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 1612:00 PM EDT
-11:59 PM EDT
You’re first to know: Save $200 on GMAT prep starting tomorrow, 4/13. Get 6 official practice tests and study with real questions. Be ready to save!
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.
25 Nov 2021, 05:15
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
32% (02:29) correct
68%
(02:48)
wrong
based on 59
sessions
History
Date
Time
Result
Not Attempted Yet
Paul, Quinn, and Richard each have 3 coins: Gold, Silver, and Bronze. They all toss each of their coins once and record the results. Two results are the same if two gold coins land on the same face, two silver coins land on the same face, and two bronze coins land on the same face. What is the probability that Paul's result is the same as at least one of the other two results?
A. 15/32
B. 15/64
C. 31/64
D. 3/8
E. 1/2
A. 15/32
B. 15/64
C. 31/64
D. 3/8
E. 1/2
21 Dec 2021, 08:40
Kudos
Bookmarks
Regardless of what Paul gets, there will be a 1/2 chance Quinn's gold coin matches Paul's, a 1/2 chance Quinn's silver matches Paul's, and a 1/2 chance Quinn's bronze matches Paul's, so a (1/2)(1/2)(1/2) = 1/8 probability that all three of Quinn's coins match Paul's. Similarly there is a 1/8 probability all three of Richard's coins match Paul's. If you wanted to now solve directly, you might notice that the answer will be very close to 1/8 + 1/8, from which you can already pick the right answer (only B is plausible), but we're double-counting the case when both Quinn and Richard's coins match Paul's, which will happen (1/8)(1/8) = 1/64 of the time, and the answer is 1/4 - 1/64 = 15/64.
But in questions like this, where we need at least one thing to happen, it's usually easiest to reverse the problem (work out the probability we do not get the result the question asks about), since that will leave us with only one case, and we won't need to worry about the double-counting issue. The probability is 7/8 that Quinn's coins do not match Paul's, and 7/8 that Richard's do not match Paul's, and thus the probability is (7/8)(7/8) = 49/64 that neither of their coins match Paul's, and 1 - 49/64 = 15/64 that at least one does match.
But in questions like this, where we need at least one thing to happen, it's usually easiest to reverse the problem (work out the probability we do not get the result the question asks about), since that will leave us with only one case, and we won't need to worry about the double-counting issue. The probability is 7/8 that Quinn's coins do not match Paul's, and 7/8 that Richard's do not match Paul's, and thus the probability is (7/8)(7/8) = 49/64 that neither of their coins match Paul's, and 1 - 49/64 = 15/64 that at least one does match.
General Discussion
21 Dec 2021, 07:22
Kudos
Bookmarks
Hi,
the question is looking for the probability of:
Only Paul Richard Same (OPRS)+OPQS(Only Paul Quinn Same)+PQRS(...)=Result.
OPRS=PRS and PQ NOT same. We have 2 possibilities per coin toss, meaning for a 3 combination, there are 8 unique values. This means that Paul and Richard have the same in 8/64 cases. However, we now have to make sure that Paul and Quinn DON'T have the same. Now, for every of the 8 cases which were valid for PRS, we can combine 7 out of the 8 possible coin tosses of Quinn (because we can't take same results in!). This means that we are left with 56/512 possibilities.
OPQS= ... -> This case is just the same as above, we have the same result
PQRS: There are only 8 in 8^3 cases in which all tosses are the same. In total we get
2*(56/512)+(8/512)=120/512=60/256=...=15/64
Bunuel ?
the question is looking for the probability of:
Only Paul Richard Same (OPRS)+OPQS(Only Paul Quinn Same)+PQRS(...)=Result.
OPRS=PRS and PQ NOT same. We have 2 possibilities per coin toss, meaning for a 3 combination, there are 8 unique values. This means that Paul and Richard have the same in 8/64 cases. However, we now have to make sure that Paul and Quinn DON'T have the same. Now, for every of the 8 cases which were valid for PRS, we can combine 7 out of the 8 possible coin tosses of Quinn (because we can't take same results in!). This means that we are left with 56/512 possibilities.
OPQS= ... -> This case is just the same as above, we have the same result
PQRS: There are only 8 in 8^3 cases in which all tosses are the same. In total we get
2*(56/512)+(8/512)=120/512=60/256=...=15/64
Bunuel ?
