Ron--Please help.Probability!!

[dddanny2006]

Source-Magoosh
A four digit safe code does not contain the digits 1 and 4 at all. What is the probability that it has at least one even digit?

a) ¼
b) ½
c) ¾
d) 15/16
e) 1/16

Here's my method to do it.

Digits allowed-0,2,3,5,6,7,8,9

No of possible arrangements=8*8*8*8=2096 -----Order matters since its a code but I dont understand what if the code contains the same numbers? ---Why??

Now back to finding the desired possibilities
There are 4 conditions possible
1.All numbers are even =4*4*4*4=256
2.3 numbers are even,one is odd =4*4*4*4=256
3.2 are even,2 are odd=4*4*4*4=256
4.1 is even,3 are odd=4*4*4*4=256

256+256+256+256=1024

Therefore 1024/4096 equals 1/4.

Whenever Ive used the slot method if order matters I dont divide it by the number of interchangeable slots,if order doesnt matter I divide by the number of interchangeable slots.All the odd slots above are bolded and hence for each condition we've got to check the number of ways in which those slots can be further arranged--I was told this from another forum,I didnt know this.Im finding it hard to identify problems where we have to further multiply by the number of ways the slots can be further arranged .Isnt the permutation,supposed to get me all those possible conditions,without further multiplying?I otherwords cant those 256include all the possible arrangements?


Please help me with the above.

Also please clear the mystery around the 8*8*8*8=4096------Order shouldn't matter if the numbers are same,at the same time matter if the numbers are different.How do I go about this?

Also in the 4 conditions that Ive mentioned above where I got 256 each time,what arrangements are included in the 256?,since Ive got to multiply them further by the number of ways they got to be arranged.Also how do I do this?I mean multiplying it further?


We don't do anything such in the problem below
Lets look at another problem below-

How many four digit numbers that do not contain the digits 3 or 6 are there?
a 2401 b 3584 c 4096 d 5040 e 7200


7.8.8.8 = 3584

My approach has got to be correct since most problems that Ive solved using the slot method have got me the right answer.



Thank you Ron

[RonPurewal]

Probably the most important lesson you can take from this problem is this: Use common sense, whenever it's possible to do so.

In this problem, you want the probability that at least one digit -- absolutely anywhere in the number -- is even.
The chance that the first digit is even is already 50%, and, even if that digit isn't even, you still have three more digits to go. So, obviously, the overall percentage is substantially greater than 50%.
This isn't much of an observation, but it's enough to narrow the choices down to just two options. (It's also good enough to eliminate the answer 1/4 -- meaning that you shouldn't have picked that answer, even if you had ostensibly "calculated" it).

[RonPurewal]

Another important lesson: Official GMAC problems don't mix probability and combinatorics/counting. If you're thinking about probability, you won't also have to think about combinations/permutations/whatever, and vice versa.

You can just approach this problem by combining the probabilities for 4 consecutive choices.

There are lots and lots and lots of ways to have at least one even digit -- but there's only one way for that NOT to happen: all odd digits. So, find the probability of getting all odd digits, and then subtract it from 1.
Since you can re-use digits, the chance that each individual digit is odd is 4/8, or 1/2. So the chance of getting all odd digits is
(1/2)(1/2)(1/2)(1/2) = 1/16.
So, the probability of anything else -- i.e., the probability of getting at least one even digit -- is 1 - 1/16, or 15/16.

[RonPurewal]

Source-Magoosh
A four digit safe code does not contain the digits 1 and 4 at all.

This problem is not written correctly. The intention of the problem writers is clear enough, but the purple word should say "or", not "and". (As literally written, the problem only excludes codes that contain both 1 and 4.)

No of possible arrangements=8*8*8*8=2096 -----Order matters since its a code but I dont understand what if the code contains the same numbers? ---Why??

It's irrelevant whether any of the slots have matching digits, since matching digits are allowed here. If you're confused by that idea, just imagine that each digit is written in a different font; maybe that will help.

In a situation like this one, "order matters" means "the slots are distinct". I.e., being in one slot is not equivalent to being in another slot.
Since "the first digit", "the second digit", "the third digit", and "the fourth digit" are clearly distinct entities, this is an "order matters" situation.
(By contrast, if you have a three-person committee, then there's no meaningful distinction among being "first", "second", or "third" on that committee. So, "order dosen't matter" in that situation.)

[RonPurewal]

Now back to finding the desired possibilities
There are 4 conditions possible
1.All numbers are even =4*4*4*4=256
2.3 numbers are even,one is odd =4*4*4*4=256
3.2 are even,2 are odd=4*4*4*4=256
4.1 is even,3 are odd=4*4*4*4=256

This calculation is flawed because you have to take into account the different ways in which your situations #2, #3, and #4 can happen.
Situation #2 can happen in four different ways (the odd digit could be first; it could be second; and so on).
Situation #3 can happen in six different ways. (If you don't know how to figure that out, just list them.)
Situation #4, like situation #2, can happen in four different ways.

If you do what you did here, but multiply situations #2 and #4 by four and #3 by six, you'll arrive at the correct probability.

[RonPurewal]

Whenever Ive used the slot method if order matters I dont divide it by the number of interchangeable slots,if order doesnt matter I divide by the number of interchangeable slots.All the odd slots above are bolded and hence for each condition we've got to check the number of ways in which those slots can be further arranged--I was told this from another forum,I didnt know this.Im finding it hard to identify problems where we have to further multiply by the number of ways the slots can be further arranged .Isnt the permutation,supposed to get me all those possible conditions,without further multiplying?I otherwords cant those 256include all the possible arrangements?

The best way to get a handle on these concepts is to try the calculations with smaller, more manageable numbers, and then actually make exhaustive lists to test whether your calculations are correct.
E.g., try the same calculations for a three-digit code whose digits have to be 1, 2, 3, 4, 5, or other such. That way the numbers will be small enough that you can easily check your results just by making lists.

As stated above, though, NEVER pick an answer that violates common sense.
After only one digit is chosen, you're already at a 50% probability of having an even number -- and that probability is only going to increase as you choose the other digits. So, cross out 1/4 (along with the other choices that are 1/2 or less).

[RonPurewal]

Also please clear the mystery around the 8*8*8*8=4096------Order shouldn't matter if the numbers are same,at the same time matter if the numbers are different.How do I go about this?

"Order" has nothing to do with the identities of the chosen digits. It has to do with whether the "places" or "slots" are distinguishable or indistinguishable.

How many four digit numbers that do not contain the digits 3 or 6 are there?
a 2401 b 3584 c 4096 d 5040 e 7200


7.8.8.8 = 3584

This is correct.
(Here, you're not worrying about the possibility of repeated digits. So, you shouldn't worry about that possibility in the original problem here, either.)

[dddanny2006]

Thank you so much Ron.That was clear and succinct.Sorry about posting the subject in an inappropriate manner.These problems get me frustrated and anxious at times and make me wonder if Im ever going to break the 700+ barrier.Is 700+restricted just to brainy students?,can a person with average intelligence ,right dedication and hardwork get there?

Thank you Ron

Also please clear the mystery around the 8*8*8*8=4096------Order shouldn't matter if the numbers are same,at the same time matter if the numbers are different.How do I go about this?

"Order" has nothing to do with the identities of the chosen digits. It has to do with whether the "places" or "slots" are distinguishable or indistinguishable.

How many four digit numbers that do not contain the digits 3 or 6 are there?
a 2401 b 3584 c 4096 d 5040 e 7200


7.8.8.8 = 3584

This is correct.
(Here, you're not worrying about the possibility of repeated digits. So, you shouldn't worry about that possibility in the original problem here, either.)

[RonPurewal]

Thank you so much Ron.That was clear and succinct.

You're welcome.

Sorry about posting the subject in an inappropriate manner.

No need for an apology, just follow the forum rules from now on. (And recycle, and clean up after your dog. And stuff like that.)

[RonPurewal]

Is 700+restricted just to brainy students?,can a person with average intelligence ,right dedication and hardwork get there?

The math folders aren't the right place for something like this, but, hey, it's Christmas, so here's a (rather long-winded) answer.

"700+" isn't the end of the world. Even at the HBS's and GSB's of the world, something like 3 of every 10 admitted students have a GMAT score below 700. At other schools, even more of them. That's not just that one freaky student who had already started three charities in Africa by his/her 18th birthday. No, that's at least thirty percent of the entire entering class.

Also, 700 is the worst number to set as a goal, because it's a "round" number.
That sounds like a weird statement if you've never heard it before, but think about it for a bit.

* If someone really wanted to bench-press 300 pounds but was stuck at 295, or really wanted to make $100,000 a year but had been pulling in $95,000 for a few years, how would that feel (psychologically)? Probably not very good. Most likely, your own words ("frustrated and anxious") would be an apt description.

* Now, imagine the threshold were 285 pounds, and so-and-so was temporarily stuck at 275. Or $90,000 and $80,000. These are objectively bigger shortfalls -- twice as far off the goal in absolute terms, even more in percentage terms -- but, weirdly enough, they're not nearly as psychologically oppressive for most people. Instead, the most common reaction tends to be optimistic disappointment, i.e., "I'm almost there".

The same is true in marketing, where it's well known that customers will (completely irrationally) perceive a discount from $2.00 to $1.75 as more significant than a discount from, say, $1.80 to $1.50.

So, yeah. You may be able to kill a big fraction of that anxiety by just making your goal something like, say, 680 or 690. Weird but true.

[dddanny2006]

Well,that's really convincing.For the reduced target ,like 680 or 690,do I need to be among the best brains out there?Can I do it with hard work?Or do I need to have it in born in me?I consider my self an average student.Please tell me Ron,Im beginning to doubt my ability at times.This will be my last post that's not related to General Math in this folder.
Thanks

Is 700+restricted just to brainy students?,can a person with average intelligence ,right dedication and hardwork get there?

The math folders aren't the right place for something like this, but, hey, it's Christmas, so here's a (rather long-winded) answer.

"700+" isn't the end of the world. Even at the HBS's and GSB's of the world, something like 3 of every 10 admitted students have a GMAT score below 700. At other schools, even more of them. That's not just that one freaky student who had already started three charities in Africa by his/her 18th birthday. No, that's at least thirty percent of the entire entering class.

Also, 700 is the worst number to set as a goal, because it's a "round" number.
That sounds like a weird statement if you've never heard it before, but think about it for a bit.

* If someone really wanted to bench-press 300 pounds but was stuck at 295, or really wanted to make $100,000 a year but had been pulling in $95,000 for a few years, how would that feel (psychologically)? Probably not very good. Most likely, your own words ("frustrated and anxious") would be an apt description.

* Now, imagine the threshold were 285 pounds, and so-and-so was temporarily stuck at 275. Or $90,000 and $80,000. These are objectively bigger shortfalls -- twice as far off the goal in absolute terms, even more in percentage terms -- but, weirdly enough, they're not nearly as psychologically oppressive for most people. Instead, the most common reaction tends to be optimistic disappointment, i.e., "I'm almost there".

The same is true in marketing, where it's well known that customers will (completely irrationally) perceive a discount from $2.00 to $1.75 as more significant than a discount from, say, $1.80 to $1.50.

So, yeah. You may be able to kill a big fraction of that anxiety by just making your goal something like, say, 680 or 690. Weird but true.

[RonPurewal]

When you're studying, don't worry about scores. Don't think about them at all.
If you were a competitive dancer, would you constantly obsess about judges' scores while you were trying to learn new dance routines? Hopefully not (and, if you did, you know how well you'd learn those new routines).