Proof of probability formula involving AuB

rpatel · #1 $Old$ November 3rd, 2009, 04:02 PM

I am stuck in proving that if two events A and B are independent then :

P(AuB)=P(A)+P(A')P(B)

I am stuck with this question i know that if two events are independent then P(AnB)= P(A)P(B)

Can someone please help

thanks

matheagle · #2 $Old$ November 3rd, 2009, 04:17 PM

P(AUB)=P(A)+P(B)-P(AB)

use indep....

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

=P(A)+P(B)[1-P(A)]

=P(A)+P(B)P(A').

rpatel · #3 $Old$ November 3rd, 2009, 04:20 PM

OK i think i have got the answer

P(AuB)=P(A)+P(B)-P(A)P(B)
P(AuB)=P(A)+P(B)(1-P(A)
P(AuB)=P(A)+P(B)P(A')

rpatel · #4 $Old$ November 3rd, 2009, 04:21 PM

ok thank you for your help as i was typing my answer in you had already replied with your answer. Thanks for helping me out though.