• NEW! FREE Beat The GMAT Quizzes
    NEW! FREE Beat The GMAT Quizzes
    NEW! FREE Beat The GMAT Quizzes
    Hundreds of Questions Highly Detailed Reporting Expert Explanations TAKE A FREE GMAT QUIZ
  • 7 CATs FREE!
    If you earn 100 Forum Points

    Engage in the Beat The GMAT forums to earn
    100 points for $49 worth of Veritas practice GMATs FREE

    Veritas Prep
    VERITAS PRACTICE GMAT EXAMS
    Earn 10 Points Per Post
    Earn 10 Points Per Thanks
    Earn 10 Points Per Upvote
    REDEEM NOW

beat this probability Qs

This topic has 7 expert replies and 5 member replies

beat this probability Qs

Post
A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
1/3
3/8
1/2
5/8
11/16

Source : Gmat Club Tests

_________________
Regards,
Sach

  • +1 Upvote Post
  • Quote
  • Flag
Junior | Next Rank: 30 Posts Default Avatar
Joined
09 Feb 2013
Posted:
26 messages
Upvotes:
4
Post
We have 2 possibilities:
1-Mary picks a Withe ball and then John picks a Blue ball
3/8 * 5/7 = 15/56
2-Mary picks a Blue ball and then John picks a Blue ball
5/8 * 4/7 = 20/56

so (15+20)/56= 35/56 ==> 5/8 is the answer!

  • +1 Upvote Post
  • Quote
  • Flag

GMAT/MBA Expert

Post
sachindia wrote:
A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
1/3
3/8
1/2
5/8
11/16
No math is needed here if we understand the following concept:
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
Thus, P(blue on the 2nd pick) = P(blue on the first pick) = 5/8.

The correct answer is D.

The answer would be the same even if the problem were as follows:

Quote:
A basket contains 3 white and 5 blue balls. Mary will extract TWO BALLS at random and keep them. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
P(blue on the 3rd pick) = P(blue on the 1st pick) = 5/8.

Other problems that test the same concept:

http://www.beatthegmat.com/probablity-ques-t60161.html
http://www.beatthegmat.com/manhattan-probability-t89481.html (2 posts)
http://www.beatthegmat.com/a-box-contains-some-blue-balls-and-9-red-balls-if-the-t151368.html

_________________
Mitch Hunt
Private Tutor for the GMAT and GRE
GMATGuruNY@gmail.com

If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.

Available for tutoring in NYC and long-distance.
For more information, please email me at GMATGuruNY@gmail.com.
Student Review #1
Student Review #2
Student Review #3

  • +1 Upvote Post
  • Quote
  • Flag
Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.
Newbie | Next Rank: 10 Posts Default Avatar
Joined
09 Jan 2013
Posted:
2 messages
Post
These are two mutually exclusive events and occurrence of one is not dependent on occurrence of the other as the ball is placed back into bag after first pick up.
Since Mary placed the ball back again so when John will be picking up the ball he will have same number of balls in bag to choose from as there were initially i.e 5 blue and 3 white.

Hence probability of John picking up blue ball will be 5/8.

  • +1 Upvote Post
  • Quote
  • Flag
Master | Next Rank: 500 Posts Default Avatar
Joined
22 Jun 2012
Posted:
258 messages
Followed by:
3 members
Upvotes:
6
GMAT Score:
640
Post
GMATGuruNY wrote:
sachindia wrote:
A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
1/3
3/8
1/2
5/8
11/16
No math is needed here if we understand the following concept:
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
Thus, P(blue on the 2nd pick) = P(blue on the first pick) = 5/8.

The correct answer is D.

The answer would be the same even if the problem were as follows:

Quote:
A basket contains 3 white and 5 blue balls. Mary will extract TWO BALLS at random and keep them. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
P(blue on the 3rd pick) = P(blue on the 1st pick) = 5/8.

Other problems that test the same concept:

http://www.beatthegmat.com/probablity-ques-t60161.html
http://www.beatthegmat.com/manhattan-probability-t89481.html (2 posts)
http://www.beatthegmat.com/a-box-contains-some-blue-balls-and-9-red-balls-if-the-t151368.html
hey , thanks a lot mitch for this. Does this work for problems with and without replacement?

_________________
Regards,
Sach

  • +1 Upvote Post
  • Quote
  • Flag

GMAT/MBA Expert

Post
sachindia wrote:
GMATGuruNY wrote:
sachindia wrote:
A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
1/3
3/8
1/2
5/8
11/16
No math is needed here if we understand the following concept:
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
Thus, P(blue on the 2nd pick) = P(blue on the first pick) = 5/8.

The correct answer is D.

The answer would be the same even if the problem were as follows:

Quote:
A basket contains 3 white and 5 blue balls. Mary will extract TWO BALLS at random and keep them. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
P(blue on the 3rd pick) = P(blue on the 1st pick) = 5/8.

Other problems that test the same concept:

http://www.beatthegmat.com/probablity-ques-t60161.html
http://www.beatthegmat.com/manhattan-probability-t89481.html (2 posts)
http://www.beatthegmat.com/a-box-contains-some-blue-balls-and-9-red-balls-if-the-t151368.html
hey , thanks a lot mitch for this. Does this work for problems with and without replacement?
Yes.
If a bag contains 4 blue marbles and 5 white marbles, and 3 marbles are selected at random -- WITH OR WITHOUT REPLACEMENT -- the probability that the THIRD marble selected is blue is equal to the probability that the FIRST marble selected is blue: 4/9.

_________________
Mitch Hunt
Private Tutor for the GMAT and GRE
GMATGuruNY@gmail.com

If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon.

Available for tutoring in NYC and long-distance.
For more information, please email me at GMATGuruNY@gmail.com.
Student Review #1
Student Review #2
Student Review #3

  • +1 Upvote Post
  • Quote
  • Flag
Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now.

GMAT/MBA Expert

Post
sachindia wrote:
A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
1/3
3/8
1/2
5/8
11/16
This question reminds me of my childhood, when my friends and I would sometimes "draw straws" to randomly select one person to do something (often either work, like getting wood for the fire, or dumb, like eating something that shouldn't be eaten).

So, someone would hold up n pieces of grass (for n guys), and one of those pieces was very short. The person who selected the shortest piece was the one who had to perform the task.

There was always one guy who wanted to choose his piece last. His reasoning was that his chances of drawing the shortest piece were minimized since every person before him had a chance of drawing the short piece before it got to his turn.

The truth of the matter is that each of the n guys had a 1/n chance of selecting the shortest piece, regardless of the order in which they selected.

The same applies to the original question here. 5 of the 8 balls are blue, so P(John selects a blue ball) = 5/8

Cheers,
Brent

_________________
Brent Hanneson – Creator of GMATPrepNow.com
Use my video course along with Beat The GMAT's free 60-Day Study Guide

Sign up for free Question of the Day emails
And check out all of these free resources

  • +1 Upvote Post
  • Quote
  • Flag
GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!
Junior | Next Rank: 30 Posts Default Avatar
Joined
22 May 2017
Posted:
10 messages
Post
GMATGuruNY wrote:
sachindia wrote:
A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
1/3
3/8
1/2
5/8
11/16
No math is needed here if we understand the following concept:
The probability of selecting X on the NTH pick is equal to the probability of selecting X on the FIRST pick.
Thus, P(blue on the 2nd pick) = P(blue on the first pick) = 5/8.

The correct answer is D.

The answer would be the same even if the problem were as follows:

Quote:
A basket contains 3 white and 5 blue balls. Mary will extract TWO BALLS at random and keep them. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
P(blue on the 3rd pick) = P(blue on the 1st pick) = 5/8.

Other problems that test the same concept:

http://www.beatthegmat.com/probablity-ques-t60161.html
http://www.beatthegmat.com/manhattan-probability-t89481.html (2 posts)
http://www.beatthegmat.com/a-box-contains-some-blue-balls-and-9-red-balls-if-the-t151368.html
This is interesting. Where does this concept come from? I've not seen it used anywhere else. Thanks.

  • +1 Upvote Post
  • Quote
  • Flag
Junior | Next Rank: 30 Posts Default Avatar
Joined
22 May 2017
Posted:
10 messages
Post
Brent@GMATPrepNow wrote:
sachindia wrote:
A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
1/3
3/8
1/2
5/8
11/16
This question reminds me of my childhood, when my friends and I would sometimes "draw straws" to randomly select one person to do something (often either work, like getting wood for the fire, or dumb, like eating something that shouldn't be eaten).

So, someone would hold up n pieces of grass (for n guys), and one of those pieces was very short. The person who selected the shortest piece was the one who had to perform the task.

There was always one guy who wanted to choose his piece last. His reasoning was that his chances of drawing the shortest piece were minimized since every person before him had a chance of drawing the short piece before it got to his turn.

The truth of the matter is that each of the n guys had a 1/n chance of selecting the shortest piece, regardless of the order in which they selected.

The same applies to the original question here. 5 of the 8 balls are blue, so P(John selects a blue ball) = 5/8

Cheers,
Brent
But Mary keeps the ball...How is it the same if she didn't keep it? Thank you.

  • +1 Upvote Post
  • Quote
  • Flag

GMAT/MBA Expert

Post
sachindia wrote:
A basket contains 3 white and 5 blue balls. Mary will extract one ball at random and keep it. If, after that, John will extract one ball at random, what is the probability that John will extract a blue ball?
1/3
3/8
1/2
5/8
11/16
Let's solve this question a different way.

P(John gets blue ball) = P(Mary gets white ball and John gets blue ball OR Mary gets blue ball and John gets blue ball
= P(Mary gets white ball and John gets blue ball) + P(Mary gets blue ball and John gets blue ball)
= (3/8 x 5/7) + (5/8 x 4/7)
= 15/56 + 20/56
= 35/56
= 5/8

Answer: still D Smile

_________________
Brent Hanneson – Creator of GMATPrepNow.com
Use my video course along with Beat The GMAT's free 60-Day Study Guide

Sign up for free Question of the Day emails
And check out all of these free resources

  • +1 Upvote Post
  • Quote
  • Flag
GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!

GMAT/MBA Expert

Post
In an ordinary deck of cards, 1/4 of the cards are hearts. If a magician spreads out a deck of cards on a table, and asks you to pick one, and you pick the second card from the top, I think everyone would agree the probability that card is a heart is 1/4.

If instead, the magician picks the top card off the deck and keeps it for herself, then asks you to pick the second card from the deck, you're picking exactly the same card as in the situation above. The probability it is a heart must be the same as before - it must be 1/4.

That's the principle Mitch is using above. The probability the top card in a deck is a heart is the same as the probability the second card, or the eleventh card, or the 41st card is a heart. It doesn't matter if you reach into the deck and take a card from the middle, or if someone removes the earlier cards one-by-one before you pick your card.

The key here is that you have no information about the cards that have been removed, so they neither make it more nor less likely that subsequent cards are hearts. If you knew something about the removed cards ,then that would change everything - then you'd be able to use that knowledge to adjust the probabilities for subsequent card selections. But in the question in the OP, you know nothing about Mary's selection, so the probability John picks a blue marble is still 5/8.

_________________
If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

  • +1 Upvote Post
  • Quote
  • Flag

GMAT/MBA Expert

GMAT Instructor Default Avatar
Joined
12 Sep 2012
Posted:
2635 messages
Followed by:
117 members
Upvotes:
625
Target GMAT Score:
V51
GMAT Score:
780
Post
Another way to demonstrate this:

Suppose that there are x marbles that I want in a jar that contains y marbles in all. My friend gets to pick and keep a marble first, then it's my turn.

If my friend gets one of the nice marbles, then my chance of getting one on my pick is (x - 1)/(y - 1).

If my friend gets one of the crummy marbles, then my chance getting a nice on on my pick is x/(y - 1).

My friend's chance of getting a nice marble is x/y, and my friend's chance of getting a crummy marble is (y - x)/y, so my chance of getting a nice marble is

(x/y)*(x-1)/(y-1) + (y - x)/y * x/(y - 1) => x/y

  • +1 Upvote Post
  • Quote
  • Flag
Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now!

GMAT/MBA Expert

GMAT Instructor Default Avatar
Joined
12 Sep 2012
Posted:
2635 messages
Followed by:
117 members
Upvotes:
625
Target GMAT Score:
V51
GMAT Score:
780
Post
As a fun aside, this problem is a nice primer on the perils of insider trading. Let's say that I'm a bookmaker and decide to take bets on this marble drawing. Prior to Mary's draw, John is 62.5% to draw blue, so the fair odds on that are 3 to 5: if you want to bet that John will draw blue, I'll wager 60¢ against every $1 that you're willing to risk.

After Mary's draw, however, *someone* -- Mary, at least (unless she doesn't look at the marble), and anyone she signals -- knows that the true odds have changed. If Mary drew white, then John's chances of blue are HIGHER, and if Mary drew blue, then John's chances of blue are lower. In either case, Mary now has better information than I do, and can outfox me in the betting! From John's perspective, or from any unknowing observer's, the best guess of his chances is still 62.5%, but anyone taking bets at this price can be defeated.

  • +1 Upvote Post
  • Quote
  • Flag
Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now!
  • Kaplan Test Prep
    Free Practice Test & Review
    How would you score if you took the GMAT

    Available with Beat the GMAT members only code

    MORE DETAILS
    Kaplan Test Prep
  • Veritas Prep
    Free Veritas GMAT Class
    Experience Lesson 1 Live Free

    Available with Beat the GMAT members only code

    MORE DETAILS
    Veritas Prep
  • Varsity Tutors
    Award-winning private GMAT tutoring
    Register now and save up to $200

    Available with Beat the GMAT members only code

    MORE DETAILS
    Varsity Tutors
  • e-gmat Exclusive Offer
    Get 300+ Practice Questions
    25 Video lessons and 6 Webinars for FREE

    Available with Beat the GMAT members only code

    MORE DETAILS
    e-gmat Exclusive Offer
  • The Princeton Review
    FREE GMAT Exam
    Know how you'd score today for $0

    Available with Beat the GMAT members only code

    MORE DETAILS
    The Princeton Review
  • Magoosh
    Magoosh
    Study with Magoosh GMAT prep

    Available with Beat the GMAT members only code

    MORE DETAILS
    Magoosh
  • PrepScholar GMAT
    5 Day FREE Trial
    Study Smarter, Not Harder

    Available with Beat the GMAT members only code

    MORE DETAILS
    PrepScholar GMAT
  • Economist Test Prep
    Free Trial & Practice Exam
    BEAT THE GMAT EXCLUSIVE

    Available with Beat the GMAT members only code

    MORE DETAILS
    Economist Test Prep
  • EMPOWERgmat Slider
    1 Hour Free
    BEAT THE GMAT EXCLUSIVE

    Available with Beat the GMAT members only code

    MORE DETAILS
    EMPOWERgmat Slider
  • Target Test Prep
    5-Day Free Trial
    5-day free, full-access trial TTP Quant

    Available with Beat the GMAT members only code

    MORE DETAILS
    Target Test Prep

Top First Responders*

1 Ian Stewart 41 first replies
2 Brent@GMATPrepNow 40 first replies
3 Scott@TargetTestPrep 39 first replies
4 Jay@ManhattanReview 32 first replies
5 GMATGuruNY 26 first replies
* Only counts replies to topics started in last 30 days
See More Top Beat The GMAT Members

Most Active Experts

1 image description Scott@TargetTestPrep

Target Test Prep

159 posts
2 image description Max@Math Revolution

Math Revolution

92 posts
3 image description Brent@GMATPrepNow

GMAT Prep Now Teacher

60 posts
4 image description Ian Stewart

GMATiX Teacher

50 posts
5 image description GMATGuruNY

The Princeton Review Teacher

37 posts
See More Top Beat The GMAT Experts