A Briefcase Lock is A Combination of 4 Digits Where each Digit Varies GMAT Problem Solving

Question: A briefcase lock is a combination of 4 digits where each digit varies from 0 to 9. If a person tries any combination randomly then what is the probability that he puts atleast 2 digits right but is unable to unlock the briefcase?

1. 0.0544
2. 0.0562
3. 0.0522
4. 0.0564
5. None of these

Answer:
Approach Solution (1):

Case 1: 2 digits correct:
----: Is the code. We want 2 digits at the correct place, but there are 4 places available for these 2 digits. So, we have to select two places, i.ie. 4C2 = 6
For the remaining two digits we have 9 digits to choose form. As the 10th digit will definitely be correct.
So the total combinations = 6 * 9 * 9 = 486

Case 2: 3 digits correct:
----: Is the code. We want 3 digits at the correct place, but there are 4 places available for these 3 digits. So, we have to select 3 places, i.ie. 4C3 = 4
For the remaining two digits we have 9 digits to choose form. As the 10th digit will definitely be correct.
So the total combinations = 4 * 9 = 36
Total favorable cases = 486 + 36 = 522
Total cases = 10,000
Probability = 0.0522

Correct option: C

Approach Solution (2):

Total number of ways in which the 4 digits can be put = 10 * 10 * 10 * 10 =\(10^4\)
There are two cases:
(1) The one in which 2 are correct and 2 are incorrect:
2 correct: 1 * 1
2 incorrect: 9 * 9
Selecting 2 places out of 4: 4C2 = 6
Therefore probability =\(\frac{6*1*1*9*9}{10^4}=0.0486\)

(2) The one in which 3 are correct and 1 is incorrect:
3 correct: 1 * 1 * 1
1 incorrect: 9
Selecting 3 places out of 4 for incorrect or 1 place out of 4 correct: 4C3 or 4C1 = 4
Therefore probability = \(\frac{4*1*1*1*9}{10^4}=0.0036\)
Taking sum of the above two independent probabilities = 0.0486 + 0.0036 = 0.0522

Correct option: C

Approach Solution (3):
We can deduct the probabilities of 1 right 3 wrong, 4 right and 4 wrong from 1:
(1) 1R3W =\(\frac{1}{10}*(\frac{9}{10})^3*4\)
(2) 4R =\(\frac{1}{10000}\)
(3) 4W =\((\frac{9}{10})^4\)
Adding the above 3 cases:\(\frac{9478}{10000}\)
Hence 1 -\(\frac{9478}{10000}\)=\(\frac{522}{10000}=0.0522\)

Correct option: C

“A briefcase lock is a combination of 4 digits where each digit varies from 0 to 9. If a person tries any combination randomly then what is the probability that he puts atleast 2 digits right but is unable to unlock the briefcase?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

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