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A briefcase lock is a combination of 4 digits where each digit vary [#permalink]
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
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Re: A briefcase lock is a combination of 4 digits where each digit vary [#permalink]
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Re: A briefcase lock is a combination of 4 digits where each digit vary [#permalink]
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