Probability - How many different ways to solve this?

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Q.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?
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by vikram4689 » Thu Jun 16, 2011 2:13 am
IMO [spoiler]3/10[/spoiler]
With Julia and Halle in MOvie , 3C1 is to be used to fill last role. Therefore total chances = 3
Without any restriction total ways of selection are 5C3 = 10

Prob = 3/10
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by winniethepooh » Thu Jun 16, 2011 2:16 am
Thanks Vikram.
Another way?? Any one else?

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by Anurag@Gurome » Thu Jun 16, 2011 2:28 am
winniethepooh wrote:Q.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?
Consider Julia and Halle have been already got the roles. Now, the last role can be filled by any actress from the rest of the three, i.e. in 3C1 = 3 ways. As the roles are not same, for each combination of Julia, Halle, and the third actress we can assign them in the three roles in 3! = 6 ways.

Hence, total number of possible castings where Julia and Halle are starring together = 3*6 = 18

Total number of possible casting = (5C3)*(3!) = 60

Hence, required probability = 18/60 = 3/10
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by GMATGuruNY » Thu Jun 16, 2011 3:26 am
winniethepooh wrote:Q.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?
Another approach:

Let J = Julia, H = Hallie, N = neither Julia nor Hallie.

P(J or H chosen first) = 2/5. (5 actresses, among them both J and H.)
P(J or H chosen second) = 1/4. (4 actresses left, among them either J or H but not both.)
P(N is chosen third) = 3/3. (3 actresses left, neither J nor H among them, since J and H have already been chosen.)

Since the events above must happen together, we multiply the fractions:
2/5 * 1/4 * 3/3 = 1/10.

Since N can be chosen first, second, or third -- giving us three possible variations -- the result above must be multiplied by 3:
3 * 1/10 = 3/10.
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by Brent@GMATPrepNow » Thu Jun 16, 2011 5:52 am
I'll throw out one more approach, but keep in mind that it's a strategy to be used when:
A) You can't solve the question using traditional approaches
and/or
B) You are pressed for time

The strategy is the "gut feeling" approach.

Probability lends itself to this approach, because most of us have an innate feeling about probabilities. Without using any mathematical techniques, we can often get a good idea of the likelihood of an event. For example, if a coin is tossed 3 times, we probably have a rough idea of the likelihood that the coins will turn up heads all 3 times. We know that it's "pretty" unlikely (perhaps the probability is less than 3/10 ??), but we also know that it's not an exceedingly rare event (perhaps the probability is greater than 1/20 ??). Given this, we might be able to eliminate 2 or 3 answer choices in a matter of seconds and then make a reasonable guess.

So, given the actress question, how likely is it that Julia and Hallie will both star in the film ?

Well, does the probability feel greater than 0.5 or less than 0.5? Use this gut feeling to eliminate answers.

Does the probability feel greater than 0.1 or less than 0.1? Eliminate answers. Etc.

Now I'm not suggesting that this is an exact method. However, I know that A LOT of students struggle with probability questions, so it's important that they don't simply abandon a probability question altogether before guessing. In most cases, you should be able to eliminate some answer choices first.

Aside: For this reason, it's always important for posters to include the answer choices.

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by stuffstuff » Thu Jun 16, 2011 5:17 pm
probability of the first event * the probability of the second conditional on the first occuring.

or 3/5 * 2/4

= 6/20
=3/10

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by winniethepooh » Thu Jun 16, 2011 6:12 pm
Beautiful, Brent!!!
Awesome answer!
Actually, I myself don't have the choices. I have direct answer in my book!