## Manhattan Guide 4 Probability, Chapter 5 Q11

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zhanpingjin
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### Manhattan Guide 4 Probability, Chapter 5 Q11

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?

I got the same result by solving this problem this way. Am I just lucky or is this way also ok?

My way of doing it: Out of 5 actresses, the probability of Julia getting picked is 3/5. Then out of the remaining 4 actresses, the probablity of Hallie getting picked is 2/4. So the probability that Julia and Hallie will star together is 3/5 * 2/4 = 3/10.
sieb2004
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### Re: Manhattan Guide 4 Probability, Chapter 5 Q11

Your approach looks good and I'm stuck in the same boat.

Total # of ways 3 actresses can be picked out of 5 = 5C3 = 10 ways

If J & H have to be a part of the successful outcomes, then the remining spot can be filled in (5-2) = 3 ways.

Prob (JHx) = # of successful outcomes / Total # of outcomes = 3/10

Please Note: Since the question is about J & H to star together, I treated this as is a combination-probablity problem and did not consider the order of picking J & H relevant.
jnelson0612
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### Re: Manhattan Guide 4 Probability, Chapter 5 Q11

sieb2004 wrote:Your approach looks good and I'm stuck in the same boat.

Total # of ways 3 actresses can be picked out of 5 = 5C3 = 10 ways

If J & H have to be a part of the successful outcomes, then the remining spot can be filled in (5-2) = 3 ways.

Prob (JHx) = # of successful outcomes / Total # of outcomes = 3/10

Please Note: Since the question is about J & H to star together, I treated this as is a combination-probablity problem and did not consider the order of picking J & H relevant.

sleb, your approach is perfect! That is exactly how I teach this problem.
Jamie Nelson
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jnelson0612
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### Re: Manhattan Guide 4 Probability, Chapter 5 Q11

zhanpingjin 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?

I got the same result by solving this problem this way. Am I just lucky or is this way also ok?

My way of doing it: Out of 5 actresses, the probability of Julia getting picked is 3/5. Then out of the remaining 4 actresses, the probablity of Hallie getting picked is 2/4. So the probability that Julia and Hallie will star together is 3/5 * 2/4 = 3/10.

That looks good! That is a very creative way to conceptualize this problem. Nice work!
Jamie Nelson
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mjschlotterbeck
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### Re: Manhattan Guide 4 Probability, Chapter 5 Q11

The second explanation for this problem (5th ed., pg 126) starts by calculating the number of casting combinations using the anagram method: 5!/3!2! = 10 combinations. Then it lists the 3 possible combinations that feature Julia and Hallie to arrive at the probability of 3/10.

How can you calculate there are 3 possible combinations that include Julia and Hallie without listing them out?

I tried building a new anagram using the 3 spots, assuming that two of the spots need to go to either Julia or Hallie in order to get a desired outcome: YYN. In which case 3!/2! = 3. Is this the correct way to do the calculation? Is there an alternative way to calculate the 3 desired outcomes?
jnelson0612
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### Re: Manhattan Guide 4 Probability, Chapter 5 Q11

mjschlotterbeck wrote:The second explanation for this problem (5th ed., pg 126) starts by calculating the number of casting combinations using the anagram method: 5!/3!2! = 10 combinations. Then it lists the 3 possible combinations that feature Julia and Hallie to arrive at the probability of 3/10.

How can you calculate there are 3 possible combinations that include Julia and Hallie without listing them out?

I tried building a new anagram using the 3 spots, assuming that two of the spots need to go to either Julia or Hallie in order to get a desired outcome: YYN. In which case 3!/2! = 3. Is this the correct way to do the calculation? Is there an alternative way to calculate the 3 desired outcomes?

Regarding the three possible combinations involving Julie and Halle, you just need to think to yourself that if you choose those two you still need one more actress for three total. You can then combine the pair of J and H with any one of the remaining three actresses; thus, there are three combinations that include J and H.
Jamie Nelson
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mjschlotterbeck
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### Re: Manhattan Guide 4 Probability, Chapter 5 Q11

Thanks Jamie*. Guess I was just trying to make it much harder than needed.
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### Re: Manhattan Guide 4 Probability, Chapter 5 Q11

Joe Lucero
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arjun.malhotra
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### Re: Manhattan Guide 4 Probability, Chapter 5 Q11

Out of curiosity, the way I attempted to solve this was through chain. Though I seem to have erred right at the end.

3x chains,

J or H (assume J) * H * Other (i)
Other, J or H , H (ii)
J or H , Other, H (iii)

2/5 * 1/4 * 3/3 = 6/60
3/5 * 2/5 * 1/5 = 6/60
2/5 * 3/4 * 1/3 = 6/60

added = 18/60 = 3/10 (as per answ in book)

The mistake I made was to multiply each chain by 2, as there are two possibilities for order in each of the above chains,
So on chain (i) J or H (assume J), H, Other OR J or H (assume H), J, Other
Hence (6/60*2)*3 which led to the original answer I got, 3/5

Is there any reason why order H, J or J, H along the same probability chain does not matter here as it does in the answer solution where chance of success is broken down so as to identify each actress individually, 1/5 * 1/4 instead of probability of 'success', selecting either one of them 2/5 * 1/4.

Thank you
jnelson0612
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### Re: Manhattan Guide 4 Probability, Chapter 5 Q11

arjun.malhotra wrote:Out of curiosity, the way I attempted to solve this was through chain. Though I seem to have erred right at the end.

3x chains,

J or H (assume J) * H * Other (i)
Other, J or H , H (ii)
J or H , Other, H (iii)

2/5 * 1/4 * 3/3 = 6/60
3/5 * 2/5 * 1/5 = 6/60
2/5 * 3/4 * 1/3 = 6/60

added = 18/60 = 3/10 (as per answ in book)

The mistake I made was to multiply each chain by 2, as there are two possibilities for order in each of the above chains,
So on chain (i) J or H (assume J), H, Other OR J or H (assume H), J, Other
Hence (6/60*2)*3 which led to the original answer I got, 3/5

Is there any reason why order H, J or J, H along the same probability chain does not matter here as it does in the answer solution where chance of success is broken down so as to identify each actress individually, 1/5 * 1/4 instead of probability of 'success', selecting either one of them 2/5 * 1/4.

Thank you

Hi arjun,
It appears that you've identified something important here, which is that order does NOT matter. We just need to have combinations of J, H, and someone else. I don't care of J is picked first for the movie, if H is picked first, or if the third member of the group is picked first. I need all three together as a group.

You are right that multiplying 1/5 * 1/4 * 1 gives the probability of a particular order. By doing that I'm using the slot method:
1/5 chance of picking Julia
(I pick Julia. Four people are left)
1/4 chance of picking Halle
(I pick Halle. Three people are left)
3/3 chance of picking a third person (we don't care about who this is)

1/5 * 1/4 * 1 = 1/20

When you use the slot method you are assuming that order matters, and the slot method does give you the probability of a specific order. What if I want to use the slot method but order actually does NOT matter? If that's the case, I need to adjust my number of outcomes. I do that by multiplying the result by the factorial of the number of slots. In this case I have three slots, so I multiply by 3!, or 6. Thus 1/20 * 6 = 3/10, the answer.

I hope that this makes more sense. Please let us know if you have more questions.
Jamie Nelson
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KaranS496
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### Re: Manhattan Guide 4 Probability, Chapter 5 Q11

I understand that we can use 5!/3!2! to get the total number of combinations possible without any restrictions (= 10).

Then, this is where I get it wrong - I try using two anagram to calculate the number of combinations with J and H. In one, J and H are grouped, and another, H and J are grouped as one option.

(H+J) M S L =========> 4!/2!2! = 6.
Y Y N N

(J+H) M S L =========> 4!/2!2! = 6.
Y Y N N

Add them together: 6+6 = 12. So, probability of picking J+H together is 12/10 (which, of course, is wrong! ).

Can someone please tell me where I'm going wrong? What is the correct way of doing this whole problem with combinatorics?

Thank you in advance - I'll appreciate the help!

- Karan
RonPurewal
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### Re: Manhattan Guide 4 Probability, Chapter 5 Q11

what you're calculating appears to be the number of ways of picking ANY pair of things out of those groups—whether "h+j" is actually in that pair or not.
in other words, you are not limiting yourself to choices THAT ACTUALLY CONTAIN THE THING YOU WANT.

__

but... MUCH MORE IMPORTANTLY

um...

you are using ridiculously complex formulas, to try to calculate the number of outcomes IN A SITUATION WHERE THERE ARE FEWER THAN 10 OUTCOMES.

...is that an intelligent way to set priorities / make decisions?

of course it isn't.

there are only 10 possibilities in the entire universe! with such a small number of possibilities, it's a total waste of your time to do anything other than simply MAKING A LIST, and COUNTING the elements in the list.
(not to mention a thousand million zillion times harder... consider the fact that you got totally stuck on the reasoning when you tried.)

here are all the ways to pick 3 people:
JHM
JHS
JHL
JMS
JML
JSL
HMS
HML
HSL
MSL
...and that's it.
10 possibilities. (if it takes you more than 20 seconds to make this list... practice until it doesn't.)

of those 10 possibilities, exactly 3—just the first three in the list—have both "J" and "H" in them. so, 3/10.
you can see that it's rather silly to throw formulas at something like this.