Math Probability HELP! proof union intersection?

I will use the correct symbol for intersection, ∩

We already know for events A and B

P(A U B) = P(A) + P(B) - P(A ∩ B)

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Show that

P(A U B U C)

= P(A) + P(B) + P(C)

- P( A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)

Proof

Let W = B U C

P(A U W) = P(A) + P(W) - P(A ∩ W)

= P(A) + [ P(B) + P(C) - P(B ∩ C) ] - P( A ∩ (B U C) )

= P(A) + P(B) + P(C) - P(B ∩ C)- P( (A ∩ B) U (A ∩ C) )

= P(A) + P(B) + P(C) - P(B ∩ C)

- [ P(A ∩ B) + P( A ∩ C) - P(A ∩ B ∩ A ∩ C) ]

= P(A) + P(B) + P(C) - P(A ∩ B) - P(A ∩ C) - P(B ∩ C) + P(A ∩ B ∩ C)

QED.

Probability Union

well for the first part you wanna use the Multiplication Rule 2 of probability. this equation is P(A and B)= P(A) x P(B|A). Using this equation we now plug in the numbers we have. P(A)=.4 and P(B|A)=.25/ So now we multiply the two numbers and get .1 as the answer. For the second part P(A union B) is .4 + .3=.7 P(A intersection B) is .4 x .3= .12 P(A/B) is not sure if they just want u to divide here if so then it will be .3/.4=.75 but im not sure about this one

I will consider interseccion A D B just as AB.!!

First prove that P(A u B)= P(A) + P(B) -P(A B)

Remember P of disjoint union is the sum!!

Since A u B= A B^c u B A^c u AB then:

P(A u B)=P(A B^c)+P(B A^c)+P(AB)

Decompose A=A B^c u A B then:

P(A)=P(A B^c)+P(AB)

The same for B:

P(B)=P(B A^c)+P(AB)

Adding you get:

P(A)+P(B)-P(AB)=P(A B^c)+P(B A^c)+P(AB)=P(A u B)

Now use it for A u (B u C) and you get:

P(A u B u C)=P(A)+P(B u C) - P(A (B u C))

Now use it for B u C and you get:

P(A u B u C)=P(A)+P(B)+P(C)-P(BC) - P(A (B u C))

But A(B u C)=AB u AC so use it again:

P(A u B u C)=P(A)+P(B)+P(C)-P(BC) - P(AB)-P(AC)+

P(ABC).

Ready for 3.