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# Five A-list actresses are vying for the three leading roles

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Manager
Joined: 08 Jun 2011
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Updated on: 04 Sep 2015, 09:34
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68% (01:25) correct 32% (01:43) wrong based on 414 sessions

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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

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 ENGRTOMBA2018 on 04 Sep 2015, 09:34, edited 1 time in total.
Edited the question, renamed the topic, and added the options and OA
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Re: Five A-list actresses are vying for the three leading roles  [#permalink]

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18 Oct 2012, 07:45
7
6
$$\frac{C_2^2*C_1^3}{C_3^5} = \frac{3}{10}$$

$$C_2^2$$ - the number of ways to choose Julia and Hallie

$$C_1^3$$ - the number of ways to choose 1 actress (neither Julia nor Hallie) out of those who left

$$C_3^5$$ - the total number of ways to choose 3 actresses out of 5.
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Re: Five A-list actresses are vying for the three leading roles  [#permalink]

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18 Oct 2012, 05:45
6
I have got the following solution which seems different from the ones above but results in the same answer:
3 roles for 5 actresses
3/5 chances of the first actress to get a role
2/4 chances of the second one to get a role
3/5*2/4 = 6/20 = 3/10 chances of both actresses
##### General Discussion
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Re: Five A-list actresses are vying for the three leading roles  [#permalink]

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29 Nov 2011, 05:30
2
1
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?

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

For more on combinations, check out this post:
http://www.veritasprep.com/blog/2011/11 ... binations/
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Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! Manager Joined: 25 May 2011 Posts: 126 Re: Five A-list actresses are vying for the three leading roles [#permalink] ### Show Tags 12 Dec 2011, 12:22 2 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: $$\frac{3}{10}$$ Manager Status: Disbelief! The Countdown Begins Affiliations: CFA Joined: 29 Jul 2011 Posts: 217 Concentration: Finance, Economics Schools: Johnson '15 (M) GMAT 1: 600 Q45 V28 GMAT 2: 670 Q44 V39 GMAT 3: 750 Q49 V42 GPA: 3.5 WE: Accounting (Insurance) Re: Five A-list actresses are vying for the three leading roles [#permalink] ### Show Tags 12 Dec 2011, 12:33 1 2 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. Intern Joined: 02 Sep 2010 Posts: 40 Location: India Re: Five A-list actresses are vying for the three leading roles [#permalink] ### Show Tags 20 Oct 2012, 22:16 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. _________________ The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done! Intern Joined: 10 Dec 2013 Posts: 20 Location: India Concentration: Technology, Strategy Schools: ISB '16 (S) GMAT 1: 710 Q48 V38 GPA: 3.9 WE: Consulting (Consulting) Re: Five A-list actresses are vying for the three leading roles [#permalink] ### Show Tags 28 Jan 2014, 10:41 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? Intern Joined: 05 Dec 2013 Posts: 30 Concentration: Technology GMAT Date: 02-01-2014 GPA: 3.95 WE: Information Technology (Venture Capital) Re: Five A-list actresses are vying for the three leading roles [#permalink] ### Show Tags 28 Jan 2014, 19:47 5C3 total combinations. Three combinations where J and H are together. JHS JHL JHM 3/10 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8188 Location: Pune, India Re: Five A-list actresses are vying for the three leading roles [#permalink] ### Show Tags 28 Jan 2014, 21:33 1 2 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. _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >

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Re: Five A-list actresses are vying for the three leading roles  [#permalink]

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19 May 2015, 05:57
VeritasPrepKarishma wrote:
Rohan_Kanungo wrote:
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.

Quote:
I still don't understand why the Probability of Picking Hallie and Julia is 3*3! and why no of ways of picking any 3 actresses is 10*3!. Could you explain this to me?

Probability = Favorable cases/All cases

Favourable cases are those in which Hallie and Julia are picked. In how many different ways can you pick 3 actresses such that they include Hallie and Julia?
You pick Hallie, Julia and then any one of the remaining 3 actresses. So the third actress can be picked up in 3 ways. There are 3 different roles and 3 actresses now. So they can be arranged in 3 roles in 3! ways. So our favourable cases are 3*3!

How many total cases are there? From the 5 actresses, we have to choose any 3. We can do this in 5C3 ways.
5C3 = 5*4*3/3*2*1 = 10 (Using the combination formula)
Now we arrange the 3 actresses in 3 different roles in 3! ways.
So total ways = 10*3!

Probability = 3*3!/10*3! = 3/10
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Re: Five A-list actresses are vying for the three leading roles  [#permalink]

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30 May 2015, 07:09

I calculated this as 1/5 * 1/5* 3/5 ... but this is wrong in official solution they said when first actress is selected it impacts probability for second actreess to get selected.
But in question stem they have not said "one by one selected" ... So i am assuming all three are selected in one stroke .
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Re: Five A-list actresses are vying for the three leading roles  [#permalink]

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30 May 2015, 15:33
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.
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Re: Five A-list actresses are vying for the three leading roles  [#permalink]

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04 Sep 2015, 08:40
Hi all,

When I read the Guide 5 – Number Properties – Fifth edition of Manhattan GMAT, chapter 7 “Extra Combinatorics & Probability”, I found the problem below (in which I have many confusions):

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?

In the problem above: the answer from Manhattan is 3/10. Their explanation is below:

The number of different combinations in which the actresses can be cast in the roles, assuming we are not concerned with which actress is given which role, is 5!/(3!2!) = 5 x2 = 10.
There are 3 possible combinations that feature both Julia and Hallie:
(1) Julia, Hallie, Sally
(2) Julia, Hallie, Meryl
(3) Julia, Hallie, Lauren
Therefore, the probability that Julia and Hallie will star together is 3/10.

My other solving method is below:
If we don’t consider which role for which actress, the probability (prob) for Julia is 3/5 and then for Hallie is 2/4 (Domino-effect). Thus the resulting prob = 3/5 x 2/4 = 3/10
But I am still not convinced at this stage, it should be some kinds of symmetrical situation: that either Julia or Hallie may be chosen first for the role (any of the 3 roles), so the prob should be 3/10 x 2 = 3/5

So anyone could advise me whether my thinking is correct and the answer of Manhattan for this problem is insufficient?
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Re: Probability - one problem from Manhattan word translations problem set  [#permalink]

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01 Jan 2017, 02:03
FortMI wrote:
Hi There,

I am trying to solve below problem and am getting it wrong. However, i dont know what's the problem of my approach. Hope get your help.

Five actresses are vying for the three leading roles in the new film. The actresses are JR, MS, SF, LB, and HS. Assuming that no actress has any advantave in getting any rold, what is the probability that JR and HS wil star in the fim together?

My approach is like below.

First Actress Second Actress Third Actress Probability
JH 1/5 HS 1/4 1/20
HS 1/5 JH 1/4 1/20
JH 1/5 HS 1/3 1/15
HS 1/5 JH 1/3 1/15
JH 1/4 HS 1/3 1/12
HS 1/4 JH 1/3 1/12

Probability = 2*1/20+2*1/15+2*1/12=4/10

Hi

This problem is based on combinatorics. We need to choose different actresses to fill 3 possible roles.

“JR and HS will star in the film together” two role positions are already occupied and we need to choose only one actress from remaining 3 to fill one last vacancy.

3C1

Total number of possibilities to choose 3 actresses from 5 – 5C3

$$\frac{3C1}{5C3} = \frac{3}{10}$$

Hope it helps.

Regards
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Re: Five A-list actresses are vying for the three leading roles  [#permalink]

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08 Jul 2017, 23:39
This is how i solved this problem. Can someone please let me know if i am correct??
So there are a total of 5 stars out of which we have to select any 3.Therefore that can be done in 10 ways. Then since Hallie and Julia must be in the cast the rest one remaining position can be filled by any of the remaining 3 actress. This can be done in 3 ways (1*1*3). Therefore the total probability is 3/10
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Re: Five A-list actresses are vying for the three leading roles  [#permalink]

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09 Jul 2017, 09:36
1
longhaul123 wrote:
This is how i solved this problem. Can someone please let me know if i am correct??
So there are a total of 5 stars out of which we have to select any 3.Therefore that can be done in 10 ways. Then since Hallie and Julia must be in the cast the rest one remaining position can be filled by any of the remaining 3 actress. This can be done in 3 ways (1*1*3). Therefore the total probability is 3/10

Yes, your solution is perfect - it's the solution Karishma posted at the end of the second post in this thread.
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Re: Five A-list actresses are vying for the three leading roles  [#permalink]

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08 Oct 2017, 02:00
Hi all,

I tried to solve the question this way: The probability of choosing Julia or Hallie as the first actress is 2/5. The probability of choosing Julia or Hallie as the second actress is 1/4. So the probability of having them both is 2/5 x 1/4 = 1/10.

Can anyone explain why my answer is wrong? Thanks!
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Re: Five A-list actresses are vying for the three leading roles  [#permalink]

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31 Jan 2018, 00:32
Total number of ways in which actresses could be selected is 5C3 = 10 ways
Keeping two actresses constant, remaining 1 from 3 could be selected in 3C1 = 3 ways
Probability = 3/10
Re: Five A-list actresses are vying for the three leading roles &nbs [#permalink] 31 Jan 2018, 00:32
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