Problem Solving

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?


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.

Why am I wrong?
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 I have got the right answer.

dddanny2006 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

P(at least 1 digit is even) = 1 - P(all 4 digits are odd).

Digit options: 0, 2, 3, 5, 6, 7, 9

P(all 4 digits are odd):
P(1st digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
P(2nd digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
P(3rd digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
P(4th digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
Since selecting 4 odd digits requires that the 1st digit be odd AND the 2nd digit be odd AND the 3rd digit be odd AND the 4th digit be odd, we MULTIPLY the fractions:
4/8 * 4/8 * 4/8 * 4/8 = 1/16.

Thus:
P(at least 1 digit is even) = 1 - 1/16 = 15/16.

The correct answer is D.

Can you please help me out with the Combinatorics method here.I want to know where I went wrong,as it can help me conceptually too.Please Mitch.

GMATGuruNY wrote:

dddanny2006 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

P(at least 1 digit is even) = 1 - P(all 4 digits are odd).

Digit options: 0, 2, 3, 5, 6, 7, 9

P(all 4 digits are odd):
P(1st digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
P(2nd digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
P(3rd digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
P(4th digit is odd) = 4/8. (Of the 8 digits above, 4 are odd.)
Since selecting 4 odd digits requires that the 1st digit be odd AND the 2nd digit be odd AND the 3rd digit be odd AND the 4th digit be odd, we MULTIPLY the fractions:
4/8 * 4/8 * 4/8 * 4/8 = 1/16.

Thus:
P(at least 1 digit is even) = 1 - 1/16 = 15/16.

The correct answer is D.

dddanny2006 wrote:Can you please help me out with the Combinatorics method here.I want to know where I went wrong,as it can help me

Please see my corrections/notes in red:

dddanny2006 wrote:
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=4096

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
There are FOUR ways Case 2 could happen:
The 1st digit could be odd.
The 2nd digit could be odd.
The 3rd digit could be odd.
The 4th digit could be odd.
Thus, the result above must be multiplied by 4:
4*256.

3. 2 are even,2 are odd=4*4*4*4=256
There are SIX ways Case 3 could happen.
The following pairs of digits could be even:
1st and 2nd
1st and 3rd
1st and 4th
2nd and 3rd
2nd and 4th
3rd and 4th.
Thus, the result above must be multiplied by 6:
6*256.

4.1 is even,3 are odd=4*4*4*4=256
There are FOUR ways Case 4 could happen:
The 1st digit could be even.
The 2nd digit could be even.
The 3rd digit could be even.
The 4th digit could be even.
Thus, the result above must be multiplied by 4:
4*256.

Total good cases = 256 + (4*256) + (6*256) + (4*256) = 15*256.

P(good outcome) = (15*256)/4096 = 15/16.

Thanks Mitch.Lets consider case 2 of the problem where we have 3even and one odd.

I had it as a 4*4*4*4 ---what would this have given me? I was of the opinion that when order doest matter we divide it by the number of interchangeable slots.But here since order is important I left it alone thinking that it would give me all the possible arrangements,but No,we have to firther multiply it by 4.Im a little confused here.Since I treated it as a permutation I should have got all possible arrangements.

Also when deciding the number of possible arrangements ie.8*8*8*8,why dont we do the samething thats in red?Here too we dont know if the order matters because it could be that all the 4 numbers could be the same?Right??Why is that a permutation?

For example in the 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

Here too the number positions could get interchanged right.

We could have 4598 at the same time we may also have a 9485.How do we know if these two numbers are included in those 3584 combinations?Please clear my doubt Mitch,these problems are giving me a hard time.

GMATGuruNY wrote:

dddanny2006 wrote:Can you please help me out with the Combinatorics method here.I want to know where I went wrong,as it can help me

Please see my corrections/notes in red:

dddanny2006 wrote:
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=4096

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
There are FOUR ways Case 2 could happen:
The 1st digit could be odd.
The 2nd digit could be odd.
The 3rd digit could be odd.
The 4th digit could be odd.
Thus, the result above must be multiplied by 4:
4*256.

3. 2 are even,2 are odd=4*4*4*4=256
There are SIX ways Case 3 could happen.
The following pairs of digits could be even:
1st and 2nd
1st and 3rd
1st and 4th
2nd and 3rd
2nd and 4th
3rd and 4th.
Thus, the result above must be multiplied by 6:
6*256.

4.1 is even,3 are odd=4*4*4*4=256
There are FOUR ways Case 4 could happen:
The 1st digit could be even.
The 2nd digit could be even.
The 3rd digit could be even.
The 4th digit could be even.
Thus, the result above must be multiplied by 4:
4*256.

Total good cases = 256 + (4*256) + (6*256) + (4*256) = 15*256.

P(good outcome) = (15*256)/4096 = 15/16.