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.

probability!!!!!!!!!!

Expert replies
by siddarthd2919 » Sat Mar 08, 2008 9:16 am
hi can some one help me with this one

a bag contains 3 red and 7 blue marbles.......2 marbles are drawn at random wat is the probability that ATLEAST 1 is blue?

a. 21/50
b. 3/13
c. 1/6
d. 1/12
e. 1/3

prob:2

a certain deck of cards has 2 blue,2 red,2 green and 2 yellow cards. if 2 cards are drawn at random wat is the probability that both are not blue?

a. 15/28
b. 1/4
c. 9/16
d. 1/32
e. 1/16



i dont have the qa........can some one post a brief explanation of these problems.......
:D
Join the discussion
Source: — Problem Solving |

Re: probability!!!!!!!!!!

by Stuart@KaplanGMAT » Sat Mar 08, 2008 3:21 pm
siddarthd2919 wrote:hi can some one help me with this one

a bag contains 3 red and 7 blue marbles.......2 marbles are drawn at random wat is the probability that ATLEAST 1 is blue?

a. 21/50
b. 3/13
c. 1/6
d. 1/12
e. 1/3
There are two ways we can use math to solve this: the fast way and the slow way!

The slow way is to recognize that there are 3 scenarios that match the "at least one blue" requirement:

R B
B R
B B

So, one way we could solve the problem is to calculate the probability of each scenario occoring and then, since we want scenario 1 OR scenario 2 OR scenario 3, to add those probabilities together.

Accordingly:

R B = (3/10)(7/9) = 21/90
B R = (7/10)(3/9) = 21/90
B B = (7/10)(6/9) = 42/90

So, the probability of getting AT LEAST 1 blue is (21+21+42)/90 = 84/90 = 14/15

(So either you've stated the question wrong or the choices are messed up. In fact, just by estimating we should be able to conclude that if the probabiliy that 1 marble is blue is > 50%, the probability that at least one out of two being blue should be even higher, so none of those choices could possibly be correct.)

The quicker method to solve this type of question is to know that the sum of all possible events is 1 and to recognize that if we want AT LEAST 1 blue, the only event we don't want is:

R R

So, we instead of adding up the 3 events we do want to happen, we can instead use the formula:

Prob (things we want) = 1 - Prob (things we don't want)

or

Prob (at least 1 blue) = 1 - Prob (RR)
Prob (at least 1 blue) = 1 - (3/10)(2/9) = 1-6/90 = 84/90 = 14/15


a certain deck of cards has 2 blue,2 red,2 green and 2 yellow cards. if 2 cards are drawn at random wat is the probability that both are not blue?

a. 15/28
b. 1/4
c. 9/16
d. 1/32
e. 1/16
The first problem with this question is ambiguity. "What is the probability that both are not blue" could be interpreted as "neither is blue" or as "at least one isn't blue". Technically, by placing "both" where it is in the sentence, the question is asking for the probability that NEITHER is blue. The latter interpretation would be more accurate if the question had asked "What is the probability that not both are blue?"

Since the "neither" interpretation is the correct one, let's solve it that way and see what happens.

As we saw above, we can attack this two ways - we can either add up all the things we do want (which would take insanely long in this question) or use the 1 - (what we don't want) method.

The only thing we don't want is BB, so:

Probability (not BB) = 1 - Prob (BB) = 1 - (2/8)(1/7) = 1 - 2/56 = 54/56 = 27/28

Which again isn't one of the choices. The fact that we don't have a match leads me to a few subsidiary conclusions:

(1) the writer may have intended the second possible interpretation;
(2) the writer may have intended the draw to be "with replacement", which would change the results (by not specifying with or without replacement, we should assume that the draws are taking place simultaneously, i.e. you pull out 2 cards at once, so there's no replacement); and/or
(3) the answer choices provided are wrong

and, coupled with the first question (which I assume comes from the same place), one big conclusion:

(1) the source of these questions is NOT to be trusted.

What is the source? It's always a good idea to post that information, so we can better evaluate the validity of the questions. These two are definitely GMAT-style questions, they just seem to be horribly miswritten (or possibly misquoted).
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Join the discussion

by siddarthd2919 » Sat Mar 08, 2008 5:12 pm
THANKS STUART ........ I FOUND THIS QUESTION IN THE YAHOO GROUPS....... I TOO HAD THE FEELING THAT THE QUESTIONS WHERE WRONG......BUT DIDNT HAVE THE CONFIDENCE, TO LEAVE THEM AS WRONG. :D
Join the discussion

The probability of not drawing a blue card on the first draw is

= (number of cards that are not blue) / (total number of cards)
= 6/8

The probability of not drawing a blue card on the second draw is
= (number of cards that are not blue) / (total number of cards)
= 5/7

= (6/8) * (5/7)
= 30/56
= 15/28
Join the discussion

Sorry some smileys got clicked...

by sssmoorthy » Sat Aug 16, 2008 6:54 am
The probability of not drawing a blue card on the first draw is

= (number of cards that are not blue) / (total number of cards)
= 6/8

The probability of not drawing a blue card on the second draw is
= (number of cards that are not blue) / (total number of cards)
= 5/7

= (6/8) * (5/7)
= 30/56
= 15/28 :?
Join the discussion

by ronsom » Tue Apr 27, 2010 12:11 am
I think getting 27/28 for neither of the cards being blue is a little ambiguous!
It is not possible for such an event to occur with a probability of 1 (27/28 is almost equal to 1)
.. so i think the 15/28 answer is more convincing..

however... now i'm confused as to which approach should be followed when similar questions are encountered!
Staurt followed an approach wherein he did:-
P(event does not occur) = 1 - P(event occurs)

others who concluded to 15/28 as the answer followed an approach wherein they calculated the Probability of an event not occurring in a different way..

which approach to follow when.. is my question

Thanks
Join the discussion