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Bob cannot completely remember his four-digit ATM pin

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Senior Manager
Senior Manager
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Joined: 30 May 2005
Posts: 374
Followers: 1

Kudos [?]: 6 [0], given: 0

Bob cannot completely remember his four-digit ATM pin [#permalink] New post 03 Jul 2005, 09:48
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A
B
C
D
E

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(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Bob cannot completely remember his four-digit 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
Director
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Joined: 18 Apr 2005
Posts: 551
Location: Canuckland
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 [#permalink] New post 03 Jul 2005, 13:37
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
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Joined: 17 Apr 2005
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Location: India
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Re: PS - ATM [#permalink] New post 04 Jul 2005, 07:01
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.
Re: PS - ATM   [#permalink] 04 Jul 2005, 07:01
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Bob cannot completely remember his four-digit ATM pin

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