From a theatre troupe of eight members, including Lou

Question

From a theatre troupe of eight members, including Lou, one person will be randomly chosen to play Abel, another person will be randomly chosen to play Barry, and a third person will be randomly chosen to play Caine. What is the probability that Lou will NOT be chosen to play either Abel or Caine?

A) \(\frac{1}{168}\)

B) \(\frac{1}{8}\)

C) \(\frac{1}{4}\)

D) \(\frac{1}{3}\)

E) \(\frac{3}{4}\)

A)  \frac{1}{168}

B)  \frac{1}{8}

C)  \frac{1}{4}

D)  \frac{1}{3}

E)  \frac{3}{4}

Bunuel · Accepted Answer

The probability that Lou plays Abel OR Caine, is simply 1/8 + 1/8 = 1/4. Therefore the probability that Lou will NOT play either Abel OR Caine is 1 - 1/4 = 3/4.

Answer: E.

sahildrift · Answer

"one person will be randomly chosen to play Abel, another person will be randomly chosen to play Barry, and a third person will be randomly chosen to play Caine"

The above statement is NOT clear

Still a solution:
For 8 students we have 3 posts
one post can be occupied by only one member
no two posts can be occupied by the same member

probability(P) for a member getting selected for particular 2 of the 3 posts= 
probable outcomes=1(Caine)+1(Abel)=2
total outcomes=1(Caine)+1(Abel)+1(Barry)+5(non post positions)=8
=2/8=1/4

Probability required for the given case= 1-P=1-1/4=3/4

AKG1593 · Answer

We could also do the following:

We have 7 choices for Abel-Selecting 1=>C(7,1)
Then 6 choices for Caine-
Selecting 1=>C(6,1)
Finally 5 choices+Lou for B=>C(6,1)
Total no. of possible outcomes=C(8,3)
P(Any event)= No. of favourable outcomes/Total no. of possible outcomes.

C(7,1)*C(6,1)*C(6,1)/C(8,3)
=3/4

Ans. E

Please point out mistakes if there are any.

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282552 · Answer

Hi AKG
I got a question in here
I guess you missed a scenario to consider in here.
For ABEL we have 7 options.
Then lets say for B we have 7 options and Lou is not selected for B.
Then in that case we will have only 5 options for C.
But In case Lou is selected for B then we have 6 options available for C
So can you please explain where am I thinking wrong.

m3equals333 · Answer

I believe the equation s/b:

C(7,1)*C(6,1)*C(6,1)/ P (8,3)

Selections (where each person is either A B C or N):
A, B, C, N, N, N, N, N

Permutations/Possible Outcomes:
8P3

xLUCAJx · Answer

You are adding the same fraction is there no requirement to subtract 1 person to make it 1/7 for purpose of no replacement?

ENGRTOMBA2018 · Answer

The formula is P(A or B) = P(A) + P(B) - P(A and B). In this scenario as Lou can not play both A and B at the same time, thus P(A and B) = 0. Probability of Lou playing Abel = P(A) = 1 person out of 8 = 1/8 = Probability of Lou playing Barry = P(B) = 1/8.

Thus Probability of Lou playing Abel or Barry = P(A) + P(B) - P (A or B) = 1/8 + 1/8 - 0 = 2/8.

Thus, the required probability = 1-2/8 = 6/8 = 3/4. Thus E is the correct answer.

mvictor · Answer

my approach..don't know if it's the fastest, but helped me to crack this one in under 2 minutes.

Probability A = 1/8. probability not A = 7/8
probability B = 1/7. probability not B = 6/7
probability C = 1/6. probability not C = 5/6

now..we are asked for not A or C, or 1 - P(A or C)
since we have 3 mutual events, and we have "conditional" probability, then:
P(A or C) = P(A) + P(C | not A and not B)
P(A) is 1/8
P(C)= P(not A) * P(not B) * P(selected for C after A and B not) = 7/8 x 6/7 x 1/6 -> basically cancelling everything to get 1/8
now, P(A or C|not A and not B) = 1/8 + 1/8 = 2/8 = 1/4
since we are asked probability of not A or C, then 1-1/4 = 3/4

E

hwgmat2015 · Answer

Not sure whether my method is right

Probability of A getting selected is 7/8 (As Lou cannot be A)
Probability of B getting selected is 6/7 ( As Lou cannot be B)
Probability of C getting selected is 6/6 ( As Lou can be C)

therefore total probability is 7/8 * 6/7 * 6/6 = 3/4

My correct if my method is wrong.

thanx.

LODD2 · Answer

Using the negation technique

1-P(A or C), where P(A or C) = P(A) + P(C) + P(A and C) 
1-(1/8 + 1/8 + 0) = 3/4 
The reason why P(A and C) is 0 is because Lou cannot play two roles at once, so it is impossible and thus has a 0% chance of happening.

Nups1324 · Answer

Hi Bunuel , But wouldn't the probability for Abel be 1/8 and Caine be 1/6.? Since you choose one for Abel and one for Barry then there are 6 left out of 8. How do you understand that it is with replacement or without replacement.? Usually if you pick an object from a bag or fill a position, we minus that object or position from the total right? Please explain me if I'm missing something. I'm eager to learn how experts and people (who are better than me at P&C) get this right. Tagging others just in case Bunuel is busy or offline. yashikaaggarwal chetan2u GMATinsight IanStewart ScottTargetTestPrep fskilnik nick1816 Thank you

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IanStewart · Answer

The selections here are definitely made without replacement -- one actor cannot play two different roles, from the wording of the question (it says "another actor will play Barry" and "a third actor will play Caine", so we can't use the same actor twice).

Bunuel's solution is still perfect. The question in this thread is the same as this one, mathematically speaking:

There are eight actors. They all line up in a row. The first and third actors in the lineup will be given the roles of Abel and Caine in a play. The rest of the actors will not get a role. If Lou is one of the eight actors, what is the probability Lou does not get a role?

and since 2/8 of the actors get a role, 6/8 = 3/4 of the actors do not, so that is the answer.

Incidentally, the probability that Lou gets the role of Caine is 1/8. Lou is just as likely as any of the other actors to get that role -- they all have a 1/8 chance to get it. Lou only has a 1/6 probability to get that role if you know Lou definitely does not get the roles of Abel or Barry. And you could divide the problem up into several cases (Lou gets the role of A, Lou gets the role of B, Lou does not get either role A or role B) and then you could end up using that "1/6" in a correct solution, but that complicates what can be a simple problem. In fact, if you notice there are 8 actors, and only 2 roles Lou wants to get, you can tell it's not very likely Lou will get one of those roles, and with these answer choices, 3/4 is the only one that is even remotely plausible, so you could even skip the work altogether.

chetan2u · Answer

As Ian has also mentioned, the word play itself should tell you that all characters will be played by different persons. But say, the play somehow allowed the characters to be played by same person, maybe these characters are never together and it is possible to manage all 3 roles, the words ‘another person’ and at third person’ do tell us that we are looking at different people for different roles.

Nups1324 · Answer

Hi IanStewart and chetan2u , Thank you so much for your respective responses. I think I have got it. A logic just struck me. Please confirm whether it's correct or not. So if there are 3 balls in a bag and 3 guys have to pick one ball each randomly without replacement. So for 1st guy the probability is out of 3 balls. For the 2nd guy the probability is out of 2 balls and so on. Right? But when a position is to be filled then the rules are bit different. 3 roles to be filled by 3 guys. So each guy has a probability out of 3 positions. Here we do not subtract the position as we do in the case of the balls because there is no fixed allotment when we talk about positions. Balls can be picked and not replaced but position can't be picked rather it is entitled and so the total outcome is the same for every guy. Each guy has a equal chance for a position. Am I right? Please let me know. Tagging others just in case any or both of you is/are busy. yashikaaggarwal Bunuel GMATinsight ScottTargetTestPrep Thank you

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ScottTargetTestPrep · Answer

From a theatre troupe of eight members, including Lou, one person will be randomly chosen to play Abel, another person will be randomly chosen to play Barry, and a third person will be randomly chosen to play Caine. What is the probability that Lou will NOT be chosen to play either Abel or Caine?

A) \(\frac{1}{168}\)

B) \(\frac{1}{8}\)

C) \(\frac{1}{4}\)

D) \(\frac{1}{3}\)

E) \(\frac{3}{4}\)

A)  \frac{1}{168}

B)  \frac{1}{8}

C)  \frac{1}{4}

D)  \frac{1}{3}

E)  \frac{3}{4}

The probability that Lou plays Abel OR Caine, is simply 1/8 + 1/8 = 1/4. Therefore the probability that Lou will NOT play either Abel OR Caine is 1 - 1/4 = 3/4.

Answer: E.

Hi  Bunuel ,

But wouldn't the probability for Abel be 1/8 and Caine be 1/6.? Since you choose one for Abel and one for Barry then there are 6 left out of 8.

How do you understand that it is with replacement or without replacement.?

Usually if you pick an object from a bag or fill a position, we minus that object or position from the total right?

Please explain me if I'm missing something. I'm eager to learn how experts and people (who are better than me at P&C) get this right.

Discussion:

The reasoning is not correct. Filling positions randomly is not at all different from drawing balls without replacement. If you had 8 balls representing the eight members and if you drew randomly to fill the positions, you would only replace the balls if it was possible for the actors to play more than one role. The question tells us that three separate people must play the three roles; therefore, we can’t make the selection with replacement.

To understand what’s going on here, I think it is helpful to consider the following questions: suppose we are randomly drawing two balls from among 8 balls, one of which is red. What is the probability that the second ball is red? Some students think that the answer should be 1/7 since the second ball is selected from a total of 7 balls; however, the answer is still 1/8. In order to be able to draw the red ball in the second draw, the first draw must be non-red. The probability of drawing non-red in the first draw is 7/8. Assuming the first draw is non-red, the probability of drawing a red in the second draw is 1/7. Thus, the probability that the second ball turns out to be red is 7/8 * 1/7 = 1/8. You can apply this idea to the third, fourth etc. draws and conclude that the probability of drawing a red ball is 1/8 in all those cases.

Let’s solve the question with the above in mind. The probability that Lou is selected for the Abel role is 1/8, and the probability that Lou is selected for the Cain role is also 1/8. Since Lou can't be selected for both roles, the probability that he is selected for the role of Abel or Cain is 1/8 + 1/8 = 1/4. Thus, the probability that he is not selected for either role is 1 - 1/4 = 3/4.

Foreheadson · Answer

I think it can be solved easier.

6 slots for Lou. Neither Abel, nor Cain. So 6C1=6. 
8 slots in total. so 8C1=8. so 6/8=3/4.

Is this way of thinking somehow incorrect?

Hacchuu · Answer

This is the first thing that came to my mind. It may not be super straightforward, though.

If Lou is Barry, he cannot be Caine or Abel
P(Lou is barry) = 7/8 * 1/7 * 6/6 = 1/8 (7 people can be Abel, Lou is Barry, 6 can be Caine)

If Lou plays no role, neither Barry, Caine or Abel
P(Lou plays no role) = 7/8 * 6/7 * 5/6 = 5/8

Total probability of Lou not playing Caine or Abel = P(Barry is Lou) + P(Lou plays no role) = 1/8 + 5/8 = 3/4

dhruva09 · Answer

Solving this case wise

--> C1: Not selected for Abel's role (But selected for role of Barry)

7/8 x 1/7 =  1/8

-->C2: Selected for none of the roles (Since there is no condition post selection of Caine, so none of the roles are allocated)

7/8 x 6/7 x 5/6 =  5/8

C1 + C2 = 1/8 + 5/8

Hence,  3/4

From a theatre troupe of eight members, including Lou

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From a theatre troupe of eight members, including Lou

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