calculate the conditional probabilities?

Question

P(a)=1/3 P(b)=1/4 P(a n b)=1/6 whats is ? P(a U b), P(b\a), P(a\b)
P(a^c U b^c ) WHERE ^c = complement
P(b^c \a), P(b\a^c)

Answer 1

Most of these things come from formulas derived directly from the axioms of probability. There are two examples where I don't use the axioms directly. I will provide explanation for those two examples.

P(a U b) = P(a) + P(b) - P(a n b) = 1/3 + 1/4 - 1/6 = 5/12

P(b|a) = P(a n b) / P(b) = (1/6) / (1/4) = 2/3

P(a|b) = P(a n b) / P(a) = (1/6) / (1/3) = 1/2


This next one is DeMorgan's law.

P(a^c U b^c) = P( (a n b)^c ) = 1 - P(a n b) = 1 - 1/6 = 5/6


You will need these next two in a moment. Rather than deriving them from the axioms of probability, I picture the probability space with event subspaces represented by circles A and B that have a small intersection. In this interpretation, probability is equal to area. The intersection of b^c with a is simply the area in the A circle that is NOT in the b^c circle. Similarly, the intersection of a^c with b is the area in the B circle that is NOT in the A circle. So,

P(b^c n a) = P(a) - P(a n b) = 1/3 - 1/6 = 1/6
P(b n a^c) = P(b) - P(a n b) = 1/4 - 1/6 = 1/12


P(a^c) = 1 - P(a) = 1 - 1/3 = 2/3

P( b^c | a ) = P( b^c n a ) / P(a) = (1/6)/(1/3) = 1/2
P( b | a^c ) = P( b n a^c ) / P(a^c) = (1/12)/(2/3) = 1/8

Answer 2

the two Bolts faulty: 2/10 for the 1st bolt, one million/9 for the 2d ( because of the fact between the faulty bolts has already been withdrawn) risk is two/ninety=one million/10*one million/9 the two bolts good : First determination 8/10 2d determination 7/9 (as there are purely 7 good bolts interior the 9 final bolts) risk 7*8/ninety ultimate of success - Mike

Answer 3

See the following web site:

http://www.mathgoodies.com/lessons/vol6/conditional.html

Answer 4

That looks creepily familiar . . wait I know where its from;
It's from a worksheet our math lecturer give out

STOP ASKING OTHERS TO DO YOUR HOMEWORK!!!

Or at least ask them to explain the answer and how they got it otherwise you'll never learn anything.

For anyone actually going to give him the answer he is at Uni he should be able to go and find out how to get the answer himself!!