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# Bob cannot completely remember his four-digit ATM pin

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

Kudos [?]: 7 [0], given: 0

Bob cannot completely remember his four-digit ATM pin [#permalink]  03 Jul 2005, 09:48
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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
Joined: 18 Apr 2005
Posts: 549
Location: Canuckland
Followers: 1

Kudos [?]: 8 [0], given: 0

[#permalink]  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
Joined: 17 Apr 2005
Posts: 375
Location: India
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Re: PS - ATM [#permalink]  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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