• Oct 31st 2007, 11:21 AM
compufatwa
1. P(A)=0.6 , P(B)=0.8 , P(A∩B)=0.5 ,compute :
i. p(A^C∩B^C) =(1-0.6)+(1-0.8) - (1-0.5)
=0.4+0.2-0.5
=0.1

i. P(B∩A^C) =[P(B)+(1-P(A))]
= 0.8+(1-0.6)
=0.8+0.4
=1.2

ii. P(A∩B^C) = P(A) – P(A∩B)
= 0.6-0.5
= 0.1

iii. P(A^CUB^C) =1-P(A∩B)
=1- 0.5
=0.5
• Oct 31st 2007, 12:06 PM
ThePerfectHacker
Do we know if these events are independent?

\$\displaystyle P(A^C\cup B^C) = P(A\cap B)^C = 1 - P(A\cap B) \$

\$\displaystyle P(B\cap A^C) = P(B^C \cup A)^C = 1 - P(B^C\cup A)\$
• Oct 31st 2007, 12:20 PM
compufatwa
is that rule is right (i mean the signs )
P(AUB) = P(A)+P(B)-P(A intesection B)
• Oct 31st 2007, 06:02 PM
kalagota
Quote:

Originally Posted by ThePerfectHacker
Do we know if these events are independent?

\$\displaystyle P(A^C\cup B^C) = P(A\cup B)^C = 1 - P(A\cup B) = 1 - P(A) - P(B) + P(A\cap B)\$

\$\displaystyle P(B\cap A^C) = P(B^C \cup A)^C = 1 - P(B^C\cup A)\$

sir, i just noticed it, for the first one, should it be

\$\displaystyle P(A^C\cup B^C) = P(A\cap B)^C\$
• Oct 31st 2007, 07:39 PM
ThePerfectHacker
Quote:

Originally Posted by kalagota
sir, i just noticed it, for the first one, should it be

\$\displaystyle P(A^C\cup B^C) = P(A\cap B)^C\$

Okay I fixed it now.