Re: Probability Question
P(A(BUC))=P(A)P(BUC)?
The anwser is no.
Here is an example.
Suppose that you have three fair coins and you flip them independently. Then we have all the outcomes (total 8 ) :
{HHH,HHT,HTT,...}, where H - head and T -tail. Therefore the probability of each outcome is 1/8.
Now, define three events:
A=the first flip is the same as the second one (i.e., both heads or both tails);
B=the second flip is the same as the third one;
C=the third flip is the same as the first one;
Obviously, A={HHH,HHT,TTH,TTT}. since the total number of outcomes is 8, P(A)=4/8=1/2. By symmetry, we have
P(B)=1/2
P(C)=1/2
BUC={HHH,HTH,HTT,THH,THT,TTT}. In other words, only HHT and TTH are excluded. Therefore,
P(BUC)=6/8=3/4.
A(BUC)={HHH,TTT}. IN other words, it is a subset of BUC where the first and the second flips are the same. Therefore, P(A(BUC))=2/8=1/4.
However,
P(A)P(BUC)=1/2 * 3/4=3/8.
In other words, P(A(BUC)) and P(A)P(BUC) are not equal.
Added after 20 minutes:
Sorry, I forgot one thing, that is, to check A and B are independent, and A and C are independent. But it can be done easily (yes, they are independent).