So when you pick the 1st key, what's the probability that it's the wrong one? Do the same for the 2nd, 3rd, ... , 7th key. Same type of logic goes for part b.

-Tom

 

[dullard]

Originally posted by: Vegitto
Assuming that once a key is tried, it is thrown away. Otherwise you'd get skewed results, and this test wasn't trying to be realistic, it was trying to test whether or not you understood the chapter. 😉

Well, you'd have to include that information for any of us to confirm your answers. It would be pretty stupid NOT to go one by one through the keychain and to instead randomly mix the keys after every use. But I guess it is possible for someone to be that stupid. But, you really have to tell us to be that stupid when you post the question above. Otherwise, how are we to know what chapter you are in and in what book?

Lets say we did randomly mix the keys each and every try. On the first 7 trys you have to fail: that is you have a 7/8 chance on all of them. On the 8th try you have a 1/8 chance of guessing right. Total odds: (7/8)^7*(1/8) ~= 0.049.

 

[tfinch2]

Originally posted by: dullard

Originally posted by: tfinch2
Wouldn't it be more since he tried the first key and it doesn't fit, so the odds are a little higher now that it could be the fourth or last key?

BEFORE he starts trying the keys, if we know at least one of the 8 keys fits, then his chance that the 8th key fits is 1/8.

If he uses 7 keys, and they all fail, then the chance of the last key fitting is 100%. Of course, that is a silly question, and obviously this case is not what was being asked about.

It never says "What is the chance that the 8th and last key fits, BEFORE he tried the first key" though.

What if there is a master key too? 😛

 

[Vegitto]

Originally posted by: dullard

Originally posted by: tfinch2
Wouldn't it be more since he tried the first key and it doesn't fit, so the odds are a little higher now that it could be the fourth or last key?

BEFORE he starts trying the keys, if we know at least one of the 8 keys fits, then his chance that the 8th key fits is 1/8.

If he uses 7 keys, and they all fail, then the chance of the last key fitting is 100%. Of course, that is a silly question, and obviously this case is not what was being asked about.

Only one key is correct. The question was the odds of the last key fitting, and the other ones not fitting, with the last key being tried last.

Originally posted by: Soccer55

Originally posted by: Vegitto
Hi, you guys. I just got home from a test. It went very well, but I didn't quite understand one question. Or rather, I did understand, but I didn't know the correct answer. Here's the question, translated into English of course, and my answer 🙂.

A janitor, standing in front of a door, has a keychain with 8 (eight) keys on it. He tries to open the door, but it is locked. Assume that only one of 8 keys fits.
A:What is the chance that the 8th and last key fits, and the other ones don't? Assume that a value of 1 equals a 100% chance.
B: What is the chance that the 4th key fits? Assume that a value of 1 equals a chance of 100%.

My answers were:
A: (1/8)x(1/7)x(1/6)x(1/5)x(1/4)x(1/3)x(1/2)x(1/1)
B: (1/8)x(1/7)x(1/6)x(1/5)

Are these the correct answers? If not, which answers would be correct?

Well, this depends on what the question asking. If you are supposed to start with the 1st key and try every key until you get the right one, then you're on the right track (but with the wrong answer). If you are supposed to randomly pick a key and want to know the probability that it's the right one, then dullard is correct.

-Tom

No, you're supposed to try every key until you get the correct one.

 

[Soccer55]

Originally posted by: dullard

Originally posted by: Vegitto
Assuming that once a key is tried, it is thrown away. Otherwise you'd get skewed results, and this test wasn't trying to be realistic, it was trying to test whether or not you understood the chapter. 😉

Well, you'd have to include that information for any of us to confirm your answers. It would be pretty stupid NOT to go one by one through the keychain and to instead randomly mix the keys after every use. But I guess it is possible for someone to be that stupid. But, you really have to tell us to be that stupid when you post the question above.

If they were to randomly mix the keys after every use, wouldn't that be a geometric series or something like that? I don't remember the name for it, but I remember that a formula for finding the probability of success on the nth try exists. Been too long since I took probability 😛

-Tom

 

[Soccer55]

Originally posted by: Vegitto

Originally posted by: dullard

Originally posted by: tfinch2
Wouldn't it be more since he tried the first key and it doesn't fit, so the odds are a little higher now that it could be the fourth or last key?

BEFORE he starts trying the keys, if we know at least one of the 8 keys fits, then his chance that the 8th key fits is 1/8.

If he uses 7 keys, and they all fail, then the chance of the last key fitting is 100%. Of course, that is a silly question, and obviously this case is not what was being asked about.

Only one key is correct. The question was the odds of the last key fitting, and the other ones not fitting, with the last key being tried last.

Originally posted by: Soccer55

Originally posted by: Vegitto
Hi, you guys. I just got home from a test. It went very well, but I didn't quite understand one question. Or rather, I did understand, but I didn't know the correct answer. Here's the question, translated into English of course, and my answer 🙂.

A janitor, standing in front of a door, has a keychain with 8 (eight) keys on it. He tries to open the door, but it is locked. Assume that only one of 8 keys fits.
A:What is the chance that the 8th and last key fits, and the other ones don't? Assume that a value of 1 equals a 100% chance.
B: What is the chance that the 4th key fits? Assume that a value of 1 equals a chance of 100%.

My answers were:
A: (1/8)x(1/7)x(1/6)x(1/5)x(1/4)x(1/3)x(1/2)x(1/1)
B: (1/8)x(1/7)x(1/6)x(1/5)

Are these the correct answers? If not, which answers would be correct?

Well, this depends on what the question asking. If you are supposed to start with the 1st key and try every key until you get the right one, then you're on the right track (but with the wrong answer). If you are supposed to randomly pick a key and want to know the probability that it's the right one, then dullard is correct.

-Tom

No, you're supposed to try every key until you get the correct one.

Look at my post below that 😉 I posted that while you were responding to dullard's post and instead of editing, I just posted again.

-Tom

 

[Vegitto]

Originally posted by: dullard

Originally posted by: Vegitto
Assuming that once a key is tried, it is thrown away. Otherwise you'd get skewed results, and this test wasn't trying to be realistic, it was trying to test whether or not you understood the chapter. 😉

Well, you'd have to include that information for any of us to confirm your answers. It would be pretty stupid NOT to go one by one through the keychain and to instead randomly mix the keys after every use. But I guess it is possible for someone to be that stupid. But, you really have to tell us to be that stupid when you post the question above. Otherwise, how are we to know what chapter you are in and in what book?

Lets say we did randomly mix the keys each and every try. On the first 7 trys you have to fail: that is you have a 7/8 chance on all of them. On the 8th try you have a 1/8 chance of guessing right. Total odds: (7/8)^7*(1/8) ~= 0.049.

Ah. Okay. Thanks.

I don't think you know the book, though. It's a Dutch book. I believe it's called "Netwerk VWO4 2".

 

[QED]

Originally posted by: tfinch2

Originally posted by: dullard
Assuming all keys can be mistaken for each other:

A: 1/8
B: 1/8

Your odds are way, way too low.

Wouldn't it be more since he tried the first key and it doesn't fit, so the odds are a little higher now that it could be the fourth or last key?

Nope.

One easy way to visuasize this is to let the keys be lettered A through H. Assume the key labeled A is the one that fits, while keys B through H do not.

Ordering the keys in any random fashion is akin to ordering the letters A through H in any random fashion. Asking what are the odds that the 8th key tried fits is akin to asking what are the odds that a random ordering of the letters A through H has "A" last.

Well, the odds that the letter "A" is last are the same that "A" is first. In fact, the letter A is equally likely to appear in any position. Simililary, the the key that fits is just as likely to occur first as it last, or as likely at any other position from first through last.

Mathematically, it can be worked out as follow:

What are the odds the first key fits? That's easy: 1/8.

What are the odds the second key fits assuming the first did not? The odds the first key did not fit are 7/8. And since there are seven keys left, one of which actually fits, the odds the second key fits is 1/7. So the overall odds are 7/8 * 1/7 = 1/8.

What are the odds the third key fits but the first two did not? There are 8x7x6 different possible 3-key selections. There are 7 x 6 x 1 ways to choose them such that the first two keys do not fit but the third one does-- so the odds are (7x6)/(8x7x6) = 1/8.

What are the odds the fourth key fits but the first three did not? There are 8x7x6x5 different ways to select 4 keys from the 8. There are 7x6x5x1 ways to choose them such that the first three keys do not fit but the fourth one does-- again, the odds are 1/8.


Etc, etc.

 

[Jeraden]

I think you guys are overthinking it. Given the way the problem is stated, both are 1/8th of a chance. There are 8 keys, 1 fits (doesn't matter which one). Each key has a 1/8th of a chance of being the key that opens it. The only way to get an answer that is less than 1/8th would be if the correct key was randomly assigned (or mixed together) after each attempt, which just isn't supported by the way the problem is stated.

 

[tfinch2]

Originally posted by: MathMan

Originally posted by: tfinch2

Originally posted by: dullard
Assuming all keys can be mistaken for each other:

A: 1/8
B: 1/8

Your odds are way, way too low.

Wouldn't it be more since he tried the first key and it doesn't fit, so the odds are a little higher now that it could be the fourth or last key?

Nope.

One easy way to visuasize this is to let the keys be lettered A through H. Assume the key labeled A is the one that fits, while keys B through H do not.

Ordering the keys in any random fashion is akin to ordering the letters A through H in any random fashion. Asking what are the odds that the 8th key tried fits is akin to asking what are the odds that a random ordering of the letters A through H has "A" last.

Well, the odds that the letter "A" is last are the same that "A" is first. In fact, the letter A is equally likely to appear in any position. Simililary, the the key that fits is just as likely to occur first as it last, or as likely at any other position from first through last.

Mathematically, it can be worked out as follow:

What are the odds the first key fits? That's easy: 1/8.

What are the odds the second key fits assuming the first did not? The odds the first key did not fit are 7/8. And since there are seven keys left, one of which actually fits, the odds the second key fits is 1/7. So the overall odds are 7/8 * 1/7 = 1/8.

What are the odds the third key fits but the first two did not? There are 8x7x6 different possible 3-key selections. There are 7 x 6 x 1 ways to choose them such that the first two keys do not fit but the third one does-- so the odds are (7x6)/(8x7x6) = 1/8.

What are the odds the fourth key fits but the first three did not? There are 8x7x6x5 different ways to select 4 keys from the 8. There are 7x6x5x1 ways to choose them such that the first three keys do not fit but the fourth one does-- again, the odds are 1/8.


Etc, etc.

While true, the question never states if you calculate the probability based off of 7 or 8 keys. I assumed you start with 7 because it already said the first key was tried. You know it's not going to fit so why count it? The first key was tried, then the question was asked, so the first key does not count.

It's like saying 1 out of these 2 lottery tickets has won a million dollars. The first one was not a winner. What is the probability that the second one is the winner?

 

[dullard]

Originally posted by: tfinch2
While true, the question never states if you calculate the probability based off of 7 or 8 keys. I assumed you start with 7 because it already said the first key was tried. You know it's not going to fit so why count it? The first key was tried, then the question was asked, so the first key does not count.

It's like saying 1 out of these 2 lottery tickets has won a million dollars. The first one was not a winner. What is the probability that the second one is the winner?

You misread the problem. Read it over, and over again. It NEVER states that the first key was tried.

 

[QED]

Originally posted by: tfinch2
While true, the question never states if you calculate the probability based off of 7 or 8 keys. I assumed you start with 7 because it already said the first key was tried. You know it's not going to fit so why count it? The first key was tried, then the question was asked, so the first key does not count.

I think that might be where you are getting confused. Nowhere does it say the first key was tried and didn't fit. I think you implied too much into the line "he tries to open the door but it is locked".

Even if the problem does states that one key was tried and didn't work, it doesn't make the problem any more difficult-- the answer for both questions becomes 1/7th.

 

[tfinch2]

Originally posted by: dullard

Originally posted by: tfinch2
While true, the question never states if you calculate the probability based off of 7 or 8 keys. I assumed you start with 7 because it already said the first key was tried. You know it's not going to fit so why count it? The first key was tried, then the question was asked, so the first key does not count.

It's like saying 1 out of these 2 lottery tickets has won a million dollars. The first one was not a winner. What is the probability that the second one is the winner?

You misread the problem. Read it over, and over again. It NEVER states that the first key was tried.

LOL, you're right. Oops 😛

"He tries to open the door, but it is locked" - I misinterpreted that as he tried a key and it was still locked.

 

[Vegitto]

Okay. So, ultimately, my answers were wrong?

 

[QED]

Originally posted by: Vegitto
Okay. So, ultimately, my answers were wrong?

Yep... sorry, but your answers were WAYYYY off.

 

[Vegitto]

Originally posted by: MathMan

Originally posted by: Vegitto
Okay. So, ultimately, my answers were wrong?

Yep... sorry, but your answers were WAYYYY off.

S[/i]hit.