P(A)=1/3 P(B)=0.25 P(AUB)=0.5. Show that A and B are independent and find P(AIB')?

We have P(AΛB)=P(A)+P(B)-P(AUB)<-->

P(AΛB)=1/3+1/4-1/2=1/12

But P(A)*P(B)=1/3*1/4=1/12

So P(A)*P(B)=P(AΛB)<-->

A,B independent

Now P(A/B)=P(A)=1/3

Hi

To prove that A and B are independent, that is no intersection between the two sets 'A' and 'B', you need to prove that their intersection does not exist. So use the following formula:

P(AUB) = P(A) + P(B) + P(A∩B)

So, P(A∩B) = P(AUB) - P(A) - P(B) where [ U = Union and ∩ = intersection ]

Then: P(A∩B) = (1/2) - (1/3) - (1/4) where [ 0.25 = 1/4 , 0.5 = 1/2 ]

...................+6 - 4 - 3...........-1

P(A∩B) = ---------------------- = -------- < 0 , So negative probability is impossible.

.........................12..............12

So P(A∩B) does not exist,so, A and B are independent.

[ Note: Their are some phd degrees in Mathematics concerning negative probability ]

>>> P(AIB') is finding the probability of ' A ' given that ' B' ':

The formula for this conditional probability is:

.................P(A∩B')

P(AIB') = ----------------

....................P(B')

Where: >> P(A∩B') = P(AUB') - P(A) - P(B') [ Use formula as above ]

and P(B') = 1 - P(B) = 1 - 0.25 = 0.75 = 3/4

Sorry i can not complete your question, it is urgent, i need to go, Good Luck Dude.

P(A n B) = P(A) + P(B) - P(AUB) = 4/12 + 3/12 - 6/12 = 1/12 = P(A)P(B) which proves independence.

Since A & B are independent, so are A & B' ==> P(AIB') = P(A)