A Four digit safe code does not contain the digits 1 and 4..

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 = \(\frac{6^4 - 3^4}{6^4} = 1 - \frac{3^4}{6^4} = 1 - (1/2)^4 = 15/16\)

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

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.

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 =\(1-\frac{1}{16} = \frac{15}{16}\)

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

Permutation
(_)(_)(_)(_)

Note: A safe code can have 0 as the thousands digit; a 4-digit number cannot.


Bag of 8 choices w/ replacement.


"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.



ANS: D

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

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,

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

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 ?

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 = 1-1/16=15/16

Solution

Given

• A four-digit safe code without 1 and 4.

To Find

• Probability it has at least one even digit.

Approach and Working Out

• The digit presents can be,

o 0, 2, 3, 5, 6, 7, 8, 9
• Total number of codes possible = 8 × 8 × 8 × 8

o As there are 8 choices for each place.
• Number of codes with zero even digits, 4 × 4 × 4 × 4

o As there are 4 odd numbers available.
• Probability of getting no even digits,

= (4 × 4 × 4 × 4) / (8 × 8 × 8 × 8)
= 1/16
• The probability of getting at least one even digit = 1 – (1/16) = 15/16

Correct Answer: Option D

total digits which are part of this combination ; ( 0,2,3,5,6,7,8,9) ; viz 8
and target is to find P atleast 1 even digit
total odd ; (3,5,7,9) ; 4
P of getting all odd (4/8)^4
and atleast 1 even (1-odd)
1-(4/8)^4 ; 16-1/16 ; 15/16
OPTION D

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.