Finite Mathematics

Question

Assume that there are 5 rats in the cage: 3 trained and 2 untrained. A rat is removed from the cage and it is noted whether or not it is trained. It is then placed in a different cage. 2 more rats are removed and treated the same way. How many outcomes are possible for this experiment?

Answer 1

It depends on how you look at the problem.

Suppose all 5 rats have different names.

You are choosing 3 rats from a group of 5 rats.

5C3=5!/3!(5-3)!

5C3=5*4/2*1

5C3=20/2=10 outcomes

 

 

If the rats are only trained or untrained, then...

If you are only concerned with how many are trained and how many are untrained, then there are 3 outcomes.

TTT,  TTU,  TUU, 

 

If you are concerned with the order in which a trained rat and an untrained rat is selected, then there are 7 outcomes.

TTT,  TTU,  UTT,  TUT,  TUU,  UUT,  UTU

Answer 2

Remember that this is WITHOUT replacement, so the rats that are removed are not placed back into the original cage.  There is a 3/5 chance that the first rat is trained, 2/5 that it is untrained.  Remember that you have removed a rat, so now there are 4 rats left in the original cage.  If you have a 3/5 chance that the first rat is trained, what is the probability that the second rat is trained, now that there are 4 total rats now in the original cage?  What you must do is to take each "round" and separately calculate the probability and then add them together.  Remember that a rat can be trained OR untrained, but not both, therefore it is additive probability.  It might be good to do a decision tree for this one to show all the outcomes starting with rats being split into trained and untrained rats and then go from there.