P(A)=1/3 P(B)=3/4 P(C)=3/5 is there a difference between

Question

P(A)=1/3
P(B)=3/4
P(C)=3/5

is there a difference between

What is the P(A+B)
What is the P(A+B) but ~P(C)

maratikus · Answer

I'm interpreting your last formula as P(A combined with B but not C) = P((A+B)*(~C)) - let me know if I'm misinterpreting you.

A all students who play football
B all students who play baseball
C all students who play basketball

Then A+B - all students who either play football or baseball P(A+B) = P(A)+P(B)-P(A*B)
(A+B)*(~C) - all students who either play football or baseball but not play basketball

I hope that helps.

bmwhype2 · Answer

ok, just wanted to make sure

What is the P(A+B) = 1/3 * 3/4 = 1/4
What is the P(A+B) but ~P(C) = 1/3 * 3/4 * 2/5 = 1/10

maratikus · Answer

No, P(A+B) = P(A)+P(B) - P(A*B)

In order to calculate a probability that a student plays either football (A) or baseball (B) we should add all those who play football but not baseball (P(A)-P(A*B)), baseball but not football (P(B)-P(A*B)) and both (P(A*B)). Therefore, P(A+B) = P(A)-P(A*B)+P(B)-P(A*B)+P(A*B) = P(A)+P(B)-P(A*B). If we don't subtract P(A*B), we double count those who play both sports.

The second formula would be more difficult, there are some assumptions there that should be considered. For example, if C is independent of A+B, then P((A+B)*(~C))=P(A+B)*P(~C) = P(A+B)*(1-P(C)).

bmwhype2 · Answer

how do we calculate P(A*B) if it is not given?

maratikus · Answer

In general we can't calculate P(A*B) just knowing P(A) and P(B). Only if we know that A and B are independent, we can say that P(A*B) = P(A)*P(B). Let me give you a counter-example if A and B are not independent. Let's say A = B, then P(A*A) = P(A) <> P(A)*P(A) unless P(A) = 1 or P(A) = 0 which is a trivial case.

Let me show you once again why we can't calculate P(A*B) just knowing P(A) and P(B). Let's say you are given the following problem. There are 24 students in a class. 4 play football, 6 play baseball. How many play both football and baseball? Unfortunately, you can't answer such question because information is insufficient. It can be any number from 0 to 4. However, if there is independence of events that a student plays football and baseball, then probability of playing both is 4/24*6/24 = 1/24 which means there is one student who plays both football and baseball.

bmwhype2 · Answer

thanks fior clarifying it

both questions were variations of OG questions. they were both solved the way i solved it, so i assume ETS implies all probabilities are independent events, just as ETS assumes "no replacement" in their probability questions, unless it is otherwise stated

P(A)=1/3 P(B)=3/4 P(C)=3/5 is there a difference between

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