Statistics

[Linnyk]

Hi!

I have a question about probablility. I don't know what the name of this type of statistics is, but here goes:

 

John has three exams coming up, one in Math, one in Physics and one in Biology. He thinks his chances of passing are 0.8 for Math, 0.7 for Physics and 0.3 for Biology. His result on each exam is independent of the others.

 

a) What is the probablility that he passes at least two of the exams?

b ) Given that he has passed two of the exams, what is the probability that he has passed the exam in Math.

 

I would love some help with this, either in the form of a direction to online pages that has examples of this kind of problems with solutions, or if you could aid me with what type of formula I could use. (Sorry, might be writing in "Swenglish" here, but I hope I hope you understand what I mean.)

 

The way I see it there are three possible ways he could pass two exams (M+P, M+B and P+B ). Could I for example multiply the probabilility M*P*B-complement (and for the other two respectively, and then add the answers toghether (for a))?

 

Any help would be much appreshiated!

Edited September 13, 2010 by Linnyk

[timo]

The way I see it there are three possible ways he could pass two exams (M+P, M+B and P+B).

That is the three possible ways to pass exactly two exams. You're asked about the probability to pass at least two exams.

 

Could I for example multiply the probabilility M*P*B-complement (and for the other two respectively, and then add the answers together (for a))?

In principle yes, that's how to do it. But take into account my previous comment.

 

 

I would love some help with this, either in the form of a direction to online pages that has examples of this kind of problems with solutions, or if you could aid me with what type of formula I could use.

I don't think online pages will help you. You are already given an example problem and there's no way of learning that competes with solving problems yourself. I don't think you need aid with what type of formula to use: you are already doing pretty good so far. Just ask for more advice the same way you just did (i.e. also include your ideas how to solve the problem) in case you get stuck with b.

 

Your English is fine, btw (for me as a non-native speaker myself, that is). It's certainly more readable than what some native speakers write. Perhaps consider using a spell-checker for your browser if you want to get rid of some typos.

Edited September 13, 2010 by timo

[Linnyk]

Hi Timo,

thanks for your reply!

YES, I realized now there are four ways to pass at least two exams: he may also pass them all

 

This is what I've got now (I've renamned to math to A, physics to B and biology to C): P(passing at least two exams) = A * B * C^c + A * C * B^c + B * C * A^c + A* B * C = 0.8 * 0.7 * 0.7 + 0.8 * 0.3 * 0.3 + 0.7 * 0.3 * 0.2 + 0.8 * 0.7 *0.3 = 0.674

 

And for b ) (Given that he has passed two of the exams, what is the probability that he has passed the exam in Math?) I'm thinking the anser is 1 - ( B * C * A^c ) = 1 – 0.042 = 0.958 (This seems a high number though.)

 

Does this seem right? (fingers crossed )

[timo]

The answer for b does not look right. It's a bit more tricky but I think you can figure it out with a bit more thinking so I'll not give any real hints.

 

I'll give you a counter-example showing that your result went wrong, though (a somewhat silly one but it'll do the trick, I think):

 

Assume the tossing of two coins:

A ) What is the probability to the heads in the first toss?

B ) Provided you got heads in the first toss, what's the probability to get heads in the 2nd toss.

 

The way you solved your question the answers would be:

A ) the two possibilities to get heads in the first toss are HH and HT, each with 0.25 probability. So the probability for A is 0.5. Perfectly fine so far.

B ) By your calculation, the probability to get heads in the 2nd toss (provided you got heads in the 1st toss) would be [math]1 - {\rm HB}^c = 1 - 0.75 = 0.25[/math], which is obviously wrong.

[Linnyk]

Ok. I see now that I've obviously missed one important thing, and that is using my results from a) - the probability of passing at least two exams, as that is "given" in B... But I'm not sure what to do. I need to divide it I think.

 

 

I'm thinking the answer is 1 - (( B * C * A^c ) / 0.674) = 0.938

 

(Calculating the probability that he's not passed the math exam and subtracting this from 1 (the sum of all possible outcomes)

 

My head is beginning to spin... Am I getting it right?

[timo]

Hi Linnyk,

I'd have to read up a few first principles to give a definite answer which I don't have the time for at the moment. It does look ok in principle. Two comments:

- I am not convinced that the 0.674 is correct. Question B seems to ask for having passed exactly two exams, not at least two exams, which is what you calculated the 0.674 for.

- The way I had done it would have been to take the ratio of the probabilities to pass two exams where one of them is math, and the probability to pass any two exams. In your language [math]\frac{AB^cC+ABC^c}{A^cBC+AB^cC+ABC^c}[/math] (I am assuming exactly two exams here; 2 or 3 would look different but similar, of course). With simple rearranging this does directly lead to what you wrote, which is why I think it should be correct (except possibly for the 0.674, depending how you interpret the question).