1. ## 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,………..

2. Hello, compufatwa!

Here's the second one . . .

17. If $\displaystyle A$ and $\displaystyle B$ are disjoint, and $\displaystyle P(A) = 0.37,\;P(B) = 0.44$, compute:
$\displaystyle (a)\;P(A)\;=\;0.37$

$\displaystyle (b)\;P(B^c)\;=\;1-0.44\;=\;0.56$

$\displaystyle (c)\;P(A \cup B)\;=\;P(A) + P(B) - P(A \cap B) \;=\;0.37 + 0.44 - 0 \;=\;0.81$

$\displaystyle (d)\;P(A \cap B)\;=\;0$

$\displaystyle (e)\;P(A \cap B^c)\;=\;P(A) \;=\;0.38$

$\displaystyle (f)\;P(A^c \cap B^c)\;= \;P(A \cup B)^c \;=\;1-0.81\;=\;0.19$

3. In any probability space for every event X, $\displaystyle 0 \le P(X) \le 1$.
So $\displaystyle 0 \le P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) \le 1\quad \Rightarrow P(X) + P(Y) - 1 \le P(X \cap Y)$.

And $\displaystyle 0 \le P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)\quad \Rightarrow \quad P(X \cap Y) \le P(X) + P(Y)$.

4. Originally Posted by compufatwa
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)=
$\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.59 + 0.3 - 0.21$
for the second, i haven't solved yet.. Ü, but i think, it's just the same with the previous post..
or $\displaystyle P(A \cap B^c) = P(A) - P(A \cap B)$ .... not sure..
$\displaystyle P(A^c \cup B^c) = P(A \cap B)^c = 1 - P(A \cap B)$
same process with the fourth, apply de morgan's law..

Originally Posted by compufatwa
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,………..
$\displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)}$
but since $\displaystyle P(\cdot) \geq 0$ and P(B)>0,
therefore $\displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)} \geq 0$

$\displaystyle P(B|B) = \frac{P(B \cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1$
i dont understand the code for the third one..