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Posts: 16
Updated on: 22 Jul 2021, 23:12
Originally posted by sidchandan on 22 Jul 2021, 22:20.
Last edited by Bunuel on 22 Jul 2021, 23:12, edited 1 time in total.
Last edited by Bunuel on 22 Jul 2021, 23:12, edited 1 time in total.
Renamed the topic and edited the question.
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
41% (02:34) correct
59%
(02:51)
wrong
based on 116
sessions
History
Date
Time
Result
Not Attempted Yet
A briefcase lock is a combination of 4 digits where each digit vary from 0 to 9. If a person tries any combination randomly then what is the probability that he puts at least 2 digits right but is unable to unlock the briefcase?
A. 0.0544
B. 0.0562
C. 0.0522
D. 0.0564
E. None of these
A. 0.0544
B. 0.0562
C. 0.0522
D. 0.0564
E. None of these
ScottTargetTestPrep Bunuel - request your kind help in figuring out the approach to this problem.
My Approach: (Wrong)
_ _ _ _ --> there are 4 digits in the lock and there is no constraint on repetitions. So digits can be repeated.
Thus the total number of possibilities are 10 X 10 X 10 X10 = 10000
Now, as at least 2 digits have to be right; there can be 2 cases:
Case 1: 2 digits correct: So there is only 1 digit correct for the 1st and second place. There are 9 digits incorrect for the 3rd and 4th place.
(as there will be 1 digit that is also correct for the 3rd and 4th place)
1 1 9 9 Therefore, there are 1X1X9X9 = 81 possibilities
Case 2: 3 digits correct: So there is only 1 digit correct for the 1st place. There are 9 digits incorrect for the 2nd, 3rd and 4th place.
(as there will be 1 digit that is also correct for the 2nd, 3rd and 4th place)
1 1 1 9 Therefore, there are 1X1X1X9 = 9 possibilities
Therefore total possibilities are 81 +9 = 90 /10000 which does not lead to any of the above answers.
Request your help in bridging the gap in my approach or understanding.
My Approach: (Wrong)
_ _ _ _ --> there are 4 digits in the lock and there is no constraint on repetitions. So digits can be repeated.
Thus the total number of possibilities are 10 X 10 X 10 X10 = 10000
Now, as at least 2 digits have to be right; there can be 2 cases:
Case 1: 2 digits correct: So there is only 1 digit correct for the 1st and second place. There are 9 digits incorrect for the 3rd and 4th place.
(as there will be 1 digit that is also correct for the 3rd and 4th place)
1 1 9 9 Therefore, there are 1X1X9X9 = 81 possibilities
Case 2: 3 digits correct: So there is only 1 digit correct for the 1st place. There are 9 digits incorrect for the 2nd, 3rd and 4th place.
(as there will be 1 digit that is also correct for the 2nd, 3rd and 4th place)
1 1 1 9 Therefore, there are 1X1X1X9 = 9 possibilities
Therefore total possibilities are 81 +9 = 90 /10000 which does not lead to any of the above answers.
Request your help in bridging the gap in my approach or understanding.
rocky620
Retired Moderator
Joined: 10 Nov 2018
Last visit: 11 May 2023
Posts: 476
Location: India
Concentration: General Management, Strategy
Schools: IIMA PGPX'22 IIMB EPGP'22 IIMC PGPEX'22 ISB'22
GMAT 1: 590 Q49 V22
WE:Other (Retail: E-commerce)
Kindly check whether it helps
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.e. 4C2 = 6
For the remaining two digits we have 9 digits to choose from. As the 10th digit will definitely be correct.
So 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.e. 4C3 = 4
For the last digit we have 9 digits to choose from, because the 10th digit will definitely be correct.
So total combinations = 4*9 = 36
Total Favorable cases = 486 + 36 = 522
Total Cases = 10000
Probability = 0.0522, Option C
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.e. 4C2 = 6
For the remaining two digits we have 9 digits to choose from. As the 10th digit will definitely be correct.
So 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.e. 4C3 = 4
For the last digit we have 9 digits to choose from, because the 10th digit will definitely be correct.
So total combinations = 4*9 = 36
Total Favorable cases = 486 + 36 = 522
Total Cases = 10000
Probability = 0.0522, Option C
Posts: 12
Total number of ways in which the 4 digits can be put= 10*10*10*10= (10)^4
There are two cases-
(i) 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= (6*1*1*9*9)/[(10)^4] = 0.0486
(ii) 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 for correct: 4C3 or 4C1= 4
Therefore probability= (4*1*1*1*9)/[(10)^4] = 0.0036
Taking the sum of the above two independent probabilities= 0.0486 + 0.0036= 0.0522
THERE IS ONE MORE WAY TOO
We can deduct the probabilities of 1 right 3 wrong, 4 right and 4 wrong from 1:
(i) 1R3W= 1/10* (9/10)^3 *4
(ii) 4R= 1/10000
(iii) 4W= (9/10)^4
adding the above 3 cases= 9478/10000
hence 1-9478/10000=522/10000=0.0522
There are two cases-
(i) 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= (6*1*1*9*9)/[(10)^4] = 0.0486
(ii) 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 for correct: 4C3 or 4C1= 4
Therefore probability= (4*1*1*1*9)/[(10)^4] = 0.0036
Taking the sum of the above two independent probabilities= 0.0486 + 0.0036= 0.0522
THERE IS ONE MORE WAY TOO
We can deduct the probabilities of 1 right 3 wrong, 4 right and 4 wrong from 1:
(i) 1R3W= 1/10* (9/10)^3 *4
(ii) 4R= 1/10000
(iii) 4W= (9/10)^4
adding the above 3 cases= 9478/10000
hence 1-9478/10000=522/10000=0.0522
The question is not very clear:
1. Puts 2 digit correctly means only digits are correct or the digit and position both are correct?
2. What if the correct code has numbers getting repeated?
1. Puts 2 digit correctly means only digits are correct or the digit and position both are correct?
2. What if the correct code has numbers getting repeated?
3 digits correct means:
RRRW in any order, equals:
4!/3! = 4*(1/10)^3*(9/10) =
36/10000
2 digits correct means:
RRWW in any order, or:
4!/2! = 6*(1/10)^2*(9/10)^2 =
81*6/10000 = 486/10000
Total probability =
(486+36)/10000= 522/10000=
.0522
RRRW in any order, equals:
4!/3! = 4*(1/10)^3*(9/10) =
36/10000
2 digits correct means:
RRWW in any order, or:
4!/2! = 6*(1/10)^2*(9/10)^2 =
81*6/10000 = 486/10000
Total probability =
(486+36)/10000= 522/10000=
.0522
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