BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
Vote for Target Test Prep, Newsweek Readers’ Choice Awards 2026
NEWSWEEK READERS’ CHOICE 2026

BIG NEWS! Target Test Prep has been nominated, and they’d love your vote!

TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

Vote for TTP
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Probability Problems (Confused!)

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 48
Joined: Sat Apr 03, 2010 8:35 pmFollowed by: 1 member
Thanked: 1 times
by thp510 » Sun Oct 17, 2010 9:44 am
I'm having problems with "Probability Strategy". Can someone help explain why the answers are what they are?

Question 1:
"If a fair coin is tossed three times, what is the probability that it will turn up heads exactly twice?"
Book Answer: Draw it out. = 3/8.
How did they get this number/answer? I originally thought (1/2)(1/2)(1/2)= 1/8 chances. However, this is wrong because it's only asking for EXACTLY twice. So what's the math equation instead of just drawing it all out and spending too much time?

Question 2:
"What is the probability that, on three rolls of a single fair die, at least one of the rolls will be a six?"
Book Answer: Use the 1-x method (probability of success + probability of failure = 1). The probability of failure or the chance that the dice WON'T roll a six is 5/6. So we're rolling three times, therefore: (5/6)(5/6)(5/6) = 125/216. Now subtract that from 1 and you have 1-(125/126) = 91/216
Question: If this is the case, why can't I just go with multiplying three times the chance that the roll WILL BE a six, so it's (1/6)(1/6)(1/6) = 1/216. This isn't correct. Why?

Question 3:
"Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour. Jack picks two candies simultaneously and at random. What is the chance that exactly 1 of the candies he has picked is sour?"
Book Answer: =6/30. Draw out a probability tree. I understood how they got it this way, but this takes WAY TOO MUCH time.
Question: What's the computation look like? I thought I would take 4 over 6 (sweet candies) and multiply it to 1 over 2 (sour candy) to equal 4 over 12 or (1/3)


Thanks for all the help! I'm glad I can tap into this awesome resource.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Sun Oct 17, 2010 10:10 am
thp510 wrote: Question 2:
"What is the probability that, on three rolls of a single fair die, at least one of the rolls will be a six?"
Book Answer: Use the 1-x method (probability of success + probability of failure = 1). The probability of failure or the chance that the dice WON'T roll a six is 5/6. So we're rolling three times, therefore: (5/6)(5/6)(5/6) = 125/216. Now subtract that from 1 and you have 1-(125/126) = 91/216
Question: If this is the case, why can't I just go with multiplying three times the chance that the roll WILL BE a six, so it's (1/6)(1/6)(1/6) = 1/216. This isn't correct. Why?
If you calculate 1/216 that implies all the three rolls MUST have 6 as a result. Or we can say u will get a definite (6,6,6).
But here question asks at least one of them will be 6. So the cases are
(x,x,6), (x,6,x), (6,x,x) and as it is mentioned AT LEAST this x can be anything from 1,2,3,4,5,6.

If u wanna try your way then it will be very much clumsy. Still let me show you how it works then its upto you if u wanna follow 1-x or not ;)

Take (x,x,6) possibility:
(two failures AND one success) OR (one failure AND two success ) OR (Three Success) [Failure not 6, Success 6]
= 5/6 * 5/6 * 1/6 + 5/6 * 1/6 * 1/6 + 1/6 * 1/6 * 1/6 = 31/216 ...... (1)

Take (x,6,x) possibility: (Here the [6,6,6] possibilty we will discard as it has been taken care in (1))
(failure AND success AND failure) OR (failure AND success AND success) [Discard Three scucesses here as it has been taken care of]
= 5/6 * 1/6 * 5/6 + 5/6 * 1/6 * 1/6 = 30/216 .... (2)

Take (6,x,x) possibilty again discarding (6,6,6) = 30/216 ..... (3)

So answer will be 31/216 + 30/216 + 30/216 = 91/216

So, when asked AT LEAST then calculate the number of FAILURES and SUBTRACT it from 1 .

Hope this helps.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Sun Oct 17, 2010 10:28 am
thp510 wrote:
Question 1:
"If a fair coin is tossed three times, what is the probability that it will turn up heads exactly twice?"
Book Answer: Draw it out. = 3/8.
How did they get this number/answer? I originally thought (1/2)(1/2)(1/2)= 1/8 chances. However, this is wrong because it's only asking for EXACTLY twice. So what's the math equation instead of just drawing it all out and spending too much time?
I would always prefer the Drawing method as it gives you a crystal clear idea of what you are dealing with. As coins have only two possibilities and if it is 3 or 4 coins there is no harm in following so.

However, if you want to solve it in equation way, still you need a skeletal view of the same.

Question is asking exact two Heads.
It can be (H,H,T) or (H,T,H) or (T,H,H)
= [P(H)*P(H)*P(T)] + [P(H) * P(T) * P(H)] + [P(T) * P(H) * P(H)]
= [1/2 * 1/2 * 1/2] + [1/2 * 1/2 * 1/2] + [1/2 * 1/2 * 1/2]
= 1/8 + 1/8 + 1/8
= 3/8
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Sun Oct 17, 2010 10:39 am
thp510 wrote: Question 3:
"Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour. Jack picks two candies simultaneously and at random. What is the chance that exactly 1 of the candies he has picked is sour?"
Book Answer: =6/30. Draw out a probability tree. I understood how they got it this way, but this takes WAY TOO MUCH time.
Question: What's the computation look like? I thought I would take 4 over 6 (sweet candies) and multiply it to 1 over 2 (sour candy) to equal 4 over 12 or (1/3)
Exactly 1 has to be Sour from the pick.

So the possibilities are (Sweet,Sour) or (Sour,Sweet) [These two cases Jack will have EXACTLY 1 sour candy and also This sequence of "Sweet and Sour" or "Sour and Sweet" matters as for the first pick total possibility is 6 and for the second the total possibilty is 5 as from the first pick one candy has been in Jack's Hand]
= [P(Sw)*P(So)] + [P(So)*P(Sw)]
= (4/6 * 2/5) + (2/6 * 4/5)
= 16/30

Please clarify the solution as you have mentioned it as 6/30.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 48
Joined: Sat Apr 03, 2010 8:35 pmFollowed by: 1 member
Thanked: 1 times

by thp510 » Sun Oct 17, 2010 12:58 pm
You're awesome! Thanks for the replies.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Sun Oct 17, 2010 1:03 pm
thp510 wrote:You're awesome! Thanks for the replies.
My pleasure :)
If the problem is Easy Respect it, if the problem is tough Attack it
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Sun Oct 17, 2010 3:28 pm
Question 1:
"If a fair coin is tossed three times, what is the probability that it will turn up heads exactly twice?"

P(HHT) = 1/2 * 1/2 * 1/2 = 1/8.
Since T can occur on the 1st, 2nd or 3rd flip, we multiply by 3:
3*(1/8) = 3/8.

Question 3:
"Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour. Jack picks two candies simultaneously and at random. What is the chance that exactly 1 of the candies he has picked is sour?"

P(sweet,sour) = (4/6)*(2/5) = 4/15.
Since (sour,sweet) also is a good outcome, we multiply by 2:
2*(4/15) = 8/15.

Hope this helps!
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 48
Joined: Sat Apr 03, 2010 8:35 pmFollowed by: 1 member
Thanked: 1 times

by thp510 » Sun Oct 17, 2010 8:45 pm
GMATGuruNY wrote:Question 1:
"If a fair coin is tossed three times, what is the probability that it will turn up heads exactly twice?"

P(HHT) = 1/2 * 1/2 * 1/2 = 1/8.
Since T can occur on the 1st, 2nd or 3rd flip, we multiply by 3:
3*(1/8) = 3/8.

Question 3:
"Renee has a bag of 6 candies, 4 of which are sweet and 2 of which are sour. Jack picks two candies simultaneously and at random. What is the chance that exactly 1 of the candies he has picked is sour?"

P(sweet,sour) = (4/6)*(2/5) = 4/15.
Since (sour,sweet) also is a good outcome, we multiply by 2:
2*(4/15) = 8/15.

Hope this helps!

Mitch thanks! Your approach is very short and fast. Is this based on some universal equation that wasn't hit upon or just your ability to know math like the back of your hand? I'm amazed.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1