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# What are the odds?

Mon 27 Jan, 2014 04:56 am
There are 1000 students in a college. 100 of the students are in the Honors Program. Of the ten students that you know, what are the odds that at least one of them is in the honors program?

Thanks!
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Type: Question • Score: 2 • Views: 3,284 • Replies: 28
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markr

2
Mon 27 Jan, 2014 01:01 pm
@Nerms,
There are C(1000,10)-C(900,10) ways to select 10 students such that at least one is in the honors program. There are C(900,10) ways to select 10 students such that none are in the honors program. Therefore the odds that at least one of the ten is in the honors program are C(1000,10)-C(900,10) to C(900,10).
C(n,r) is n! / (r! * (n-r)!)
markr

2
Mon 27 Jan, 2014 01:24 pm
@markr,
Nerms

1
Mon 27 Jan, 2014 01:43 pm
@markr,
Thank you so much for the explanation! It was tremendously helpful. I've been out of school for over a decade, and I was unable to figure this out by sifting through probability websites.

Thanks!!!
0 Replies

Don Fox

1
Thu 20 Mar, 2014 02:41 pm
@Nerms,
This question is incorrectly stated. We need to know if the selection of students you know is random and mutually independent. If not, then we cannot say what the odds are. Your friends may be selected on the basis that they are similar in ability to you, for example.
0 Replies

HesDeltanCaptain

1
Thu 13 Aug, 2015 06:25 am
@markr,
Unless he really likes people who study hard, or dislikes them in which case he may know all of them, or none at all. (flips and catches his monkey wrench impish grin on his face)
0 Replies

puzzledperson

0
Thu 31 Dec, 2015 04:21 am
@Nerms,
100 out of 1000 is 1 out of 10. That means if you pick just one student at random out of a thousand, you have a 1 in 10 chance of picking an honor student. That is equivalent to rolling one 10-sided die with the hope of rolling the numeral "1" (say) instead of any of the other nine numerals.

If you have ten, 10-sided dice, each of which has a 1 in 10 probability of rolling the numeral "1", and you roll all ten dice simultaneously, the probability of at least one die rolling a "1" is 100%. Which is not to say that any of the dice will roll a "1", only that the mathematical probability is 100%.

puzzledperson

0
Thu 31 Dec, 2015 04:54 am
@puzzledperson,
P.S. To make this even clearer, imagine you have a lotto machine with 1000 balls, of which 900 are white and 100 are red. If you set the machine to draw off 10 random balls, through 10 individual holes simultaneously, what is the probability of getting at least one red ball?

1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 + 1/10 +1/10 + 1/10 = 1

The question as asked says that you already know ten students. That is equivalent to a simultaneous draw.
0 Replies

ekename

1
Wed 24 Feb, 2016 10:05 pm
@puzzledperson,
Quote:
If you have ten, 10-sided dice, each of which has a 1 in 10 probability of rolling the numeral "1", and you roll all ten dice simultaneously, the probability of at least one die rolling a "1" is 100%. Which is not to say that any of the dice will roll a "1", only that the mathematical probability is 100%.

Are you a betting man?
0 Replies

maxdancona

2
Wed 24 Feb, 2016 10:54 pm
Markr and Don are each correct (depending on how one reads the question). PuzzledPerson is clearly incorrect.
maxdancona

1
Wed 24 Feb, 2016 11:03 pm
@Nerms,
There is an easier way to look at this... we can calculate the odds that NO friends are in the Honors program and then subtract from 1.

The odds that friend number one is not in the honors program is 900/1000.
The odds that friend number two is not in the honors program is 899/1000.
The odds that friend number three is not in the honors program is 898/1000.
... and so on and ...
The odds that friend number ten is not in the honors program is 891/1000

So the odds that none of the 10 friends are in the honors program is the product of all is ...

I get 0.332 meaning that there is a 65.8 % chance that at least one of them is in the honors programs.

ekename

1
Wed 24 Feb, 2016 11:50 pm
@maxdancona,
The short cut you've identified is useful when you want to calculate the probability of one or more successes.

P(X≥1)=1−P(X=0) where

P(X = 0) = C(100, 0)C(900,10)/C(1000,10) = 0.3469

or

(900/1000)x(899/999)x(898/998)x ...(891/991) = 0.3469

"Therefore the odds that at least one of the ten is in the honors program are C(1000,10)-C(900,10) to C(900,10)."

1.88:1 becomes 1/(1+1.88) or 0.3469 after adjusting for rounding error.
0 Replies

puzzledperson

0
Thu 25 Feb, 2016 12:19 am
@maxdancona,
maxdancona wrote: "PuzzledPerson is clearly incorrect."

That's an assertion, not a refutation.

You have 1000 cents, of which 100 are dated before the Second World War, and you mix them thoroughly, then lay them out. Now you extend ten fingers and lay them simultaneously and blindly on 10 of the coins. What are the odds that at least one of them is pre-war?

Each finger, considered individually, has a 1 in 10 chance of resting on a pre-war coin, since out of 1000, 100 (or one-tenth) are pre-war. Since you have ten fingers, you have ten chances, each with a 1/10 probability of resting on a pre-war cent. That gives you a 10/10 = 100% probability that at least one of your fingers will select a pre-war cent. You may not pick one in actuality, but those are the odds.

Now what's wrong with that, specifically?
Kolyo

1
Thu 25 Feb, 2016 02:39 am
@puzzledperson,
The ten events you're talking about are not mutually exclusive and therefore, even though each has a probability of .1, they do not exhaust the sample space. That, specifically, is what's wrong with your argument.
maxdancona

1
Thu 25 Feb, 2016 07:12 am
@puzzledperson,
PuzzledPersion, the refutation to your flawed logic is trivial. You are obviously wrong, and it is very simple to see that you are wrong there are millions of ways to go about showing that. Let's take just one.

By your own logic there is a non-zero chance that if you randomly put down 9 fingers they will all be on non-pre-war cents. Now there is one finger left. According you your logic... with the 891 non-pre-war cents left you claim there is a 0% chance that the 10th finger will randomly land on one of them. This is clearly wrong.

I also gave you the correct answer, which in itself is a refutation to your incorrect answer.
ekename

1
Fri 26 Feb, 2016 10:28 pm
@maxdancona,
Quote:
I also gave you the correct answer, which in itself is a refutation to your incorrect answer.

Did you?

Although you are a lot closer to the correct answer than puzzledperson.
maxdancona

1
Fri 26 Feb, 2016 11:00 pm
@ekename,
Yes, you are right. I made a mistake. You already fixed it for me. Thanks.

puzzledperson

0
Sat 27 Feb, 2016 10:53 am
@maxdancona,
I claim nothing of the sort. Your misrepresentation of my argument is a silly straw-man argument, prefaced by ad hominem whose sole purpose is to distract the reader from the error you made and which you refuse to admit. Your English is also ungrammatical (..."according you your logic ") -- which is another sign of your defective and malfunctioning state (as is your irrational hostility).

maxdancona wrote: "By your own logic there is a non-zero chance that if you randomly put down 9 fingers they will all be on non-pre-war cents."

Not true. By my logic there is a 9/10 chance that at least one finger will rest on a pre-war cent. That's all.

maxdancona continued: "Now there is one finger left. According you your logic... with the 891 non-pre-war cents left you claim there is a 0% chance that the 10th finger will randomly land on one of them."

Who says that there are 891 pre-war cents left? Not I: nor is it implied by my argument. Obviously there is a non-zero chance that an additional finger will land on a pre-war cent.

You see, I can patiently dissect the nonsensical blather you've employed in an attempt to obscure and confuse the matter.

0 Replies

puzzledperson

0
Sat 27 Feb, 2016 11:34 am
@Kolyo,
Kolyo wrote: "The ten events you're talking about are not mutually exclusive and therefore, even though each has a probability of .1, they do not exhaust the sample space."

Of course the events are not "mutually exclusive". Two events are mutually exclusive if they cannot occur at the same time. Though strictly speaking, the culmination of the procedure I described could just as easily be called one event, since the fingers are part of a single hand and the touching of the cents is simultaneous.

The sample space of an experiment is the set of all possible outcomes of that experiment. Your use of the term appears to be a meaningless non sequitur.

This doesn't appear to be a refutation of my argument at all, but rather nonsensical double-talk. Are you capable of providing a lucid and coherent refutation of my argument which eschews the use of obscure technical lingo?
puzzledperson

0
Sat 27 Feb, 2016 11:45 am
@ekename,
ekename wrote: "Although you are a lot closer to the correct answer than puzzledperson."

I don't remember you offering a refutation of my argument. Merely asserting that I'm wrong does not constitute a refutation.

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