probability

[zule]

As I am not happy with my current job, I have decided to take a competitive examination to become a state high school teacher.

 

For one of the test you have to pass, there are 75 units to study. The day of the exam, 5 balls with 5 different numbers will be extracted from a bag. Each number corresponds to a unit and you have to choose from these 5 which one you are going to develop.

 

It is advised to prepare around 30 units to know almost for sure you will have studied one of the 5 between which you have to choose. However, I know several people who say they studied 30 or even more units for previous official announcements and they didn’t get any of those units chosen.

 

So, I wanted to work out the probability that you get one of your chosen units if you study 30 of them, but, as my field is biology and not maths, I am not sure how to do it.

 

It would be possible to know the number of ball combinations you can get from the 75 lessons: combinations of 75 elements taken from 5 in 5 = 17,259,390. But now, I don’t know how to guess how many of these options I would have covered by studying 30 units.

 

Can you help me?

Edited November 14, 2009 by zule
ortography

[Sisyphus]

So you want the probability that at least one of the 5 picked is one of the 30 you have studied? Ok. That's the same as 1-(probability that none of the 30 will be picked). So:

 

first ball: 45/75 chance of not being picked

second: 44/74 chance, assuming the first wasn't one of the 30 (there is one fewer not studied, and one fewer total)

third: 43/73, same reason

fourth: 42/72, same reason

fifth: 41/71, same reason

 

The probability that each will happen in turn is (45/75)*(44/74)*(43/73)*(42/72)*(41/71) = 146611080/2071126800 = about 7%. So if you study 30 there will be about a 93% chance that at least one of them will be picked.

 

Good luck.

[timo]

I am a bit worried that an ongoing high school teacher cannot solve a typical school exercise. But well, perhaps it's actually realistic to assume that few teachers (math teachers aside, of course) are capable of knowing the stuff demanded from their pupils.

 

For the question: One statement that is helpful very often (like in this question) is that the probability of something happening plus the probability of this something not happening is 100%. This then means that the probability P that you know at least one of the units drawn is 100% minus the probability that you know none of the questions drawn: P = 100% - Q. The statement is so practical because the probability that you know none of the units drawn is rather easy to calculate. Assume the units are drawn one-by-one. The probability that you do not know the 1st one is (75-30)/75, i.e. just equal to the ratio of units you have not prepared. Since for calculating Q, you know that the 1st unit drawn must have been one you did not prepare the probability that you then also do not know the 2nd one is (74-30)/74 - the ratio of unprepared units in the remaining pool.

The total probability that one event happens and then another (uncorrelated) event happens is the product of the individual probabilities for the even to happen (e.g. the probability to throw a 1 on a dice twice in a row is 1/6 (for throwing a 1 on the 1st throw) times 1/6 (probability on the 2nd throw) equals 1/36). Same goes for more than two, five in your case, events.

 

That should be enough to calculate it. Sorry for not just giving the results or complete calculation; it really is a typical homework question. Anyways, if you just want the rough number: The probability that when having prepared 30 units you know at least one of the ones drawn is about 90%.

[zule]

Thank you, Sisyphus. I hadn’t though about getting it out this way, but it makes sense.

 

So, I think 7% is too much for letting it to luck. I will try to study more units.