Is the Formula for the Probability of Symmetric Difference Accurate?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
3 replies · 4K views
GreenPrint
Messages
1,186
Reaction score
0

Homework Statement



Show that the formula: P(AΔB) = P(A) + P(B) - 2P(A[itex]\bigcap[/itex]B)

Homework Equations





The Attempt at a Solution



P(AΔB) = P(A) + P(B) - 2P(A[itex]\bigcap[/itex]B)

P(AΔB) = P(A[itex]\bigcap B^{c}[/itex])[itex]\bigcup (A^{c}\bigcap B)[/itex]

I don't know where to go from here. Thanks for any help.
 
Physics news on Phys.org
GreenPrint said:

Homework Statement



Show that the formula: P(AΔB) = P(A) + P(B) - 2P(A[itex]\bigcap[/itex]B)
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[itex]\bigcap[/itex]B)

P(AΔB) = P(A[itex]\bigcap B^{c}[/itex])[itex]\bigcup (A^{c}\bigcap B)[/itex]

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[itex]\bigcap[/itex]B) ? :wink:
 
GreenPrint said:

Homework Statement



Show that the formula: P(AΔB) = P(A) + P(B) - 2P(A[itex]\bigcap[/itex]B)

Homework Equations





The Attempt at a Solution



P(AΔB) = P(A) + P(B) - 2P(A[itex]\bigcap[/itex]B)

P(AΔB) = P(A[itex]\bigcap B^{c}[/itex])[itex]\bigcup (A^{c}\bigcap B)[/itex]

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

Use Venn diagrams; that's their purpose.