Hello, compufatwa!

Here's the second one . . .

17. If and are disjoint, and , compute:

Results 1 to 4 of 4

- November 2nd 2007, 07:36 AM #1

- Joined
- Oct 2007
- From
- Egypt
- Posts
- 35

## help help

16 show that : p(A)+P(B)-1 ≤ P(A∩B) ≤ P(A)+P(B).

17. If A and B are disjoint P(A) =0.37 , P(B)= 0.44 , compute :

(a) P (A)=0.37

(b) P (B^c)=1-0.44=

(c) P(AUB)=

(d)P(A∩B)=

(e)P(A∩B^c)=

(f) p (A^c∩B^c)=

18. if P(A) =0.59 ,P(B)= 0.3 , P(A∩B)=0.21 compute :

P (AUB)=

P (A∩B^c)=

p (A^cUB^c)=

p (A^c∩B^c)=

19 . show that the conditional probability satisfies the following

if P(B)>0 , then

i. P(A|B)≥ 0 ,

ii. P(B|B)=1 ,

iii.P[ A|B]=∑_(i=1)^∞▒〖p(A|B)〗

for any sequence of disjoint events A1,A2,………..

- November 2nd 2007, 08:07 AM #2

- Joined
- May 2006
- From
- Lexington, MA (USA)
- Posts
- 12,026
- Thanks
- 844

- November 2nd 2007, 09:07 AM #3

- November 2nd 2007, 08:57 PM #4

for the second, i haven't solved yet.. Ü, but i think, it's just the same with the previous post..

or .... not sure..

same process with the fourth, apply de morgan's law..

but since and P(B)>0,

therefore

i dont understand the code for the third one..