5 actresses (probability)

rahul.s wrote:5 actresses A, B, C, D and E are vying for the 3 leading roles in a new film. assuming that no actress has any advantage in getting any role, what is the probability that A and B will star in the film together?

[spoiler]OA is 3/10 [/spoiler]

[spoiler]my solution: 1/5 * 1/4 = 1/20. where did i make a mistake?[/spoiler]

So we have o take 2 actress A and B and the third one will be any other
1/4 x1/3 =1/12

shashank.ism wrote:2/5

how did you get 2/5?

rahul.s wrote:5 actresses A, B, C, D and E are vying for the 3 leading roles in a new film. assuming that no actress has any advantage in getting any role, what is the probability that A and B will star in the film together?

[spoiler]OA is 3/10 [/spoiler]

[spoiler]my solution: 1/5 * 1/4 = 1/20. where did i make a mistake?[/spoiler]

Hey rahul.s,
Did u delete ur post after posting for the 1st time??


As for the soln.,
A and B will star together.Now, the remaining 1 role can be selected in 3C1 ways.
Also,3!ways in which roles can be interchanged.

Total no. of ways = 5P3 = 60
Hence, 3x6/60 = 3/10 which is the OA.

I hope you can now see your mistake.

You dont have to consider separately as you did over here.
my solution: 1/5 * 1/4 = 1/20.

TOTAL SELECTION=5C3=10

FAVOURABLE CONDITION=3C1*1(FOR AB COMBINE WE CAN SELECT IN ONE WAY AND REST 1 OUT OF 3 IN 3C1)=3

PRO=3/10

harsh.champ wrote:Hey rahul.s,
Did u delete ur post after posting for the 1st time??

yeah, i deleted it because i thought i could solve it.

now i see where i made a mistake. thanks for your solution

shashank.ism wrote:

rahul.s wrote:5 actresses A, B, C, D and E are vying for the 3 leading roles in a new film. assuming that no actress has any advantage in getting any role, what is the probability that A and B will star in the film together?

[spoiler]OA is 3/10 [/spoiler]

[spoiler]my solution: 1/5 * 1/4 = 1/20. where did i make a mistake?[/spoiler]

So we have o take 2 actress A and B and the third one will be any other
1/4 x1/3 =1/12

shashank ,how did u get 1/3 .Here we are not considering mutually exclusive events.

Though the mistake u committed is very common and can hinder ur GMAT score.

In these questions,just keep in mind to solve by
P(sample space)
P(favorable space)

Probability=P(favorable space)/P(sample space)

Over here it is ,
P(sample space)=5P3
P(favorable space)=3C1x3!

Hence,

Probability=P(favorable space)/P(sample space) = 18/60 = 3/10.


I hope you will find this method handy.

harsh.champ wrote:

rahul.s wrote:5 actresses A, B, C, D and E are vying for the 3 leading roles in a new film. assuming that no actress has any advantage in getting any role, what is the probability that A and B will star in the film together?

[spoiler]OA is 3/10 [/spoiler]

[spoiler]my solution: 1/5 * 1/4 = 1/20. where did i make a mistake?[/spoiler]

Hey rahul.s,
Did u delete ur post after posting for the 1st time??


As for the soln.,
A and B will star together.Now, the remaining 1 role can be selected in 3C1 ways.
Also,3!ways in which roles can be interchanged.

Total no. of ways = 5P3 = 60
Hence, 3x6/60 = 3/10 which is the OA.

I hope you can now see your mistake.

You dont have to consider separately as you did over here.
my solution: 1/5 * 1/4 = 1/20.

oh sorry I thought it was mutually exclusive... yeah harsh your solutin is correct and P=3/10

rahul.s wrote:

harsh.champ wrote:Hey rahul.s,
Did u delete ur post after posting for the 1st time??

yeah, i deleted it because i thought i could solve it.

now i see where i made a mistake. thanks for your solution

Well,thanks for thanking me.But can you plz also click on the "thanks " button

Just joking.

Anyways on a more serious note,
By giving thanks(thru the thanks button) instead of jsut writing in the post,it helps to boost the morale of the user who spent his time putting the problem approach.Don't take it personally.
I am just writing in the interest of the whole BTG community.
Such acts do help in encouraging the users in keeping this community alive.

So,as a gesture I will just thank you in advance for this effort.Just check your above post:P

Hey ,

Could you elaborate further on how u got the highlighted part ?

harsh.champ wrote:

rahul.s wrote:5 actresses A, B, C, D and E are vying for the 3 leading roles in a new film. assuming that no actress has any advantage in getting any role, what is the probability that A and B will star in the film together?

[spoiler]OA is 3/10 [/spoiler]

[spoiler]my solution: 1/5 * 1/4 = 1/20. where did i make a mistake?[/spoiler]

Hey rahul.s,
Did u delete ur post after posting for the 1st time??


As for the soln.,
A and B will star together.Now, the remaining 1 role can be selected in 3C1 ways.
Also,3!ways in which roles can be interchanged.

Total no. of ways = 5P3 = 60
Hence, 3x6/60 = 3/10 which is the OA.

I hope you can now see your mistake.

You dont have to consider separately as you did over here.
my solution: 1/5 * 1/4 = 1/20.

The total number of ways would be

5!/ (3! * 2!)

You would get this because the total number of actresses goes on top and then you set up the anagram

A B C D E

Y Y Y N N

The top represents the 5 actresses. The bottom represents the 3 actresses selected (the Y's) the 2! represents the two actresses not selected (the N's). You get 10 for the total number of favaorable outcomes. To see how many combinations would be winners list them out.

bhumika.k.shah wrote:Hey ,

Could you elaborate further on how u got the highlighted part ?

harsh.champ wrote:

rahul.s wrote:5 actresses A, B, C, D and E are vying for the 3 leading roles in a new film. assuming that no actress has any advantage in getting any role, what is the probability that A and B will star in the film together?

[spoiler]OA is 3/10 [/spoiler]

[spoiler]my solution: 1/5 * 1/4 = 1/20. where did i make a mistake?[/spoiler]

Hey rahul.s,
Did u delete ur post after posting for the 1st time??


As for the soln.,
A and B will star together.Now, the remaining 1 role can be selected in 3C1 ways.
Also,3!ways in which roles can be interchanged.

Total no. of ways = 5P3 = 60
Hence, 3x6/60 = 3/10 which is the OA.

I hope you can now see your mistake.

You dont have to consider separately as you did over here.
my solution: 1/5 * 1/4 = 1/20.

bhumika.k.shah wrote:Hey ,

Could you elaborate further on how u got the highlighted part ?

harsh.champ wrote:

rahul.s wrote:5 actresses A, B, C, D and E are vying for the 3 leading roles in a new film. assuming that no actress has any advantage in getting any role, what is the probability that A and B will star in the film together?

[spoiler]OA is 3/10 [/spoiler]

[spoiler]my solution: 1/5 * 1/4 = 1/20. where did i make a mistake?[/spoiler]

Hey rahul.s,
Did u delete ur post after posting for the 1st time??


As for the soln.,
A and B will star together.Now, the remaining 1 role can be selected in 3C1 ways.
Also,3!ways in which roles can be interchanged.

Total no. of ways = 5P3 = 60
Hence, 3x6/60 = 3/10 which is the OA.

I hope you can now see your mistake.

You dont have to consider separately as you did over here.
my solution: 1/5 * 1/4 = 1/20.

The italics text was given by rahul.s.
The bold-faced text by me.

Which part did you not understand??
Anyways I will explain how I did it.

3 roles(suppose R1 ,R2 and R3 ) are there.
If A and B star together , only 1 role is left .The 3rd actress can be selected among the remaining 3 left in 3C1 ways.
Now, they can also interchange the roles between them in 3! = 6 ways.
So,no. of ways = 6 x 3 =18 .

For total no. of ways ,we can select the 3 roles among 5 actresses in 5C3(5!/3! x 2!) ways.
Also they can interchange the roles betweeen them in 3! ways.
So, total no.. of ways = 5C3 x 3! = 60 ways . [5C3 x 3! = 5P3 = 5! / 2!]
So, the ans. is 18/60 = 3/10.

I hope my soln. approach is clear to you Bhumika or I can elaborate more if you want.

hey thanks i got wht u did. but i wanna know thing. hope u dont get offended .
is there any difference b/w the way u solved it and the way thephoenix solved it ?????

harsh.champ wrote:

bhumika.k.shah wrote:Hey ,

Could you elaborate further on how u got the highlighted part ?

harsh.champ wrote:

rahul.s wrote:5 actresses A, B, C, D and E are vying for the 3 leading roles in a new film. assuming that no actress has any advantage in getting any role, what is the probability that A and B will star in the film together?

[spoiler]OA is 3/10 [/spoiler]

[spoiler]my solution: 1/5 * 1/4 = 1/20. where did i make a mistake?[/spoiler]

Hey rahul.s,
Did u delete ur post after posting for the 1st time??


As for the soln.,
A and B will star together.Now, the remaining 1 role can be selected in 3C1 ways.
Also,3!ways in which roles can be interchanged.

Total no. of ways = 5P3 = 60
Hence, 3x6/60 = 3/10 which is the OA.

I hope you can now see your mistake.

You dont have to consider separately as you did over here.
my solution: 1/5 * 1/4 = 1/20.

The italics text was given by rahul.s.
The bold-faced text by me.

Which part did you not understand??
Anyways I will explain how I did it.

3 roles(suppose R1 ,R2 and R3 ) are there.
If A and B star together , only 1 role is left .The 3rd actress can be selected among the remaining 3 left in 3C1 ways.
Now, they can also interchange the roles between them in 3! = 6 ways.
So,no. of ways = 6 x 3 =18 .

For total no. of ways ,we can select the 3 roles among 5 actresses in 5C3(5!/3! x 2!) ways.
Also they can interchange the roles betweeen them in 3! ways.
So, total no.. of ways = 5C3 x 3! = 60 ways . [5C3 x 3! = 5P3 = 5! / 2!]
So, the ans. is 18/60 = 3/10.

I hope my soln. approach is clear to you Bhumika or I can elaborate more if you want.

bhumika.k.shah wrote:hey thanks i got wht u did. but i wanna know thing. hope u dont get offended .
is there any difference b/w the way u solved it and the way thephoenix solved it ?????

harsh.champ wrote:

bhumika.k.shah wrote:Hey ,

Could you elaborate further on how u got the highlighted part ?

harsh.champ wrote:

rahul.s wrote:5 actresses A, B, C, D and E are vying for the 3 leading roles in a new film. assuming that no actress has any advantage in getting any role, what is the probability that A and B will star in the film together?

[spoiler]OA is 3/10 [/spoiler]

[spoiler]my solution: 1/5 * 1/4 = 1/20. where did i make a mistake?[/spoiler]

Hey rahul.s,
Did u delete ur post after posting for the 1st time??


As for the soln.,
A and B will star together.Now, the remaining 1 role can be selected in 3C1 ways.
Also,3!ways in which roles can be interchanged.

Total no. of ways = 5P3 = 60
Hence, 3x6/60 = 3/10 which is the OA.

I hope you can now see your mistake.

You dont have to consider separately as you did over here.
my solution: 1/5 * 1/4 = 1/20.

The italics text was given by rahul.s.
The bold-faced text by me.

Which part did you not understand??
Anyways I will explain how I did it.

3 roles(suppose R1 ,R2 and R3 ) are there.
If A and B star together , only 1 role is left .The 3rd actress can be selected among the remaining 3 left in 3C1 ways.
Now, they can also interchange the roles between them in 3! = 6 ways.
So,no. of ways = 6 x 3 =18 .

For total no. of ways ,we can select the 3 roles among 5 actresses in 5C3(5!/3! x 2!) ways.
Also they can interchange the roles betweeen them in 3! ways.
So, total no.. of ways = 5C3 x 3! = 60 ways . [5C3 x 3! = 5P3 = 5! / 2!]
So, the ans. is 18/60 = 3/10.

I hope my soln. approach is clear to you Bhumika or I can elaborate more if you want.

Ofcourse not,where is the question of any offence??
This community is there so that we can share our views,help each other and hope for getting 700+ scores in the exam.
So in case of any doubt,JUST BRING IT ON. :P

Now,as for thephoenix's method :-For objective ,it is okay and short but if you are considering the proper approach,you should also conside the permutations(The fact that the 3 roles can also be exchanged between them in 3 ways).
"thephoenix" has only considered the selections not the permutations.
In this question,the 3! cancels in the numerator as well as in the denominator ,so what you see is the final product 3/10.

I had shown the permutations,also so that any community member reading the solution understands how the approach is made.Hence,my method seems lengthier but if you get the knack of the approach you can do it in any manner.

Hope you understand the difference between the methods.Just read both the solutions once again and you will get it. :)

osirus0830 wrote:The total number of ways would be

5!/ (3! * 2!)

You would get this because the total number of actresses goes on top and then you set up the anagram

A B C D E

Y Y Y N N

thanks osirus. i was looking for an mgmat type explanation because they make it seem simple