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Probability question!

qwertykeyboards

studying statistics and this probability question came up, any ideas?

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MathsManiac

This looks like a question about Bayes' theorem. See:
http://en.wikipedia.org/wiki/Bayes'_theorem

However, I think there's a mistake in the question. I strongly suspect that the introduction to part (i) intended to say: "A piece of code is selected at random..."

Do others agree?

MathsManiac

If you accept my amendment to the question, then the answer to part (i) is that the probability of each is just the proportion of the code that each has written.

To do part (ii), apply Bayes' theorem, taking, for example,
A= this code was written by programmer A
B= this code contains errors

Fremen

MathsManiac wrote: »

Do others agree?

Yep. The phrasing of part ii) seems to support this too.

qwertykeyboards

yes i think your definitely right there the question is phrased wrong. part 1 is no problem its part 2 thats giving me grief. ive read up about this Baye theorem but cant fit a formula around the problem. do you know which formula works? thanks for your help.

Fremen

Four expressions you'll need:

What is the probability of finding an error in a randomly chosen piece of code?
Call that P(E)
What is the probability of finding an error in code written by A?
call that P(E|A)
similarly for B and C. Then you should have everything you need.

Use P(E|A)P(A) = P(A|E)P(E)

MathsManiac

If you're still struggling, note that (using the labelling suggested by Fremen):
P(E) must first be caluclated, using:
P(E) = P(E|A)P(A) + P(E|B)P(B) + P(E|C)P(C)
(or equivalently using a tree diagram)

Then you can use the result Fremen mentioned or, to phrase it in the way that's usual when discussing Bayes' theorem:

P(A|E) = P(E|A)P(A) / P(E)

ramanujan

there's always a load of mistakes in these business studiesy type college maths. Makes it hard on the students, know first hand from giving grinds.

agree with the sentiments above.

however imo question is very poorly choosen i mean, say a selection of code is chosen. whats stopping that section of code from being wirtten by more than one of the programmers.

Grudaire

you're in DCU!!!

ehm- for part two multiply across to get the percentage of total code each has with errors
so .55*.02 gives .011 of total code is A's errors,
then .0105 and .005
then the prob of it being A's error is .011/(.011+.0105+.005)

...or use bayes theorem

qwertykeyboards

thanks for all your help. got it sorted!

cliste, yep i am in dcu doing quantitative finance. are u in F+AM or QF? this statistics exam is a load of arse isn't it!!?

Grudaire

qwertykeyboards wrote: »

thanks for all your help. got it sorted!

cliste, yep i am in dcu doing quantitative finance. are u in F+AM or QF? this statistics exam is a load of arse isn't it!!?

F(&A)M alright, for all the teaching the old bat did, didn't think that exam was going to bee that hard:( I mean F*ck, Q2 grand, everything else though was way harder then last years.