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
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
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 Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget 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

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

Please help with this probability

Expert replies
by Night reader » Fri Dec 10, 2010 5:01 am
The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least two of the tosses the coin will turn up tails?
Join the discussion
Source: — Problem Solving |

by diebeatsthegmat » Fri Dec 10, 2010 5:07 am
Night reader wrote:The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least two of the tosses the coin will turn up tails?
the probability that the coin is toosed 3 times =(1/2)^3=1/8
there are 4 combination of tossing time that at least of the tosses the coin will turn up tails.
TTH
THT
HTT
TTT ( becase its "at least" probability, we will have to consider that all time its tossed is tail)
so the final probability is 4*1/8=1/2
is 1/2 the right answer?
Join the discussion

by Rahul@gurome » Fri Dec 10, 2010 5:11 am
Night reader wrote:The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times, what is the probability that on at least two of the tosses the coin will turn up tails?
Probability of getting a head in a toss = 1/2
Thus, probability of getting a tail in a toss = (1 - 1/2) = 1/2

A coins is tossed three times,
  • Probability of getting at least two tails = Probability of getting exactly two tails + Probability of getting exactly three tails
Possible exactly two tails: (TTH), (THT) and (HTT)
Each have a probability = (1/2)*(1/2)*(1/2) = 1/8
Probability of getting exactly two tails= 3*(1/8) = 3/8

Possible exactly three tails: (TTT)
Probability of getting exactly three tails = (1/2)*(1/2)*(1/2) = 1/8

Therefore, Probability of getting at least two tails = (3/8) + (1/8) = 4/8 = 1/2
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Join the discussion

by Night reader » Fri Dec 10, 2010 5:40 am
Rahul, thanks for the solution. Below I have my questions related to different interpretation of the probabilities listed, and I am including the official answer and solution which I don't understand at all, as it operates with the method I am unaware
Possible exactly two tails: (TTH), (THT) and (HTT)
Each have a probability = (1/2)*(1/2)*(1/2) = 1/8
Can I say, P(not Tail-One Head)=1/2=P(two heads) above => 1/2 means P(One Head) which is enough for us => then 1/8 is equivalent to P(TTH) & P(THT) & P(HTT)

we cancel this
Probability of getting exactly two tails= 3*(1/8) = 3/8
Also, below P(Head)=0 and we must be consistent with the probability tree here, if we start with P(Head) we finish with P(Head)
Possible exactly three tails: (TTT)
Probability of getting exactly three tails = (1/2)*(1/2)*(1/2) = 1/8
Therefore, Probability of getting at least two tails = (3/8) + (1/8) = 4/8 = 1/2
summing up two probabilities P(TTH) & P(THT) & P(HTT) + P(TTT) we get 1/2 + 0 = 1/2


And below is the official explanation of which I don't understand the red color selected part, please shed some light on this
Explanation: The words "at least" signal that you'll be working with more than one probability, then adding them together later. In this case, those multiple probabilities are the probability that two of the three tosses turn up tails, and the other is the probability is that all three of the tosses turn up tails. The only ways the coin tosses can be arranged to make one of those things possible:
H T T
T H T
T T H
T T T
There are four possible outcomes, in the Â…rest three of which two tosses turn up tails, and in the last of which all three turn up tails. The total number of possibilities is the product of the number of possibilities for each toss: (2)(2)(2) = 8
Probability is, as always, the number of desired outcomes divided by possible outcomes: 4/8 = 1/2
Join the discussion

by Rahul@gurome » Fri Dec 10, 2010 12:12 pm
Night reader wrote:And below is the official explanation of which I don't understand the red color selected part, please shed some light on this
Explanation: The words "at least" signal that you'll be working with more than one probability, then adding them together later. In this case, those multiple probabilities are the probability that two of the three tosses turn up tails, and the other is the probability is that all three of the tosses turn up tails. The only ways the coin tosses can be arranged to make one of those things possible:
H T T
T H T
T T H
T T T
There are four possible outcomes, in the Â…rest three of which two tosses turn up tails, and in the last of which all three turn up tails. The total number of possibilities is the product of the number of possibilities for each toss: (2)(2)(2) = 8
Probability is, as always, the number of desired outcomes divided by possible outcomes: 4/8 = 1/2
This is a more easy and intuitive solution.
What it does is: lists all the favorable outcomes and calculates the probability by basic definition.

Remember that probability of an event = (Number of favorable outcome)/(Total number of outcomes)

In this number of favorable outcome is = 4
Namely (TTH), (THT), (HTT) and (TTT)

Total number of outcomes = 8
(The red line is a easy way to determine the total number of outcomes using basic combinatorics. Note that we have three tosses. In each tosses the there is two possible outcomes. Then total number of possible outcome = 2*2*2 = 8)

Thus probability = 4/8 = 1/2

Hope it helps.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Join the discussion

by Night reader » Fri Dec 10, 2010 9:30 pm
Thank you Rahul. I find your method as more expletive and concept backed. I will keep it for notes to tackle other probs.
Join the discussion