GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 24 Apr 2019, 21:47

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

M14-01

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 54496
M14-01  [#permalink]

Show Tags

New post 16 Sep 2014, 00:50
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

73% (01:17) correct 27% (01:37) wrong based on 85 sessions

HideShow timer Statistics

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?

A. \(\frac{1}{3}\)
B. \(\frac{3}{8}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{8}\)
E. \(\frac{11}{16}\)

_________________
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 54496
Re M14-01  [#permalink]

Show Tags

New post 16 Sep 2014, 00:50
Official Solution:

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?

A. \(\frac{1}{3}\)
B. \(\frac{3}{8}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{8}\)
E. \(\frac{11}{16}\)


There are 2 different scenarios to evaluate:

1. Mary will extract a white ball, John will extract a blue ball. The probability of this is \(\frac{3}{8} * \frac{5}{7} = \frac{15}{56}\).

2. Mary will extract a blue ball, John will extract a blue ball. The probability of this is \(\frac{5}{8} * \frac{4}{7} = \frac{20}{56}\).

The overall probability that John will extract a blue ball = \(\frac{15}{56} + \frac{20}{56} = \frac{35}{56} = \frac{5}{8}\).

Alternative Explanation

The initial probability of drawing blue ball is \(\frac{5}{8}\). Without knowing the other results, the probability of drawing blue ball will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

If Mary extracts not 1 but 7 balls at random and keep them, then even after that, the probability that the last ball left for John is blue still would be \(\frac{5}{8}\).


Answer: D
_________________
Manager
Manager
avatar
S
Joined: 24 Dec 2015
Posts: 56
Location: India
Concentration: Marketing, Technology
GMAT 1: 710 Q50 V35
GPA: 3.8
WE: Engineering (Computer Software)
GMAT ToolKit User Reviews Badge
Re: M14-01  [#permalink]

Show Tags

New post 04 Jul 2016, 19:59
I'm confused. Can someone help me here?

Probability = Favourable Outcomes/Total possible outcomes

In this case, since the first pick is not being replaced, shouldn't the Probability of the second pick be 5/7+4/7 ?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 54496
Re: M14-01  [#permalink]

Show Tags

New post 05 Jul 2016, 01:47
winionhi wrote:
I'm confused. Can someone help me here?

Probability = Favourable Outcomes/Total possible outcomes

In this case, since the first pick is not being replaced, shouldn't the Probability of the second pick be 5/7+4/7 ?


The probability that John will extract a blue ball with second pick is 5/7 only if Mary extracts a white ball with her first pick.
The probability that John will extract a blue ball with second pick is 4/7 only if Mary extracts a blue ball with her first pick.

So, you should take the first pick into the account as shown in the solution above.
_________________
Intern
Intern
avatar
B
Joined: 28 Feb 2016
Posts: 1
GMAT ToolKit User
Re: M14-01  [#permalink]

Show Tags

New post 28 May 2017, 22:56
In the question, it should be made clear whether Mary keeps it back into the basket (replacement case) or keeps it with her (no replacement case). Just saying "keeps it" might create confusion.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 54496
Re: M14-01  [#permalink]

Show Tags

New post 28 May 2017, 23:29
Intern
Intern
avatar
B
Joined: 10 Nov 2014
Posts: 15
Location: United Kingdom
WE: Information Technology (Consulting)
Re: M14-01  [#permalink]

Show Tags

New post 20 Aug 2017, 07:55
Bunuel wrote:
Official Solution:

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?

A. \(\frac{1}{3}\)
B. \(\frac{3}{8}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{8}\)
E. \(\frac{11}{16}\)


There are 2 different scenarios to evaluate:

1. Mary will extract a white ball, John will extract a blue ball. The probability of this is \(\frac{3}{8} * \frac{5}{7} = \frac{15}{56}\).

2. Mary will extract a blue ball, John will extract a blue ball. The probability of this is \(\frac{5}{8} * \frac{4}{7} = \frac{20}{56}\).

The overall probability that John will extract a blue ball = \(\frac{15}{56} + \frac{20}{56} = \frac{35}{56} = \frac{5}{8}\).

Alternative Explanation

The initial probability of drawing blue ball is \(\frac{5}{8}\). Without knowing the other results, the probability of drawing blue ball will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

If Mary extracts not 1 but 7 balls at random and keep them, then even after that, the probability that the last ball left for John is blue still would be \(\frac{5}{8}\).


Answer: D




Hi Bunnuel,

Could you please elaborate a bit on the alternative solution? Why would 1st draw not change the probability of the second draw as the set from where the second draw is made would consist just of 7 balls and we aren't sure on the count of BLUE and WHITE in this draw as we do not know the result of the first.

Did it by the first process but took me close to 3 mins.

Best-
Amit
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 54496
Re: M14-01  [#permalink]

Show Tags

New post 21 Aug 2017, 02:47
1
Amit989 wrote:
Bunuel wrote:
Official Solution:

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?

A. \(\frac{1}{3}\)
B. \(\frac{3}{8}\)
C. \(\frac{1}{2}\)
D. \(\frac{5}{8}\)
E. \(\frac{11}{16}\)


There are 2 different scenarios to evaluate:

1. Mary will extract a white ball, John will extract a blue ball. The probability of this is \(\frac{3}{8} * \frac{5}{7} = \frac{15}{56}\).

2. Mary will extract a blue ball, John will extract a blue ball. The probability of this is \(\frac{5}{8} * \frac{4}{7} = \frac{20}{56}\).

The overall probability that John will extract a blue ball = \(\frac{15}{56} + \frac{20}{56} = \frac{35}{56} = \frac{5}{8}\).

Alternative Explanation

The initial probability of drawing blue ball is \(\frac{5}{8}\). Without knowing the other results, the probability of drawing blue ball will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

If Mary extracts not 1 but 7 balls at random and keep them, then even after that, the probability that the last ball left for John is blue still would be \(\frac{5}{8}\).


Answer: D




Hi Bunnuel,

Could you please elaborate a bit on the alternative solution? Why would 1st draw not change the probability of the second draw as the set from where the second draw is made would consist just of 7 balls and we aren't sure on the count of BLUE and WHITE in this draw as we do not know the result of the first.

Did it by the first process but took me close to 3 mins.

Best-
Amit


Because we don't know the result of the first draw.

Consider this imagine 8 balls in a row. Now, what is the probability that blue ball is 1st in that row? 5/8. What is the probability that it's 2nd? Again 5/8. What is the probability that he's 8th? 5/8.

Similar questions to practice:
http://gmatclub.com/forum/a-certain-tea ... 31321.html
http://gmatclub.com/forum/a-certain-tel ... 27423.html
http://gmatclub.com/forum/a-medical-res ... 27396.html
http://gmatclub.com/forum/a-certain-cla ... 27730.html
https://gmatclub.com/forum/a-certain-cl ... 91-20.html

Hope it helps.
_________________
GMAT Club Bot
Re: M14-01   [#permalink] 21 Aug 2017, 02:47
Display posts from previous: Sort by

M14-01

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Moderators: chetan2u, Bunuel



Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.