Statistics and probability

[Letmelooktoo]

In a competition there are 20 competitors.

The winners of the competition are the competitors with the top 4 marks.



The competition is based in 2 parts A and B



Part A counts for 60% of the marks.

Part B counts for 40% of the marks.



A competitor, John achieved 83.166 % in part A and was 4th

1st place achieved a score of 87.333%



In part B the scores are unknown however John was 6th.



By my reckoning all the competitors who beat John in Part A also beat John in Part B.



What are the chances of this happening?

John came 5th overall so did not win a price. Are there any more statistics that can be offered for being 5th?



Regards

[Hymie]

Your reckoning is not necessarily correct – it is possible that none of the competitors who beat John in part A, beat him in part B – with John having finished 6th in part B, and 5th over all. This can be shown by scoring competitors with arbitrary/relevant scores to achieve this.

There is insufficient information within this homework question to calculate the probabilities of over all placings in the competition.

[wildwood]

I can not figure out why you reckon all the competitors who beat John in Part A also beat John in Part B.
The first part could for instance be about film stars and the second about singers, both entertainment but a different type of followers.

[Rev Sermon]

With no other information we must assume that the likelihood of sucess in A and B are unrelated.

The chances of any one of the people who beat John in A also doing so in B is 5/19 = 26.32% as 5 places above John and 19 not accounted for. For all 3 to beat him = 26.32*26.32*26.32=1.82% so it's highly unlikely.

This is based on a distant memory of O level maths so hopefully correct.

[factor-fiction]

Could be , Rev Sermon, although this assumes all have equal random chance of gettinga particular score and takes no account of all the % figures quoted in the question. Is the question as stated meaningless?

[Rev Sermon]

I had another think and in actual fact probability is 5/19 x 4/18 x 3/17 = 1% as for the second person there are now only 18 places unaccounted for and similar for the third person. Sorry to be pedantic.

Factor,

Not sure of the origin of the question but I think the scores achieved are red herrings, for all we know part A was guessing what the score on a dice was so the % scored would be meaningless.

[Letmelooktoo]

This was a true life scenario in relation to a tender exercise.

Part A represented quality
Part B price In my assumption the cheapest price would get 100% ie 40 marks
the 2nd cheapest 95% ie 38 marks so therefore John came 6th and got 75% ie 30 marks.

How would this affect the statistics?

Regards Richard

[factor-fiction]

I still don't think we can give a meaningful answer to this. What point are you hoping to make?

[Rev Sermon]

My earlier answer assumed no relationship between the results in parts A and B.

I would speculate that there are three main factors affecting price ( I may be over simplifying), cost of raw materials, cost of labour, profit margin. It is not unreasonable to say that better raw materials and a more skilled work force cost more but give a better quality product, profit margin is an unknown quantity. This assumption dictates that those scoring higher in A would probably be more expensive and therefore score lower in B. Hence the probability of those beating John in A also beating him in B is even lower than predicted by chance alone. It isn't really possible to put numbers to this as the variables are too complex.

[Letmelooktoo]

Thanks to everybody

The Rev Rermon has some good points. I have no found out that the 4 winners scored higher in part b bearing in mind that this is only for 40% of the marks can this give any statistical meaning?