If Jack has two 2$ coins and two 1$ coins in his pocket, then what is the probability that he will take out exactly 5$'s out of his pocket?
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
Probability
Source: Beat The GMAT — Problem Solving |
The only situation when he gets exactly $5 is when he gets two $2 coins and one $1 coincoolhabhi wrote:If Jack has two 2$ coins and two 1$ coins in his pocket, then what is the probability that he will take out exactly 5$'s out of his pocket?
And there's 15 possible combinations since we don't know how many coins he takes out.
And there's only two possible combinations where we get two $2 coins and one $1 coin, since
he only has 2 $1 coins.
So the probabily is 2/15
Is the answer 1/8
Number of ways in which 1 or more coin can be taken out = (2 + 1) * (2 + 1) - 1
(Because the 2 2$ and the 2 1$ coins are indistinguishable)
favorable = 2 2$ + 1 1$ = 1 case
(Since we have taken the coins to be indistinguishable, taking out 1 $ is 1 case)
probability = 1/8
Had the items been distinguishable, I would have gone by
total = 4C1 + 4C2 + 4C3 + 4C4
favorable = 2C2 * 2C1
probability = 2/15
Number of ways in which 1 or more coin can be taken out = (2 + 1) * (2 + 1) - 1
(Because the 2 2$ and the 2 1$ coins are indistinguishable)
favorable = 2 2$ + 1 1$ = 1 case
(Since we have taken the coins to be indistinguishable, taking out 1 $ is 1 case)
probability = 1/8
Had the items been distinguishable, I would have gone by
total = 4C1 + 4C2 + 4C3 + 4C4
favorable = 2C2 * 2C1
probability = 2/15
What's the source of this problem? It doesn't sound very GMAT-like. The general GMAT rule is that you should NEVER ASSUME ANYTHING to get to the right answer. Here, you would have to assume that it is unknown how many coins Jack takes out of his pocket. Perhaps Jack never even touches this pocket and thus the probability is zero. See what I mean?coolhabhi wrote:If Jack has two 2$ coins and two 1$ coins in his pocket, then what is the probability that he will take out exactly 5$'s out of his pocket?
I dont really remember the source but it isnt "Jack never even touches his pocket". The question is what is the probability when Jack takes out 3 coins from his pocket and their sum is 5$?jaymw wrote:What's the source of this problem? It doesn't sound very GMAT-like. The general GMAT rule is that you should NEVER ASSUME ANYTHING to get to the right answer. Here, you would have to assume that it is unknown how many coins Jack takes out of his pocket. Perhaps Jack never even touches this pocket and thus the probability is zero. See what I mean?coolhabhi wrote:If Jack has two 2$ coins and two 1$ coins in his pocket, then what is the probability that he will take out exactly 5$'s out of his pocket?
The questions does not say anything about 3 coins, nor does it provide 5 answer choices. If I were you, I would safely ignore problems like this when preparing for the GMAT. There is enough "good" material out there.coolhabhi wrote:I dont really remember the source but it isnt "Jack never even touches his pocket". The question is what is the probability when Jack takes out 3 coins from his pocket and their sum is 5$?jaymw wrote:What's the source of this problem? It doesn't sound very GMAT-like. The general GMAT rule is that you should NEVER ASSUME ANYTHING to get to the right answer. Here, you would have to assume that it is unknown how many coins Jack takes out of his pocket. Perhaps Jack never even touches this pocket and thus the probability is zero. See what I mean?coolhabhi wrote:If Jack has two 2$ coins and two 1$ coins in his pocket, then what is the probability that he will take out exactly 5$'s out of his pocket?
I agree with jaymw.jaymw wrote:The questions does not say anything about 3 coins, nor does it provide 5 answer choices. If I were you, I would safely ignore problems like this when preparing for the GMAT. There is enough "good" material out there.coolhabhi wrote:I dont really remember the source but it isnt "Jack never even touches his pocket". The question is what is the probability when Jack takes out 3 coins from his pocket and their sum is 5$?jaymw wrote:What's the source of this problem? It doesn't sound very GMAT-like. The general GMAT rule is that you should NEVER ASSUME ANYTHING to get to the right answer. Here, you would have to assume that it is unknown how many coins Jack takes out of his pocket. Perhaps Jack never even touches this pocket and thus the probability is zero. See what I mean?coolhabhi wrote:If Jack has two 2$ coins and two 1$ coins in his pocket, then what is the probability that he will take out exactly 5$'s out of his pocket?
This question has far too much ambiguity to be a real GMAT question.
If anyone is interested, I rewrote the question to be more GMAT-like, and I posted it here: https://www.beatthegmat.com/probability- ... 18256.html
Cheers,
Brent













