Is the Formula for the Probability of Symmetric Difference Accurate?

AI Thread Summary
The discussion centers on proving the formula for the probability of the symmetric difference of two sets, P(AΔB) = P(A) + P(B) - 2P(A∩B). Participants express confusion about how to proceed with the proof after stating the formula. Suggestions include using Venn diagrams to visualize the areas involved in the calculation. The conversation emphasizes the need for a clear understanding of the definitions and properties of set operations to effectively demonstrate the formula. Overall, the focus is on finding a method to prove the accuracy of the given probability formula.
GreenPrint
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Homework Statement



Show that the formula: P(AΔB) = P(A) + P(B) - 2P(A\bigcapB)

Homework Equations





The Attempt at a Solution



P(AΔB) = P(A) + P(B) - 2P(A\bigcapB)

P(AΔB) = P(A\bigcap B^{c})\bigcup (A^{c}\bigcap B)

I don't know where to go from here. Thanks for any help.
 
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GreenPrint said:

Homework Statement



Show that the formula: P(AΔB) = P(A) + P(B) - 2P(A\bigcapB)
Show that the formula does what?
GreenPrint said:

Homework Equations

Definition of P(AΔB), perhaps?
GreenPrint said:

The Attempt at a Solution



P(AΔB) = P(A) + P(B) - 2P(A\bigcapB)

P(AΔB) = P(A\bigcap B^{c})\bigcup (A^{c}\bigcap B)

I don't know where to go from here. Thanks for any help.
 
Hi GreenPrint! :smile:

Hint: how would you prove area(AΔB) = area(A) + area(B) - 2area(A\bigcapB) ? :wink:
 
GreenPrint said:

Homework Statement



Show that the formula: P(AΔB) = P(A) + P(B) - 2P(A\bigcapB)

Homework Equations





The Attempt at a Solution



P(AΔB) = P(A) + P(B) - 2P(A\bigcapB)

P(AΔB) = P(A\bigcap B^{c})\bigcup (A^{c}\bigcap B)

I don't know where to go from here. Thanks for any help.

Use Venn diagrams; that's their purpose.
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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