Probability Question

gerry87

I'm trying to work something out and it's melting my head a little.

Take a list of 300 companies (for simplicity the companies are called A, B, C... AA, AB, AC...etc)

From this list we randomly select 10.

What is the probability that company A is in the group of 10 companies selected.

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My Attempt:

If there were 5 companies and we want to select 2
A B C D E
then there are 5C2 = 15 combinations -

AB, AC, AD, AE
BC, BD, BE
CD, CE
DE

The probability of any one of those companies being in the two selected is 4/15 or (5-1)/5C2.

So generalising that up choosing r companies out of a group of N, there are NCr combinations.

So the probability of one company being in that group of r is (N-1)/NCr.

I'm not sure if this is right or not since the answer I get seems to small...

299 / (300 choose 10) = 2.13827987 × 10^-16

Intuatively I would have expected the probability of a company being within the 10 chosen to be greater than the probability of picking the company randomly (1/300)...?

Michael Collins

Am I correct in saying you're just trying to calculate the probability of getting 1 specific country out of a list of 300, when you have ten goes?

In that case it's very simple, just find the probability that the specific one you want is either the first company or the second or the third...and so on up to ten, that is:

P(1st or 2nd or 3rd or...10th) = P(1st) + P(2nd) + P(3rd) + ... + P(10th) (by simple rules of probabilities, assuming each pick is independent)

Now all these individual probabilities are equal:
P(1st) = P(2nd) = ... = P(10th) = 1/300

So the answer is simply:

10 * 1/300 = 10/300 = 1/30.

Alternatively, you seem to have gone the route of combinations, it can be done this way two:

P = (favourable combinations) / (possible combinations)
P = 299C9 / 300C10 = 1/30

gerry87

Doh, I always overcomplicate things That's what I had initially, but I thought it wasn't taking everything into account then started doing the gerry-maths, and that never ends well! Thanks.

Lexor

Are you doing this trial with or without replacement and this would be important in how the odds are calculated.

With replacement: the odds of picking out company A is 1/300 and will be for each subsequent trial as our selection will be put back into the pot.

So if your repeating the experiment 10 times, then the probability will be 10 * 1/300 which is equal to 1/30 or 0.0333333 (repeating)

Without replacement: the odds for picking out company A on the first trial will be 1/300, but on the second trial it will be 1/299, third trial 1/298 and so on for each subsequent trial.

The odds this time for 10 trials of the experiment will be:
1/300 + 1/299 + 1/298 + ... + 1/291 which is approx. 0.03384

CJC86

Lexor wrote: »

Are you doing this trial with or without replacement and this would be important in how the odds are calculated.

With replacement: the odds of picking out company A is 1/300 and will be for each subsequent trial as our selection will be put back into the pot.

So if your repeating the experiment 10 times, then the probability will be 10 * 1/300 which is equal to 1/30 or 0.0333333 (repeating)

Without replacement: the odds for picking out company A on the first trial will be 1/300, but on the second trial it will be 1/299, third trial 1/298 and so on for each subsequent trial.

The odds this time for 10 trials of the experiment will be:
1/300 + 1/299 + 1/298 + ... + 1/291 which is approx. 0.03384

Actually, when dealing with no replacement (which is fair to assume upon reading the question), using Michael Collins' notation from above, P(2nd) = P(don't pick A on first attempt)*P(pick A on second attempt) = (299/300)*(1/299) = 1/300. Similarly, P(3rd) = P(don't pick A on first attempt)*P(don't pick A on second attempt)*P(pick A on third attempt) = (299/300)*(298/299)*(1/298) = 1/300, and so on...

In fact, it's the case you've introduced (with replacement) that is the more complicated one, but then it's just the sum of 10 hypergeometric distributions with p=1/300.

I would personally have approached this problem by looking at the probability that it doesn't happen P(don't choose A) = (299/300)*(298/299)*...*(290/291) = 290/300 = 29/30, then P(choose A) = 1-P(don't choose A) = 1/30, but there's usually a few different ways to do probability questions.