Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
Updated on: 10 Jul 2019, 23:13
Originally posted by rohan2345 on 19 May 2017, 02:38.
Last edited by Sajjad1994 on 10 Jul 2019, 23:13, edited 1 time in total.
Last edited by Sajjad1994 on 10 Jul 2019, 23:13, edited 1 time in total.
Added Source
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
76% (01:51) correct
24%
(02:04)
wrong
based on 300
sessions
History
Date
Time
Result
Not Attempted Yet
Sarah cannot completely remember her four-digit ATM pin number. She does remember the first two digits, and she knows that each of the last two digits is greater than 5. The ATM will allow her three tries before it blocks further access. If she randomly guesses the last two digits, what is the probability that she will get access to her account?
(A) 1/2
(B) 1/4
(C) 3/16
(D) 3/18
(E) 1/32
(A) 1/2
(B) 1/4
(C) 3/16
(D) 3/18
(E) 1/32
Source: Nova GMAT
Difficulty Level: 600
Difficulty Level: 600
19 May 2017, 02:59
Kudos
Bookmarks
First two digits are already known. For each of the last two digits, we have 4 options: 6, 7, 8 or 9.
So the last two digits can take a total of 4 * 4 = 16 values.
For a total of 16 possibilities, Sarah has three tries. So three things could happen:
1. She guesses on the first try, for which the probability is 1/16
2. She doesn't guess on first but second try. So in first try she entered a wrong code and in second try she entered a correct code.
probability of the same = 15/16 * 1/15 = 1/16.
3. First two tries wrong code, third try correct code. Probability = 15/16 * 14/15 * 1/14 = 1/16.
So total probability = 1/16 + 1/16 + 1/16 = 3/16.
Hence answer is C.
So the last two digits can take a total of 4 * 4 = 16 values.
For a total of 16 possibilities, Sarah has three tries. So three things could happen:
1. She guesses on the first try, for which the probability is 1/16
2. She doesn't guess on first but second try. So in first try she entered a wrong code and in second try she entered a correct code.
probability of the same = 15/16 * 1/15 = 1/16.
3. First two tries wrong code, third try correct code. Probability = 15/16 * 14/15 * 1/14 = 1/16.
So total probability = 1/16 + 1/16 + 1/16 = 3/16.
Hence answer is C.
07 Aug 2019, 22:29
Kudos
Bookmarks
Sahilkm
There are 4*4 = 16 distinct ways of getting the last two digits. e.g. 68 or 97 or 79 etc. Only 1 of these 16 is correct.
On the first try, the probability of failing is 15/16 (say you punched in 68 and that was wrong)
Now you have 15 combinations remaining so logically, you will punch in one of the other 15. Probability of failing = 14/15. Say you punched in 97 and that was wrong.
Now you have sidelined 2 of the 16 cases and will punch in one of the remaining 14. Probability of failing = 13/14
Probability fo failing on all three tries = (15/16)*(14/15)*(13/14) = 13/16
Probability of succeeding = 1 - 13/16 = 3/16
