May 24 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. May 25 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. May 27 01:00 AM PDT  11:59 PM PDT All GMAT Club Tests are free and open on May 27th for Memorial Day! May 27 10:00 PM PDT  11:00 PM PDT Special savings are here for Magoosh GMAT Prep! Even better  save 20% on the plan of your choice, now through midnight on Tuesday, 5/27 May 30 10:00 PM PDT  11:00 PM PDT Application deadlines are just around the corner, so now’s the time to start studying for the GMAT! Start today and save 25% on your GMAT prep. Valid until May 30th.
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 30 Aug 2010
Posts: 13

A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
21 Sep 2010, 06:34
Question Stats:
67% (02:17) correct 33% (01:55) wrong based on 274 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.



Intern
Joined: 30 Aug 2010
Posts: 13

Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
21 Sep 2010, 06:35
As already said in a previous post ( outofaboxthatcontains4blackand6whitemice101375.html), I'm experiencing some problems with probability and combinations. So to understand better these arguments, I need to solve the problem in several ways. Unfortunately approaches 2 and 4 are missing. Could you help me? 1) probability approach:Probability of at least one even digit = Probability of one even digit + Probability of two even digits + Probability of three even digits + Probability of four even digits I have four even digits: 0,2,6,8 and four odd digits: 3,5,7,9 So the probability of even digit is \(1/2\), as well of odd digit. P1 = \(1/2\) P2 = \(1/2^2=1/4\) P3 = \(1/2^3=1/8\) P4 = \(1/2^4=1/16\) P = P1+P2+P3+P4 = \(15/16\) 2) combinatorial approach:This approach is missing. 3) reversal probability approach:Probability of at least one even digit = 1  Probability of four odd digits P = \(1  1/2^4 = 15/16\) 3) reversal combinatorial approach:This approach is missing.



Retired Moderator
Joined: 02 Sep 2010
Posts: 759
Location: London

Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
21 Sep 2010, 06:46
Combinatorial Solution No of ways to form code with atleast 1 even digit = Total Ways  No of ways to form code with only odd digits The digits available are {5,6,7,8,9,0} Total ways are 6x6x6x6 = 6^4 Ways using only odd digits = 3x3x3x3 = 3^4 Therefore Probability = \(\frac{6^4  3^4}{6^4} = 1  \frac{3^4}{6^4} = 1  (1/2)^4 = 15/16\)
_________________



Intern
Joined: 30 Aug 2010
Posts: 13

Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
21 Sep 2010, 06:58
shrouded1 wrote: Combinatorial Solution
No of ways to form code with atleast 1 even digit = Total Ways  No of ways to form code with only odd digits
The digits available are {5,6,7,8,9,0}
Total ways are 6x6x6x6 = 6^4
Ways using only odd digits = 3x3x3x3 = 3^4
Therefore Probability = \(\frac{6^4  3^4}{6^4} = 1  \frac{3^4}{6^4} = 1  (1/2)^4 = 15/16\) Your approach is correct (even if I would call it reversal combinatorial approach) but with this number: Digits available are {0,2,3,5,6,7,8,9,0} Total ways are = 8x8x8x8 = \(8^4 = 2^{12}\) Ways using only odd digits = 4x4x4x4 = \(4^4 = 2^8\) Therefore probability = \(\frac{2^{12}  2^8}{2^{12}} = 1  \frac{2^8}{2^{12}} = 1  (1/2)^4 = 15/16\) Thanks a lot!



Retired Moderator
Joined: 02 Sep 2010
Posts: 759
Location: London

Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
21 Sep 2010, 07:04
oh now i see it, i thought you said none of the digits from 1 to 4, but as you pointed out the solution remains the same. The "direct combinatorial" approach as you call it will be painful in this case. You will need to calculate each of 1 even, 2 even, 3 even, all even. The terms are themselves probably easy enough to calculate, but simplifying will be a bit tedious to get the same answer
_________________



Retired Moderator
Joined: 02 Sep 2010
Posts: 759
Location: London

Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
21 Sep 2010, 07:43
rraggio wrote: As already said in a previous post ( outofaboxthatcontains4blackand6whitemice101375.html), I'm experiencing some problems with probability and combinations. So to understand better these arguments, I need to solve the problem in several ways. Unfortunately approaches 2 and 4 are missing. Could you help me? 1) probability approach:Probability of at least one even digit = Probability of one even digit + Probability of two even digits + Probability of three even digits + Probability of four even digits I have four even digits: 0,2,6,8 and four odd digits: 3,5,7,9 So the probability of even digit is \(1/2\), as well of odd digit. P1 = \(1/2\) P2 = \(1/2^2=1/4\) P3 = \(1/2^3=1/8\) P4 = \(1/2^4=1/16\) P = P1+P2+P3+P4 = \(15/16\) I know you got to the answer, but your approach isnt quite correct. P1 = Probability of exactly 1 even digit = (1/2)^4 * C(4,1) = 4/16 P2 = Probability of exactly 2 even digits = (1/2)^4 * C(4,2) = 6/16 P3 = Probability of exactly 3 even digits = (1/2)^4 * C(4,3) = 4/16 P4 = Probability of 4 even digits = 1/16 P = P1+P2+P3+P4 = 15/16 You have calculated P1 as prob of at least 1 even digit, P2 as atleast 2 even digits etc. This is incorrect. Its only coincidence that the answer is correct.
_________________



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2565
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
21 Sep 2010, 07:57
probability of at least one even = 1(probability of none even) probability of none even = required cases/total cases total cases = 8*8*8*8 as 0 2 3 5 6 7 8 9 all 8 can be taken for each digit required cases = 4*4*4*4 as only 3 5 7 9 can be taken probability of none even = \(\frac{4*4*4*4}{(8*8*8*8)}= \frac{1}{16}\) probability of at least one even =\(1\frac{1}{16} = \frac{15}{16}\)
_________________
Fight for your dreams : For all those who fear from Verbal lets give it a fightMoney Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership  big benefits and savingsGmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Manager
Joined: 20 Jul 2010
Posts: 210

Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
21 Sep 2010, 16:55
I solved the question thinking 1,2,3 and 4 are not allowed in safecode and got same answer . My ratios were 3*3*3*3/(6*6*6*6). So by coincidence I got same answer
_________________
If you like my post, consider giving me some KUDOS !!!!! Like you I need them



Intern
Joined: 10 Oct 2010
Posts: 21
Location: Texas

Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
11 Oct 2010, 16:36
rraggio wrote: A four digit safe code
Permutation (_)(_)(_)(_) Note: A safe code can have 0 as the thousands digit; a 4digit number cannot. rraggio wrote: does not contain the digits 1 and 4 at all.
Bag of 8 choices w/ replacement. rraggio wrote: What is the probability that it has at least one even digit?
"At least" means Probability Table. Create and work backwards. **************************** # of evens: Events 0: 1: 2: 3: 4:  Total = **************************** Total = Fill in Permutation above = 8P4 = (8)(8)(8)(8) 4 evens: 4 evens pick 1 AND 4 evens pick 1 AND 4 evens pick 1 AND 4 evens pick 1 TIMES the number of ways to arrange EEEE = 4C1 * 4C1 * 4C1 * 4C1 * 4!/4! = 4*4*4*4 * 1 3 evens: 4 evens pick 1 AND 4 evens pick 1 AND 4 evens pick 1 AND 4 odds pick 1 TIMES the number of ways to arrange EEEO = 4C1 * 4C1 * 4C1 * 4C1 * 4!/3!1!= 4*4*4*4 * 4 2 evens: 4 evens pick 1 AND 4 evens pick 1 AND 4 odds pick 1 AND 4 odds pick 1 TIMES the number of ways to arrange EEOO = 4C1 * 4C1 * 4C1 * 4C1 * 4!/2!2! = 4*4*4*4 * 6 1 even: 4 evens pick 1 AND 4 odds pick 1 AND 4 odds pick 1 AND 4 odds pick 1 TIMES the number of ways to arrange EOOO = 4C1 * 4C1 * 4C1 * 4C1 * 4!/1!3! = 4*4*4*4 * 4 0 evens: 4 odds pick 1 AND 4 odds pick 1 AND 4 odds pick 1 AND 4 odds pick 1 TIMES the number of ways to arrange OOOO = 4C1 * 4C1 * 4C1 * 4C1 * 4!/4! = 4*4*4*4 * 1 **************************** # of evens: Events 0: 4*4*4*4*1 1: 4*4*4*4*4 2: 4*4*4*4*6 3: 4*4*4*4*4 4: 4*4*4*4*1  Total = 8*8*8*8 = 4*4*4*4*2*2*2*2 =4*4*4*4*16 **************************** Use the table info to answer any questions. P(evens = 0) = 1/16 P(evens > 0) = (4+6+4+1)/16 etc. rraggio wrote: a) ¼ b) ½ c) ¾ d) 15/16 e) 1/16 ANS: D



Current Student
Joined: 18 Oct 2014
Posts: 835
Location: United States
GPA: 3.98

Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
26 May 2016, 08:55
rraggio wrote: 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 Total available digits to make 4 digit code= 8 (excludes 1 and 4) Probability of having at least one even digit= 1 probability of having all odd digits Total ways to make 4 digit code= 8*8*8*8 Total ways of having only odd digits= 4*4*4*4 probability of having all odd digits= 4*4*4*4/8*8*8*8=1/16 Probability of having at least one even digit= 1 1/16= 15/16 D is the answer
_________________
I welcome critical analysis of my post!! That will help me reach 700+



Manager
Status: IF YOU CAN DREAM IT, YOU CAN DO IT
Joined: 03 Jul 2017
Posts: 190
Location: India
Concentration: Finance, International Business

Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
09 Jul 2017, 18:28
The digits to be considered are 0,5,6,7,8,9 so the total is 6c4 =15 and having at least one even number is 4c1+4c2+4c3=14 Therefore the probability is 14/15. I know my approach is wrong . Can someone please tell me as to where am i going wrong ?



Math Expert
Joined: 02 Sep 2009
Posts: 55267

Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
10 Jul 2017, 02:21
longhaul123 wrote: The digits to be considered are 0,5,6,7,8,9 so the total is 6c4 =15 and having at least one even number is 4c1+4c2+4c3=14 Therefore the probability is 14/15. I know my approach is wrong . Can someone please tell me as to where am i going wrong ? The question says that "...code does not contain the digits 1 and 4 at all", not "from 1 through 4". Meaning that we can use 2 and 3,
_________________



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 6207
Location: United States (CA)

Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
12 Jul 2017, 17:01
rraggio wrote: 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 We can use the following equation: P(at least one even digit) = 1  P(no even digits) Since 1 and 4 cannot be used, we have 8 available digits (0, 2, 3, 5, 6, 7, 8, 9), and we see that 4 of those 8 digits are odd (3,5,7,9). Thus, P(no even digits) = 4/8 x 4/8 x 4/8 x 4/8 = (1/2)^4 = 1/16. Thus, P(at least one even digit) = 1  1/16 = 15/16. Answer: D
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



Current Student
Joined: 29 Jul 2009
Posts: 25

Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
28 Nov 2017, 20:46
How can all the 4 digits be filled with 8 combination. If it has to be a 4 digit number and 1,4 are not allowed then first digit of 4 digit number can only be filled by 7 numbers ( excepting 1,4, 0) , so shouldn't it be : 7*8*8*8 What am i missing here ? cabelk wrote: rraggio wrote: A four digit safe code
Permutation (_)(_)(_)(_) Note: A safe code can have 0 as the thousands digit; a 4digit number cannot. rraggio wrote: does not contain the digits 1 and 4 at all.
Bag of 8 choices w/ replacement. rraggio wrote: What is the probability that it has at least one even digit?
"At least" means Probability Table. Create and work backwards. **************************** # of evens: Events 0: 1: 2: 3: 4:  Total = **************************** Total = Fill in Permutation above = 8P4 = (8)(8)(8)(8) 4 evens: 4 evens pick 1 AND 4 evens pick 1 AND 4 evens pick 1 AND 4 evens pick 1 TIMES the number of ways to arrange EEEE = 4C1 * 4C1 * 4C1 * 4C1 * 4!/4! = 4*4*4*4 * 1 3 evens: 4 evens pick 1 AND 4 evens pick 1 AND 4 evens pick 1 AND 4 odds pick 1 TIMES the number of ways to arrange EEEO = 4C1 * 4C1 * 4C1 * 4C1 * 4!/3!1!= 4*4*4*4 * 4 2 evens: 4 evens pick 1 AND 4 evens pick 1 AND 4 odds pick 1 AND 4 odds pick 1 TIMES the number of ways to arrange EEOO = 4C1 * 4C1 * 4C1 * 4C1 * 4!/2!2! = 4*4*4*4 * 6 1 even: 4 evens pick 1 AND 4 odds pick 1 AND 4 odds pick 1 AND 4 odds pick 1 TIMES the number of ways to arrange EOOO = 4C1 * 4C1 * 4C1 * 4C1 * 4!/1!3! = 4*4*4*4 * 4 0 evens: 4 odds pick 1 AND 4 odds pick 1 AND 4 odds pick 1 AND 4 odds pick 1 TIMES the number of ways to arrange OOOO = 4C1 * 4C1 * 4C1 * 4C1 * 4!/4! = 4*4*4*4 * 1 **************************** # of evens: Events 0: 4*4*4*4*1 1: 4*4*4*4*4 2: 4*4*4*4*6 3: 4*4*4*4*4 4: 4*4*4*4*1  Total = 8*8*8*8 = 4*4*4*4*2*2*2*2 =4*4*4*4*16 **************************** Use the table info to answer any questions. P(evens = 0) = 1/16 P(evens > 0) = (4+6+4+1)/16 etc. rraggio wrote: a) ¼ b) ½ c) ¾ d) 15/16 e) 1/16 ANS: D



Intern
Joined: 19 Dec 2017
Posts: 7

Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
Show Tags
01 Feb 2018, 00:16
Since repetition of digits is allowed, total number of ways 8 digits can be used = 2^8 Odd digits  3,5,7,9 Even digits  0,2,6,8 Ways in which all odd digits are used  2^4 Probability of all odd digits being used = (2^4)/(2^8)=1/16 Probability of at least one even digit = 11/16=15/16




Re: A Four digit safe code does not contain the digits 1 and 4..
[#permalink]
01 Feb 2018, 00:16






