Anyone good at maths? Calculating probability

[Monoceros]

Can anyone please help with the following calculation:
143 surveys per year
307 people

I know that the probability of each person being surveyed is 47% (143/ 307)

But what's the probability of each person being surveyed:
Twice
Three times
Four times
Five times per year?

Does the probability just get divided by 2, i.e.
Probability of one person being surveyed twice in a year is 47%/ 2 = 23.5%???

I would be grateful for your help with this calculation.

I think you multiply the probabilities, rather than divide. So if probability is 47% (0.47) for once a year, it would be 0.47 x 0.47 for twice in a year.And 0.47 x 0.47 x 0.47 for three times in a year etc

I hope I’m right!

Yes I agree with the answer above.

I think your base probability is wrong. It assumes that there’s only one recipient per survey. If there is that’s fine. If there are multiple then the probability of them being surveyed in the year increases

Yes that's a good point. Does only one person receive the survey each time?

Surveys are random but each survey specifically targets one person only. It seems almost impossible for anyone to achieve 3, 4 or 5 surveys per year (which - if positive - are accompanied by rewards), hence I'm trying to calculate the probability of this occurring.

Thank you for your help!
So, it seems like probability for achieving:
1 survey - 47%
2 surveys - 22%
3 surveys - 10%
4 surveys - 0.49%
5 surveys - 0.23%

No, your last two are wrong by a factor of 10. 4.9% and 2.3%

Yes decimal point is in the wrong place for the last two. 0.049 translates to 4.9%

I know that the probability of each person being surveyed is 47% (143/ 307)
You're barking up the wrong tree.

The chances of any specific person being picked for the first survey is 1/307.

The chances of the same person being picked for the second survey is also 1/307.

Etc.

Brilliant! Thank you all for your help

Oh dear! My head is spinning

I see your point. I understand that the probability for each survey is 1 in 307. However, what I'm trying to establish is how likely it is for each person to be surveyed 1, 2, 3, 4 and 5 times in a year.

1/307 is the probability of being picked for one particular survey, but there are 143 surveys so the probability of being picked for a survey (ie any survey out of all 143) is 143/307.

But the chances of being picked for exactly 1 survey over the year involves two probabilities, the chances of being picked once and the chances of not being picked 142 times.

It's a deceptively simple question.

The answers I get are slightly different, only slightly but significant. I've done this the hard way (spreadsheet table) for the numbers below, but there's a formula/theorem that you're definitely being asked to use if this is homework and I can't remember what it is.

Also, you need to include the chances of never being surveyed

Number of surveys; probability
0; (a little under 14%, but I've got a problem with my calculation somewhere)
1; 46.6%
2; 21.5%
3; 9.9%
4; 4.5%
5; 2.0%

I think what you're after is the probability that Person A (who is one of the 307 people) is surveyed 0, 1, 2, 3 etc times per year? It's a different question as to whether any of the 307 people get more than one survey.

If the question is to determine the probabilities for Person A then I think the probability of being surveyed is much lower than you might expect - in fact I think it's a 63% chance that they don't get surveyed at all - that is, that on all 143 occasions the person who is chosen to be surveyed is one of the 306 other people. Mathematically that is (306/307)^143 or 306 divided by 307 all to the power of 143.

Their chances of being surveyed just once are (306/307)^142 (1/307) 143. That is, they are part of the 306 unsurveyed on 142 occasions, but are the one chosen on a single occasion. As that could happen in any of the 143 surveys then the probability is multiplied by 143. This is an overall probability of 29% of Person A being surveyed precisely once.

The calculations for 2, 3, 4 etc instances are more complex but follow the same principles. I get to figures of 7% for 2 surveys and 1% for three. After that the probability is close to zero.

Does that help at all?

Thank you DoginATent
No I don't need the exact formula, it's not homework. It's a real life situation from my work. My championship role is to look after survey results and award those who achieved 3, 4 or 5 positive surveys per year. Our year runs from April (like a financial year) and so far no one has achieved even 2 surveys, not to mention 3, 4 or 5. That's why I'm trying to calculate probability, as I think those odds seem unachievable and I'd like to have mathematical proof of that. I think the number of surveys should be increased.

Thank you @GoldenMalicious
My head is spinning again!
I have to say that your results would be helpful to my case (as I'm trying to prove that it's almost impossible to be surveyed 3, 4 or 5 times per year), however I don't quite understand your calculations.

I think @GoldenMalicious is right. Because the surveys are independent. So you have to calculate the chance of NOT getting a survey.

If there are 307 people and 1 survey, the chance that A does not get it is 306/307.

If there are 2 surveys, the chance A does not get it is (306/307)^2

If there are 3 surveys, the chance A does not get it is (306/307)^3

And keep going to 143 surveys

Note the chance that A does get a survey is always 1-(chance she doesn't)

If A has to get 2 surveys, that is (306/307)p143 x (306/307)p142. (Not I'm using p to mean 'raise to the power of' because the Mumsnet formatting ruins * and ^

Thank you @parietal

Something along the lines of doing outbound surveys on customer satisfaction and rewarding staff who get multiple gold stars as feedback from those?

(if that's the case the stats are probably slightly more complicated)

Not quite customer satisfaction survey @DogInATent. These are surveys conducted by our mystery shoppers. They have a quota of surveys to fulfil per year and I'm looking at the probability of our staff being surveyed multiple times a year. Think transport industry/ mystery shoppers checking if announcements are made when required.

If 147 surveys have been done and absolutely no one has been surveyed more than once then there is a problem with the underlying assumptions and some bias creeping in somewhere. You would expect to see some repetition. If all 147 surveys are of unique individuals then something is steering the process away from duplication.

But your suspicions about the reward system are correct. This particular game is very heavily stacked in the house's favour.

You're getting in a muddle because of a lack of clarity about what you're trying to find. Is it the probability of
a) a particular named person
b) any one person
c) each and every person
being surveyed exactly once/twice/three times etc? Or just more than once?

What exactly is important? If it's just 'has anyone been surveyed more than once ' it's probably easier to find the opposite, ie the probability someone isn't surveyed at all or is surveyed exactly once.

If the surveys are allocated randomly, probability of:
1 survey: 143/307
2 surveys: 143/307 * 142/307
3 surveys: 143/307 142/307 141/307
etc.

Arse, the editor turned my asterisks into bold. there should be some asterisks in the 3 surveys line.