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A Four digit safe code does not contain the digits 1 and 4.. [#permalink]
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21 Sep 2010, 06:34
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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
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Re: A Four digit safe code does not contain the digits 1 and 4.. [#permalink]
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21 Sep 2010, 06:35
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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.



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Re: A Four digit safe code does not contain the digits 1 and 4.. [#permalink]
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21 Sep 2010, 06:46
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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 = \(\frac{6^4  3^4}{6^4} = 1  \frac{3^4}{6^4} = 1  (1/2)^4 = 15/16\)
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Re: A Four digit safe code does not contain the digits 1 and 4.. [#permalink]
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21 Sep 2010, 06:58
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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!



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Re: A Four digit safe code does not contain the digits 1 and 4.. [#permalink]
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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
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Re: A Four digit safe code does not contain the digits 1 and 4.. [#permalink]
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21 Sep 2010, 07:43
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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.
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Re: A Four digit safe code does not contain the digits 1 and 4.. [#permalink]
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21 Sep 2010, 07:57
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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 =\(1\frac{1}{16} = \frac{15}{16}\)
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Re: A Four digit safe code does not contain the digits 1 and 4.. [#permalink]
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21 Sep 2010, 16:55
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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
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Re: A Four digit safe code does not contain the digits 1 and 4.. [#permalink]
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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



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Re: A Four digit safe code does not contain the digits 1 and 4.. [#permalink]
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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
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Re: A Four digit safe code does not contain the digits 1 and 4.. [#permalink]
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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 ?



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



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Re: A Four digit safe code does not contain the digits 1 and 4.. [#permalink]
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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




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