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Senior Manager
Joined: 30 May 2005
Posts: 373

Bob cannot completely remember his fourdigit ATM pin [#permalink]
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03 Jul 2005, 10:48
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Bob cannot completely remember his fourdigit ATM pin number. He does remember the first two digits, and knows that each of the last two digits is either a 6 or a 7. The ATM will allow him three tries before it blocks further access. If he randomly guesses the last two digits, what is the probability that he will get access to his account?
A. 3/8
B. 3/16
C. 5/16
D. 35/64
E. 37/64



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Joined: 18 Apr 2005
Posts: 545
Location: Canuckland

I don't think any of the choices list the correct answer given ATM customer situation at hand
possible combinations 6,6 6,7 7,6 7,7
prob of getting it right on the first attempt = 1/4
prob of getting it right on second = 3/4 * 1/3 (not getting it at first * getting it on second, assuming he is smart enough that he won't repeat the wrong combination he used on his first attempt)
prob of getting it right on third is 3/4 * 2/3 * 1/2
1/4 + 3/4 * 1/3 + 3/4 * 2/3 * 1/2 = 1  prob of getting it right on the fourth = 1  3/4 * 2/3 * 1/2 *1
= 3/4
if he is 'absent minded', putting it mildly, and cannot remember what he punched in before, than probability is
1/4 + 3/4 *1/4 + 3/4 * 3/4 * 1/4 = 37/64 or E



Senior Manager
Joined: 17 Apr 2005
Posts: 373
Location: India

Re: PS  ATM [#permalink]
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04 Jul 2005, 08:01
Probability of getting access = 1  ( probability of gettting the first three wrong).
Probability of getting one try wrong is 3/4
hence P = 1  (3/4)^3
= 1  27/64
= 37/64.
HMTG.










