IanStewart wrote:
Nups1324 wrote:
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.?
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.
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 ScottTargetTestPrepThank you
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