K Kitimbo New member Joined Feb 25, 2017 Messages 29 Apr 4, 2017 #1 Given A,B and C are events in the same experiment. Show that P(AUBUC)=P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AnBnC) Forgive me for not showing any working but I don't uderstand this probability no
Given A,B and C are events in the same experiment. Show that P(AUBUC)=P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AnBnC) Forgive me for not showing any working but I don't uderstand this probability no
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,550 Apr 4, 2017 #2 Kitimbo said: Given A,B and C are events in the same experiment. Show that P(AUBUC)=P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AnBnC) Forgive me for not showing any working but I don't uderstand this probability no Click to expand... Has your class covered set-notation statements of probability at all? Have you studied how to throw out double-counted events? Thank you!
Kitimbo said: Given A,B and C are events in the same experiment. Show that P(AUBUC)=P(A)+P(B)+P(C)-P(AnB)-P(AnC)-P(BnC)+P(AnBnC) Forgive me for not showing any working but I don't uderstand this probability no Click to expand... Has your class covered set-notation statements of probability at all? Have you studied how to throw out double-counted events? Thank you!