Last visit was: 25 Apr 2026, 23:25 It is currently 25 Apr 2026, 23:25
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
User avatar
rraggio
Joined: 30 Aug 2010
Last visit: 14 Oct 2010
Posts: 13
Own Kudos:
247
 [2]
Given Kudos: 2
Posts: 13
Kudos: 247
 [2]
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,733
Own Kudos:
36,466
 [3]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,733
Kudos: 36,466
 [3]
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
General Discussion
User avatar
shrouded1
User avatar
Retired Moderator
Joined: 02 Sep 2010
Last visit: 29 Apr 2018
Posts: 608
Own Kudos:
3,232
 [2]
Given Kudos: 25
Location: London
Products:
Posts: 608
Kudos: 3,232
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
saxenashobhit
Joined: 20 Jul 2010
Last visit: 14 Nov 2013
Posts: 135
Own Kudos:
256
 [1]
Given Kudos: 9
Products:
Posts: 135
Kudos: 256
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Without bringing formula into play I use below method

_ _ _
We have to fill 3 places.


Number of ways we can pick any mice is
10 * 9 * 8
__ __ __

Number of ways we can have all 3 black mice
4 * 3 * 2
__ __ __
So 1/30

PS - I know this is another way of C(4,3) and C(10,3) :)...
User avatar
rraggio
Joined: 30 Aug 2010
Last visit: 14 Oct 2010
Posts: 13
Own Kudos:
247
 [3]
Given Kudos: 2
Posts: 13
Kudos: 247
 [3]
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Thanks, you got it.

Since I'm very weak with combinations and probability, I'm trying to solve it in the following ways.
Three of four ways (1,2,4) work, one doesn't (3).
Could anybody say what's wrong in my third approach?
Thanks!

1) probability approach:
picking first black mice = \(4/10\) = \(2/5\)
picking second black mice = \(3/9\) = \(1/3\)
picking third black mice = \(2/8\) = \(1/4\)

picking three black mice = \(2/5*1/3*1/4=1/30\)

2) combinatorial approach:
Quote:
Number of ways to choose 3 black mice = \(C(4,3)\) (Picking any three of the four)
Number of ways to choose any 3 mice = \(C(10,3)\) (Picking any three of the ten)
So probability that all three mice are black is \(4/120\) or \(1/30\)

3) reversal probability approach:
The probability of picking three black mices equals to 1 minus the probability of picking three white mices plus the probability of picking two white mices and one black plus the probability of picking one white mices and two black.
That is:
P(b,b,b) = 1 - [P(w,w,w) + P(w,w,b) + P(w,b,b)]
P(w,w,w) = \(6/10*5/9*4/8=1/6\)
P(w,w,b) = \(6/10*5/9*4/8=1/6\)
P(w,b,b) = \(6/10*4/9*3/8=1/10\)

so:
P(b,b,b) = \(1- (1/6+1/6+1/10) = 1 - 13/30 = 17/30\)

4) reversal combinatorial approach:
Same approach of the above one.

Number of ways to choose any 3 mice = \(C(10,3)\) = 120 (Picking any three of the ten)
Number of ways to choose one mice white and two mices black = \(C(6,1)\)*\(C(4,2)\) = 36
Number of ways to choose two white mices and one mice white = \(C(6,2)\)*\(C(4,1)\) = 60
Number of ways to choose three white mices = \(C(6,3)\) = 20

So probability that all three mice are black is:
\(1 - [C(6,1)*C(4,2)+C(6,2)*C(4,1)+C(6,3)]/C(10,3) = 1 - (36+60+20)/120 = 4/120 = 1/30\)
User avatar
shrouded1
User avatar
Retired Moderator
Joined: 02 Sep 2010
Last visit: 29 Apr 2018
Posts: 608
Own Kudos:
3,232
 [2]
Given Kudos: 25
Location: London
Products:
Posts: 608
Kudos: 3,232
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
rraggio
Thanks, you got it.

Since I'm very weak with combinations and probability, I'm trying to solve it in the following ways.
Three of four ways (1,2,4) work, one doesn't (3).
Could anybody say what's wrong in my third approach?
Thanks!
3) reversal probability approach:
The probability of picking three black mices equals to 1 minus the probability of picking three white mices plus the probability of picking two white mices and one black plus the probability of picking one white mices and two black.
That is:
P(b,b,b) = 1 - [P(w,w,w) + P(w,w,b) + P(w,b,b)]
P(w,w,w) = \(6/10*5/9*4/8=1/6\)
P(w,w,b) = \(6/10*5/9*4/8=1/6\)
P(w,b,b) = \(6/10*4/9*3/8=1/10\)

so:
P(b,b,b) = \(1- (1/6+1/6+1/10) = 1 - 13/30 = 17/30\)

If you calculate probability in this manner, the order of drawing is inherently important. So you need to rule out the probability of the following as well :
P(w,b,w); P(b,w,w); P(b,w,b); P(b,b,w)
These will turn out to be 1/6,1/6,1/10,1/10
Subtracting this, once again you will get 1/30
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,733
Own Kudos:
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,733
Kudos: 36,466
Kudos
Add Kudos
Bookmarks
Bookmark this Post
rraggio

Out of a box that contains 4 black and 6 white mice, three are randomly chosen. What is the probability that all three will be black?

a) 8/125.
b) 1/30.
c) 2/5.
d) 1/720.
e) 3/10.

Thanks!

P(all 3 selected mice are black) = P(1st mouse is black AND 2nd mouse is black AND 3rd mouse is black)
= P(1st mouse is black) x P(2nd mouse is black) x P(3rd mouse is black)
= 4/10 x 3/9 x 2/8
= 1/30
= B
User avatar
saraheja
Joined: 01 Jun 2015
Last visit: 08 Jul 2018
Posts: 35
Own Kudos:
31
 [1]
Given Kudos: 7
Location: India
Products:
Posts: 35
Kudos: 31
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
selection of three mice: 4C3

Total selection of mice: 10C3

Prob : 4C3/10C3 = 1/30
avatar
luffy_ueki
Joined: 23 Oct 2017
Last visit: 13 Apr 2020
Posts: 44
Own Kudos:
Given Kudos: 23
Posts: 44
Kudos: 20
Kudos
Add Kudos
Bookmarks
Bookmark this Post
rraggio
Could anyone solve the following with a combinatorial approach instead of probability one?

Out of a box that contains 4 black and 6 white mice, three are randomly chosen. What is the probability that all three will be black?

a) 8/125.
b) 1/30.
c) 2/5.
d) 1/720.
e) 3/10.

Thanks!
---------
A box contains 4 black and 6 white mice.
Using probability approach to determine the probability that all three will be black?

First selection: 4/10 (4B, 6W)
Second selection: 3/9 (3B, 6W)
Third selection: 2/8 (2B, 6W)

Combined = (4/10)*(3/9)*(2/8) = 1/30
User avatar
EMPOWERgmatRichC
User avatar
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,777
Own Kudos:
Given Kudos: 450
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert
Expert reply
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Posts: 21,777
Kudos: 13,051
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi All,

You can approach this question in a couple of different ways. If you're going to treat it as a straight-forward probability question, then here's how you would do the math:

4 Black
6 White
10 Total

It's important to remember that once you choose a mouse, there will be one FEWER mouse for the next pick. The probability of choosing 3 black mice is…

(4/10)(3/9)(2/8) = 1/30

Final Answer:

GMAT assassins aren't born, they're made,
Rich
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,988
Own Kudos:
Posts: 38,988
Kudos: 1,118
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Math Expert
109830 posts
Tuck School Moderator
852 posts