Last visit was: 26 Apr 2024, 01:04 It is currently 26 Apr 2024, 01:04

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
SORT BY:
Date
Tags:
Show Tags
Hide Tags
User avatar
Intern
Intern
Joined: 30 Sep 2010
Posts: 11
Own Kudos [?]: 18 [18]
Given Kudos: 4
Send PM
Most Helpful Reply
SVP
SVP
Joined: 17 Nov 2007
Posts: 2408
Own Kudos [?]: 10036 [25]
Given Kudos: 361
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Send PM
avatar
Intern
Intern
Joined: 16 Sep 2012
Posts: 14
Own Kudos [?]: 16 [15]
Given Kudos: 2
Concentration: Finance, General Management
GMAT 1: 710 Q47 V41
Send PM
General Discussion
Math Expert
Joined: 02 Sep 2009
Posts: 92921
Own Kudos [?]: 619087 [7]
Given Kudos: 81596
Send PM
Five A-list actresses are vying for the three leading roles in the new [#permalink]
4
Kudos
3
Bookmarks
Expert Reply
Five A-list actresses are vying for the three leading roles in the new film, "Catfight in Denmark." The actresses are Julia Robards, Merly Strep, Sally Fieldstone, Lauren Bake-all, and Hallie Strawberry. Assuming that no actresses has any advantage in getting any role, what is the probability that Julia and Hallie will star in the film together?

(A) 3/10
(B) 3/125
(C) 6/10
(D) 3/5
(E) 1/125

\(Probability=\frac{Number \ of \ favorable \ outcomes}{total \ Number \ of \ outcomes}\).

Now, total the total Number of outcomes will be \(C^3_5=10\), (number of ways to choose 3 different actresses out of 5 when order of selection doesn't matter);

As for the number of favorable outcomes: we want 2 places out of 3 to be taken by J and H: so {JH-?}, for the third spot we can choose ANY from the 3 actresses left M, S, L. So, there are 3 such favorable groups possible: {JH-M}, {JH-S}, {JH-L}. Or \(C^1_1*C^1_3\), where \(C^1_1=1\) is 1 way to choose J and H (as it's one group) and \(C^1_3=3\) is 3 ways to choose the third member;

Thus \(P=\frac{3}{10}\).

Answer: A.

Hope it helps.
User avatar
Manager
Manager
Joined: 08 Jun 2011
Posts: 62
Own Kudos [?]: 76 [0]
Given Kudos: 65
Send PM
Five A-list actresses are vying for the three leading roles [#permalink]
Why isn't this approach right?

This is how I am doing it and I end up getting 2/5 which is not right.

I am doing the following:

Julia first (1/5) and Hallie second (1/4) = (1/5 * 1/4) OR
Hallie first (1/5) and Julia second (1/4) = (1/5 * 1/4) OR
Julia first (1/5) and Hallie third (1/3)=(1/5 * 1/3) OR
Hallie first (1/5) and Juila third (1/3)= (1/5 * 1/3) OR
Julia second (1/4) and Hallie third (1/3) = (1/4 * 1/3) OR
Hallie second (1/4) and Julia third (1/3) = (1/4 * 1/3)

I add them all up which totals to 2/5

Anything wrong here?

Originally posted by Lstadt on 29 Nov 2011, 01:51.
Last edited by Bunuel on 20 Sep 2022, 01:48, edited 2 times in total.
Edited the question, renamed the topic, and added the options and OA
Tutor
Joined: 16 Oct 2010
Posts: 14823
Own Kudos [?]: 64926 [3]
Given Kudos: 426
Location: Pune, India
Send PM
Five A-list actresses are vying for the three leading roles [#permalink]
2
Kudos
1
Bookmarks
Expert Reply
Lstadt wrote:
Five A-list actresses are vying for the three leading roles in the new film, "Catfight in
Denmark." The actresses are Julia Robards, Meryl Strep, Sally Fieldstone, lauren
Bake-all, and Hallie Strawberry. Assuming that no actress has any advantage in getting
any role, what is the probability that Julia and Hallie will star in the film together?

Why isn't this approach right?

This is how I am doing it and I end up getting 2/5 which is not right.

I am doing the following:

Julia first (1/5) and Hallie second (1/4) = (1/5 * 1/4) OR
Hallie first (1/5) and Julia second (1/4) = (1/5 * 1/4) OR
Julia first (1/5) and Hallie third (1/3)=(1/5 * 1/3) OR
Hallie first (1/5) and Juila third (1/3)= (1/5 * 1/3) OR
Julia second (1/4) and Hallie third (1/3) = (1/4 * 1/3) OR
Hallie second (1/4) and Julia third (1/3) = (1/4 * 1/3)

I add them all up which totals to 2/5

Anything wrong here?


Here is your problem:
Julia first (1/5) and Hallie second (1/4) and anyone else third = (1/5 * 1/4 * 1) OR
Hallie first (1/5) and Julia second (1/4) and anyone else third = (1/5 * 1/4 * 1) OR
Julia first (1/5), someone other than Hallie second and Hallie third (1/3)=(1/5 * 3/4 * 1/3) OR
Hallie first (1/5), someone other than Julia second, and Juila third (1/3)= (1/5 * 3/4 * 1/3) OR
Someone other than Hallie and Julia first, Julia second (1/4) and Hallie third (1/3) = (3/5 * 1/4 * 1/3) OR
Someone other than Hallie and Julia first, Hallie second (1/4) and Julia third (1/3) = (3/5 * 1/4 * 1/3)

When you add them up, you get 6/20 = 3/10

Another way to do it:
Pick Hallie and Julia. Now you can pick the third actress in 3 ways so total number of ways of picking 3 actresses (including Hallie and Julia) = 3
Total number of ways of picking any 3 actresses out of 5 = 5*4*3/3! = 10
Probability that both Hallie and Julia will be picked = 3/10

Originally posted by KarishmaB on 29 Nov 2011, 05:30.
Last edited by KarishmaB on 08 Oct 2022, 03:31, edited 1 time in total.
User avatar
Manager
Manager
Joined: 25 May 2011
Posts: 60
Own Kudos [?]: 148 [3]
Given Kudos: 71
Send PM
Re: Five A-list actresses are vying for the three leading roles [#permalink]
2
Kudos
1
Bookmarks
I just count even though it may seem somehow basic.

Suppose we want to select 3 from 5 of following:

A-B-C-J-H

There are:
ABC
ABJ
ABH
ACJ
ACH
AJH
BCJ
BCH
BJH
CJH

The group which contain both J and H are 3

So the probability is: \(\frac{3}{10}\)
User avatar
Manager
Manager
Joined: 29 Jul 2011
Status:Disbelief! The Countdown Begins
Affiliations: CFA
Posts: 214
Own Kudos [?]: 96 [5]
Given Kudos: 7
Concentration: Finance, Economics
GMAT 1: 600 Q45 V28
GMAT 2: 670 Q44 V39
GMAT 3: 750 Q49 V42
GPA: 3.5
WE:Accounting (Insurance)
Send PM
Re: Five A-list actresses are vying for the three leading roles [#permalink]
1
Kudos
4
Bookmarks
combinatorics tell us there are ten possibilities of combinations of actresses.

n!/k!(n-k)!

5 possibilities of choices - n = 5
3 will be seleced - k = 3

5*4*3*2*1/(3*2*1)(2*1) which reduces to (5*4)/(2*1) = 10.

There are 10 possible combinations you can choose to fill the 3 spots.

Now that you know there are 10, you need to find out which how many combinations you can make with the 2 selected actresses.

Julia and Halle can be grouped with any of the other 3 actresses, but there are no other possible combinations. So there are 3 groups that Julia and Halle could both be in.

3/10 is the answer.

Nice work by the creator of this question with the actress names. They all sound gorgeous.
User avatar
Intern
Intern
Joined: 02 Sep 2010
Posts: 24
Own Kudos [?]: 553 [2]
Given Kudos: 17
Location: India
 Q48  V35
Send PM
Re: Five A-list actresses are vying for the three leading roles [#permalink]
1
Kudos
1
Bookmarks
Lstadt wrote:
Five A-list actresses are vying for the three leading roles in the new film, "Catfight in Denmark." The actresses are Julia Robards, Merly Strep, Sally Fieldstone, Lauren Bake-all, and Hallie Strawberry. Assuming that no actresses has any advantage in getting any role, what is the probability that Julia and Hallie will star in the film together?

Why isn't this approach right?

This is how I am doing it and I end up getting 2/5 which is not right.

I am doing the following:

Julia first (1/5) and Hallie second (1/4) = (1/5 * 1/4) OR
Hallie first (1/5) and Julia second (1/4) = (1/5 * 1/4) OR
Julia first (1/5) and Hallie third (1/3)=(1/5 * 1/3) OR
Hallie first (1/5) and Juila third (1/3)= (1/5 * 1/3) OR
Julia second (1/4) and Hallie third (1/3) = (1/4 * 1/3) OR
Hallie second (1/4) and Julia third (1/3) = (1/4 * 1/3)

I add them all up which totals to 2/5

Anything wrong here?


Very simple solution, i would say is
Total number of ways in which you can select 3 out of 5 actresses is 5C3 = 10
Total number of ways in which you can select 2 actress for 3 slots is 3C2 = 3 (as all the actresses have same probability of getting selected)

Probability = 3/10. :-D
avatar
Intern
Intern
Joined: 10 Dec 2013
Posts: 14
Own Kudos [?]: 10 [0]
Given Kudos: 7
Location: India
Concentration: Technology, Strategy
Schools: ISB '16 (S)
GMAT 1: 710 Q48 V38
GPA: 3.9
WE:Consulting (Consulting)
Send PM
Re: Five A-list actresses are vying for the three leading roles [#permalink]
Can someone please help me understand why didnt we use permutation here. Because
1. Julia, Hallie and Meryl
and
2. Hallie, Julia and Meryl

should be considered two different ways to select actresses, shouldnt it?
Tutor
Joined: 16 Oct 2010
Posts: 14823
Own Kudos [?]: 64926 [3]
Given Kudos: 426
Location: Pune, India
Send PM
Re: Five A-list actresses are vying for the three leading roles [#permalink]
1
Kudos
2
Bookmarks
Expert Reply
Rohan_Kanungo wrote:
Can someone please help me understand why didnt we use permutation here. Because
1. Julia, Hallie and Meryl
and
2. Hallie, Julia and Meryl

should be considered two different ways to select actresses, shouldnt it?


There are 3 different roles so selecting Julia, Hallie and Meryl is different from Hallie, Julia and Meryl. But we ignore the arrangements because we are looking for a probability.

P(Picking Hallie and Julia) = (No of ways in each Hallie and Julia will be picked)/(No of ways of picking any 3 actresses)
= (3*3!)/(10*3!)
Since we are calculating the probability, the arrangements of the numerator (3!) will get cancelled by the arrangements of the denominator (3!). So we can ignore them.
GMAT Tutor
Joined: 24 Jun 2008
Posts: 4128
Own Kudos [?]: 9245 [2]
Given Kudos: 91
 Q51  V47
Send PM
Re: Five A-list actresses are vying for the three leading roles [#permalink]
1
Kudos
1
Bookmarks
Expert Reply
You can also reverse the question. If we'll hire 3 actresses, we'll fire 2 of them. We can focus on the two people we're firing. We just need to be sure we don't fire Julia or Hallie. There's a 3/5 chance the first person we fire is not J or H, and then a 2/4 chance the second person we fire is not J or H. So the answer is (3/5)(2/4) = 3/10.
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 32679
Own Kudos [?]: 822 [0]
Given Kudos: 0
Send PM
Re: Five A-list actresses are vying for the three leading roles [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
GMAT Club Bot
Re: Five A-list actresses are vying for the three leading roles [#permalink]
Moderators:
Math Expert
92918 posts
Senior Moderator - Masters Forum
3137 posts

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