It is currently 29 Jun 2017, 07:39

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

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Out of a classroom of 6 boys and 4 girls the teacher picks a

Author Message
TAGS:

Hide Tags

CEO
Joined: 21 Jan 2007
Posts: 2739
Location: New York City
Out of a classroom of 6 boys and 4 girls the teacher picks a [#permalink]

Show Tags

09 Dec 2007, 11:16
3
KUDOS
11
This post was
BOOKMARKED
00:00

Difficulty:

(N/A)

Question Stats:

68% (02:30) correct 32% (00:56) wrong based on 115 sessions

HideShow timer Statistics

Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary.

What is the probability that only girls will be elected?

What is the probability that only boys will be elected?

What is the probability that at least one girl will be elected?
Senior Manager
Joined: 09 Oct 2007
Posts: 466

Show Tags

09 Dec 2007, 11:58
1) (4/10)(3/9)(2/8) = 4/105
2) (6/10)(5/9)(4/8) = 4/21
3) (4/10)(6/9)(5/8) = (1/6)*3 = 1/2
Manager
Joined: 29 Jul 2007
Posts: 182

Show Tags

09 Dec 2007, 12:33
6
KUDOS
1
This post was
BOOKMARKED
1) 4/10 * 3/9 * 2/8 = 2/5 * 1/3 * 1/4 = 2/60 = 1/30

2) 6/10 * 5/9 * 4/8 = 3/5 * 5/9 * 1/2 = 3/18 = 1/6

3) 1 - P(all boys selected) = 1- (1/6) = 5/6
Senior Manager
Joined: 09 Oct 2007
Posts: 466

Show Tags

09 Dec 2007, 12:46
I must be stupid. I was wondering how I got it wrong so I went back to my numbers and somehow I multiplied by 7 when I had 8:
4/10)(3/9)(2/8 but mutliplied by 7) = 4/105 ... WTF?

Is third right, though? 1-all boys possibility?
CEO
Joined: 21 Jan 2007
Posts: 2739
Location: New York City

Show Tags

09 Jan 2008, 08:10
Skewed wrote:
1) 4/10 * 3/9 * 2/8 = 2/5 * 1/3 * 1/4 = 2/60 = 1/30

2) 6/10 * 5/9 * 4/8 = 3/5 * 5/9 * 1/2 = 3/18 = 1/6

3) 1 - P(all boys selected) = 1- (1/6) = 5/6

very nice. all are correct

for #3, we can also do (10C3 - 6C3)/10C3 = 100/120 = 5/6
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

CEO
Joined: 29 Mar 2007
Posts: 2559

Show Tags

09 Jan 2008, 08:28
2
KUDOS
bmwhype2 wrote:
Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary.

What is the probability that only girls will be elected?

What is the probability that only boys will be elected?

What is the probability that at least one girl will be elected?

4/10*3/9*2/8 24/720 --> 12/360 --> 6/180 --> 3/90 --> 1/30

6/10*5/9*4/8 --> 120/720--> 12/72 --> 1/6

1-Probability that none are girls: --> just use the above 1-6/10*5/9*4/8 = 5/6.
Manager
Joined: 07 Jan 2008
Posts: 74

Show Tags

10 Jun 2008, 14:11
For 3Q. What is the probability that at least one girl will be elected?

Isn't it.... 1-p(no girls selected) = 1-(6/10) = 4/10 = 2/5

Can someone give, y above is wrong..pls
Director
Joined: 14 Oct 2007
Posts: 753
Location: Oxford
Schools: Oxford'10

Show Tags

10 Jun 2008, 23:56
1
KUDOS
sumanamba wrote:
For 3Q. What is the probability that at least one girl will be elected?

Isn't it.... 1-p(no girls selected) = 1-(6/10) = 4/10 = 2/5

Can someone give, y above is wrong..pls

1 - p(no girls selected) is the correct strategy, p(no girls selected) isn't as simple as 6/10

6/10 is the probability of one boy to be "pulled out of the hat", the next boy is 5/9 and so forth

p(no girls selected) is therefore p(all boys selected) = 6C3/10C3 = 1/6
as shown above
Intern
Joined: 18 Jan 2008
Posts: 1

Show Tags

12 Jun 2008, 08:01
I believe this is a permuation problem as the order does matter...

What is the probability that only girls will be elected? would be 4p3/10p3 = 1/30

Its a coincidence that the answer is 1/30 when combinations is used i.e., 4c3/10c3.
Director
Joined: 25 Oct 2008
Posts: 596
Location: Kolkata,India

Show Tags

24 Jul 2009, 05:42
3
KUDOS
1.4C3/10C3
2. 6C3/10C3
3. 1-6C3/10C3
_________________

http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

Manager
Joined: 10 Jul 2009
Posts: 126
Location: Ukraine, Kyiv

Show Tags

06 Sep 2009, 03:09
2
KUDOS
I have another opinion on the third question. Wonder, if it is right?

1) I got 1/30 too
2) got 1/6 as well

But, in my opinion, the third question should be solved this way:
the question asks what is the probability of AT LEAST one girl will be selected.

we have three mutually exclusive possibilities to select (they are all satisfy the asked condition):
GBB (girl,boy,boy)
GGB (girl,girl,boy)
GGG (girl,girl,girl)

the probability of GGG is already known - 1/30.
the probability of GGB is: 4/10*3/9*6/8 - 1/10.
the probability of GBB is: 4/10*6/9*5/8 - 1/6.

As these are mutually exclusive, we shall add them together: 1/30+1/10+1/6 = 3/10.

What do you think of this?
_________________

Never, never, never give up

Verbal Forum Moderator
Joined: 02 Aug 2009
Posts: 4575

Show Tags

06 Sep 2009, 04:03
1
KUDOS
Expert's post
I have another opinion on the third question. Wonder, if it is right?

1) I got 1/30 too
2) got 1/6 as well

But, in my opinion, the third question should be solved this way:
the question asks what is the probability of AT LEAST one girl will be selected.

we have three mutually exclusive possibilities to select (they are all satisfy the asked condition):
GBB (girl,boy,boy)
GGB (girl,girl,boy)
GGG (girl,girl,girl)

the probability of GGG is already known - 1/30.
the probability of GGB is: 4/10*3/9*6/8 - 1/10.
the probability of GBB is: 4/10*6/9*5/8 - 1/6.

As these are mutually exclusive, we shall add them together: 1/30+1/10+1/6 = 3/10.

What do you think of this?

ANS.......
u r right but only mistake u r doing is that u have to multiply two of possiblities by 3....reason:-
possiblities of taking two girls is GGB,GBG,BGG ie 1/10*3=3/10.....
possiblities of taking two girls is GBB,BGB,BBG ie 1/6*3=1/2....
As these are mutually exclusive, we shall add them together: 1/30+3/10+1/2 = 5/6.
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Manager
Joined: 27 Oct 2008
Posts: 185

Show Tags

27 Sep 2009, 00:42
Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary.

What is the probability that only girls will be elected?

What is the probability that only boys will be elected?

What is the probability that at least one girl will be elected?

Soln:
All are permutations since even though we select 3 people , they are pushed into different posts. i.e. We could have selected the same 3 people, and arrange them in 6 different ways (3!) coz the posts that they take are different. president ,vice president and secretary.

What is the probability that only girls will be elected?
Ans = 4P3/10P3

What is the probability that only boys will be elected?
Ans = 6P3/10P3

What is the probability that at least one girl will be elected?
Ans = (4P1 * 6P2 + 4P2 * 6P1 + 4P3)/10P3
Manager
Joined: 05 Jul 2009
Posts: 181

Show Tags

29 Sep 2009, 02:58
Yup, this is a permutation problem since three people will be selected for three different posts.

A) 4P3/10P3

B) 6P3/10P3

C) 1- (6P3/10P3)
Intern
Joined: 29 Nov 2009
Posts: 7

Show Tags

29 Nov 2009, 03:39
I got the answers right but have a doubt.

If the teacher selects for president, vice pres & secretary, there are 3 distinct positions unlike say getting selected for a generic team where it doesn't matter what kind of position.

So girl 1 can be selected for pres, girl 2 for vp & girl 3 for secretary is a different case than girl 3 for pres, girl 2 for vp & girl 1 for secretary.

Can someone explain why we dont consider them distinct possibilities in this case?
thanks.
Senior Manager
Joined: 22 Dec 2009
Posts: 359

Show Tags

06 Jan 2010, 13:07
IMO.. it shouldn't matter whether it is calculated with combinations or permutations. Reason being, if you consider the order in choosing the girls (numerator), you do the same in the denominator too. Hence it balances it out...

the results are coming out to be the same....

Cheers!
JT
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Senior Manager
Joined: 22 Dec 2009
Posts: 359

Show Tags

14 Feb 2010, 08:33
2
KUDOS
bmwhype2 wrote:
Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary.

What is the probability that only girls will be elected?

What is the probability that only boys will be elected?

What is the probability that at least one girl will be elected?

Total comb = 10c3

1. Only girls = 4c3/10c3 = 1/30
2. Only boys = 6c3/10c3 = 1/6
3. Atleast one gurl = 1 - no gurl = 1 - only boys = 1 - 1/6 = 5/6
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Current Student
Joined: 02 Apr 2012
Posts: 77
Location: United States (VA)
Concentration: Entrepreneurship, Finance
GMAT 1: 680 Q49 V34
WE: Consulting (Consulting)
Re: Out of a classroom of 6 boys and 4 girls the teacher picks a [#permalink]

Show Tags

13 Jul 2013, 16:12
2
KUDOS
1
This post was
BOOKMARKED
Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary.

What is the probability that only girls will be elected?

What is the probability that only boys will be elected?

---------------------------------------------------------------------------------------------------------------------------------------------------------------
(1) What is the probability that only girls will be elected?

Just for practice, since the most efficient way to solve this problem is the Probability approach.

(1) probability approach:

$$\frac{4}{10} * \frac{3}{9} * \frac{2}{8} = \frac{1}{30}$$

But I need to practice so I solved this problem with the other 3 approaches:

(2) Reversal Probability approach:
P = 1 - q (q=probability of not picking only girls, this means, picking 1, 2 or 3 boys):

--> 1 boy + 2 girls = BGG (and combinations):
$$1 - \frac{6}{10} * \frac{4}{9} * \frac{3}{8} * \frac{3!}{2!} = \frac{3}{10}$$

-->2 boys + 1 Girl = BBG (And combinations):
$$\frac{6}{10} *\frac{5}{9} * \frac{4}{8} * \frac{3!}{2!} = \frac{1}{2}$$

--> 3 boys:
$$\frac{6}{10} * \frac{5}{9} * \frac{4}{8} * \frac{3!}{3!} = \frac{1}{6}$$

Then, we add all the probabilities:

$$\frac{3}{10} + \frac{1}{2}+ \frac{1}{6} = \frac{29}{30} = q$$

$$1-q = 1- \frac{29}{30} = \frac{1}{30}$$ = P

(3) Combinatorial approach:

$$\frac{C^4_3}{C^10_3} = \frac{4}{120} = \frac{1}{30}$$

(4) Reversal combinatorial approach:

$$1 - \frac{{C^6_1 * C^4_2 + C^6_2 * C^4_1 + C^6_3}}{C^10_3} = 1- \frac{{36 + 60 + 20}}{120} = 120/120 - \frac{116}{120} = \frac{1}{30}$$

(2) What is the probability that only boys will be elected?

(1) Probability approach:

$$\frac{6}{10} * \frac{5}{9} * \frac{4}{8} = \frac{4}{24} = \frac{1}{6}$$

(2) Reversal probability approach --> P = 1 - q

1girl + 2 boys: $$\frac{4}{10} *\frac{6}{9} *\frac{5}{8} * \frac{3!}{2!} = \frac{1}{6} * 3 = \frac{1}{2}$$

2 girls + 1 boy: $$\frac{4}{10} *\frac{3}{9} * \frac{6}{8} *\frac{3!}{2!} = \frac{1}{10} * 3 = \frac{3}{10}$$

3 girls: $$\frac{4}{10} *\frac{3}{9} * \frac{2}{8} = \frac{1}{30}$$

(3) Combinatorial approach:

$$\frac{C^6_3}{C^10_3} = \frac{20}{120} = \frac{1}{6}$$

(4) Reversal Comb approach:

$$\frac{C^4_1 * C^6_2 + C^4_2 * C^6_1 + C^4_3}{C^10_3}$$

1 girl + 2 boys: $$C^4_1 * C^6_2 = 60$$
2 girls + 1 boy: $$C^4_2 * C^6_1 = 36$$
3 girls:$$C^4_3 = 4$$

60 + 36 + 4 = 100

$$\frac{120}{120} - \frac{100}{120} = \frac{20}{120} = \frac{1}{6}$$
_________________

Encourage cooperation! If this post was very useful, kudos are welcome
"It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

Math Expert
Joined: 02 Sep 2009
Posts: 39759
Re: Out of a classroom of 6 boys and 4 girls the teacher picks a [#permalink]

Show Tags

14 Jul 2013, 00:14
Expert's post
4
This post was
BOOKMARKED
Maxirosario2012 wrote:
Out of a classroom of 6 boys and 4 girls the teacher picks a president for the student board, a vice president and a secretary.

What is the probability that only girls will be elected?

What is the probability that only boys will be elected?

---------------------------------------------------------------------------------------------------------------------------------------------------------------
(1) What is the probability that only girls will be elected?

Just for practice, since the most efficient way to solve this problem is the Probability approach.

(1) probability approach:

$$\frac{4}{10} * \frac{3}{9} * \frac{2}{8} = \frac{1}{30}$$

But I need to practice so I solved this problem with the other 3 approaches:

(2) Reversal Probability approach:
P = 1 - q (q=probability of not picking only girls, this means, picking 1, 2 or 3 boys):

--> 1 boy + 2 girls = BGG (and combinations):
$$1 - \frac{6}{10} * \frac{4}{9} * \frac{3}{8} * \frac{3!}{2!} = \frac{3}{10}$$

-->2 boys + 1 Girl = BBG (And combinations):
$$\frac{6}{10} *\frac{5}{9} * \frac{4}{8} * \frac{3!}{2!} = \frac{1}{2}$$

--> 3 boys:
$$\frac{6}{10} * \frac{5}{9} * \frac{4}{8} * \frac{3!}{3!} = \frac{1}{6}$$

Then, we add all the probabilities:

$$\frac{3}{10} + \frac{1}{2}+ \frac{1}{6} = \frac{29}{30} = q$$

$$1-q = 1- \frac{29}{30} = \frac{1}{30}$$ = P

(3) Combinatorial approach:

$$\frac{C^4_3}{C^10_3} = \frac{4}{120} = \frac{1}{30}$$

(4) Reversal combinatorial approach:

$$1 - \frac{{C^6_1 * C^4_2 + C^6_2 * C^4_1 + C^6_3}}{C^10_3} = 1- \frac{{36 + 60 + 20}}{120} = 120/120 - \frac{116}{120} = \frac{1}{30}$$

(2) What is the probability that only boys will be elected?

(1) Probability approach:

$$\frac{6}{10} * \frac{5}{9} * \frac{4}{8} = \frac{4}{24} = \frac{1}{6}$$

(2) Reversal probability approach --> P = 1 - q

1girl + 2 boys: $$\frac{4}{10} *\frac{6}{9} *\frac{5}{8} * \frac{3!}{2!} = \frac{1}{6} * 3 = \frac{1}{2}$$

2 girls + 1 boy: $$\frac{4}{10} *\frac{3}{9} * \frac{6}{8} *\frac{3!}{2!} = \frac{1}{10} * 3 = \frac{3}{10}$$

3 girls: $$\frac{4}{10} *\frac{3}{9} * \frac{2}{8} = \frac{1}{30}$$

(3) Combinatorial approach:

$$\frac{C^6_3}{C^10_3} = \frac{20}{120} = \frac{1}{6}$$

(4) Reversal Comb approach:

$$\frac{C^4_1 * C^6_2 + C^4_2 * C^6_1 + C^4_3}{C^10_3}$$

1 girl + 2 boys: $$C^4_1 * C^6_2 = 60$$
2 girls + 1 boy: $$C^4_2 * C^6_1 = 36$$
3 girls:$$C^4_3 = 4$$

60 + 36 + 4 = 100

$$\frac{120}{120} - \frac{100}{120} = \frac{20}{120} = \frac{1}{6}$$

Some "at least" probability questions to practice:
leila-is-playing-a-carnival-game-in-which-she-is-given-140018.html
a-fair-coin-is-tossed-4-times-what-is-the-probability-of-131592.html
for-each-player-s-turn-in-a-certain-board-game-a-card-is-132074.html
a-string-of-10-light-bulbs-is-wired-in-such-a-way-that-if-131205.html
a-shipment-of-8-tv-sets-contains-2-black-and-white-sets-and-53338.html
on-a-shelf-there-are-6-hardback-books-and-2-paperback-book-135122.html
in-a-group-with-800-people-136839.html
the-probability-of-a-man-hitting-a-bulls-eye-in-one-fire-is-136935.html
for-each-player-s-turn-in-a-certain-board-game-a-card-is-141790.html
the-probability-that-a-convenience-store-has-cans-of-iced-128689.html
a-manufacturer-is-using-glass-as-the-surface-144642.html
a-fair-coin-is-to-be-tossed-twice-and-an-integer-is-to-be-148779.html
in-a-game-one-player-throws-two-fair-six-sided-die-at-the-151956.html
_________________
Director
Joined: 03 Aug 2012
Posts: 894
Concentration: General Management, General Management
GMAT 1: 630 Q47 V29
GMAT 2: 680 Q50 V32
GPA: 3.7
WE: Information Technology (Investment Banking)
Re: Out of a classroom of 6 boys and 4 girls the teacher picks a [#permalink]

Show Tags

06 Aug 2013, 21:46
(1). All girls are selected.

Just for argument sake, it has not been mentioned that One GIRL/BOY cannot be absorbed in multiple roles.

So why the probability P(All Girls) is not calculated as:

(4*4*4)/(10*10*10) = (8/125)

Desired outcome= You can fill each position in 4 ways or select each position in 4 ways
Total outcome = You can make 10 selections each time

Rgds,
TGC !
_________________

Rgds,
TGC!
_____________________________________________________________________
I Assisted You => KUDOS Please
_____________________________________________________________________________

Re: Out of a classroom of 6 boys and 4 girls the teacher picks a   [#permalink] 06 Aug 2013, 21:46

Go to page    1   2    Next  [ 24 posts ]

Similar topics Replies Last post
Similar
Topics:
In how many arrangements can a teacher seat 3 girls and 3 boys in a 2 17 Apr 2017, 08:14
1 Out of a classroom of 6 boys and 4 girls the teacher picks a president 3 13 Jul 2016, 13:15
6 Another variant: There are 6 girls and 6 boys. 6 03 Jul 2016, 23:27
9 There are 6 girls and 6 boys. If they are to be seated in a row 5 23 Nov 2016, 08:18
18 Three boys are ages 4, 6 and 7 respectively. Three girls are 7 13 Mar 2017, 01:48
Display posts from previous: Sort by