Just I need answers for: 2-1, 2-2, 2-3, 2-4, 2-5, 2-8, 2-9, 2-10, 2-11, 2-14, 2-15, 2-16, 2-17, 2-18, 2-21. in the uploaded Questions.

(a) According to DeMorgan’s law, we have    BAAB 
Therefore,
       BAABBABA 
Since,   BAAB  , we have
   BABA  BAAB  AASBBA  )(
Thus, we see that
    ABABA 
(b)      BABAABBA 
According to DeMorgan’s law, we have    BAAB 
   BBAABA 
BBBAABAA 
ABBA  Since }0{},0{  BBAA
Therefore,
    ABBAABBA 
The information in the given formula can be represented with the following notations:
A = {2, 3, 4, 5}
B = {3, 4, 5, 6}
A + B = {2, 3, 4, 5, 6}
AB = {2, 3, 4, 5}
(A + B)AB = {2, 3, 4, 5}
It is given that AB = {Φ}, indicating that A and B are disjoint events. Then BA  .
Therefore, P (A) ≤ P ( B )
Given that
P (A) = P(B) = P (AB)
Using Difference law, we have
ABABandBABA 
Therefore,
    0 ABBAPABBAP
AB is the intersection of two sets. Therefore, if P (A) = P (B) = 1, then
P (AB) = P (A) * P (B) = 1 * 1 = 1
We know that P (A + B) = P (A) + P (B) – P (AB)
P (A + B + C) = P ((A + B) + C)
= P (A + B) + P (C) – P ((A+B)C)
= P (A) + P (B) – P (AB) + P (C) – P ((A+B)C)
= P (A) + P (B) – P (AB) + P (C) – [(P (A) + P (B) – P (AB)) P (C)]
= P (A) + P (B) – P (AB) + P (C) – [(P (AC) + P (BC) – P (ABC)]
= P (A) + P (B) + P (C) – P (AB) – (P (AC) - P (BC) + P (ABC)
Given that BA , we have
P (AB) = P (A) = 1/4
P (B) = 1/3
 
 
4
3
31
41
)(
| 
BP
ABP
BAP
 
 
1
4
4
41
41
)(
| 
AP
ABP
ABP
 
 
)(
\
CP
ABCP
CABP 

 
)(
)(\
CP
BCPBCAP

 BCAP
CP
BCP
\
)(
)(

 CABP \    CBPBCAP \\ --------------------------- (1)
Now, consider
P (ABC) = P (AB\C) P (C)
Using (1), we have
P (ABC)     )(\\ CPCBPBCAP
We know that
 
 
)(
|
1
12
12
AP
AAP
AAP 
 
 
)(
|
21
321
213
AAP
AAAP
AAAP 
Now, expand the conditional probabilities P(An | A1 · · ·An−1), we have
 11121 ...|.).........\()( AAAPAAPAP nn 
  ...