Statistics 630 Assignment 2: Coin Flip Probabilities, Baseball Hits, and Urn Ball Colors

In this assignment we will discuss about statistics and below are the summaries point:- In problem 1.5.3, the probabilities of different events involving coin flips are calculated. Problem 1.5.7 deals with the probabilities of a baseball batter hitting a home run based on different pitches thrown by the pitcher. Problem 1.5.10 involves calculating the probabilities of drawing balls of specific colors from different urns.  

1
Statistics 630 - Assignment 2
(due Thursday, January 31, 2019, 8am CST)
Name ______________________________________________________________
Email Address ______________________________________________________
Chapter 1 Problems
1.5.3
Let S be the sample space. Then, S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
(a) Let A be the event that all three coins are heads.
Then A= { HHH }
|A| = 1
Therefore, P (A) = A / S = 1/8
(b) Let B be the event that the number of heads is odd, Then B = { HHH, HTT, THT, TTH}
|B| = 3
P(B) = 4/8 = 1/2
Therefore, the conditional probability, P(A|B) =  
 BP
BAP  = 2/1
8/1 = 4
1
Where,     8/1}{  HHHPBAP
(c) Let C be the event that the number of heads is even, then C = {HHT,HTH, THH)
Then, P(C) = 3/8
But, 0)(  CAP
Therefore, the conditional probability, P(A|C) = 0
1.5.7
Baseball pitcher: Fastballs, P(F) = 80% ; curve balls, P(C) = 20%
Batter: home run; P(H ∩ F) = 8% ; P(H ∩C) = 5% .

2
(a) The probability that the batter will hit a home run on the pitchers next pitch is given by:
P(H) = 074.001.0064.0)2.005.0()8.008.0( 
P(home run) = 7.4% of the time
(b) P(curve ball | home run ) = 6757.0
074.0
)05.0(
)(
)( 

HP
HCP
(c) P(Curve ball | No home run) =     162.0
926.0
15.0
074.01
05.02.0
)()(
)( 





 cc
c
HP
HCPCP
HP
HCP
1.5.9
|S| = 36
A = {11, 22,33,44,55,66} => |A| = 6
B = {66} => |B| = 1
C = {41,42,43,44,45,46} => |C| = 6
D = {14,24,34,44,54,64} => |D| = 6
(a) 6/1
36
6)( AP ; 36/1
36
1)( BP
36/1})66({)(  PBAP ;   216
1
36
1
6
1
)(  BPAP
=> Events A and B are dependent ( not independent) because      BPAPBAP 
(b) 6/1
36
6)( AP ; 6/1
36
6)( CP ; 36/1})44({)(  PCAP
  36
1
6
1
6
1
)(  CPAP
=> Events A and C are independent because      CPAPCAP 
(c) 6/1
36
6)( AP ; 6/1
36
6)( DP
36/1})44({)(  PDAP ;   36
1
6
1
6
1
)(  DPAP
=> Events A and D are independent because      DPAPDAP 

3
(d) 6/1
36
6)( CP ; 6/1
36
6)( DP
36/1})44({)(  PDCP ;   36
1
6
1
6
1
)(  DPCP
=> Events C and D are independent because      DPCPDCP 
(e) 6/1
36
6)( AP ; 6/1
36
6)( CP ; 6/1
36
6)( DP
36/1})44({)(  PDCAP ;
    216
1
6
1
6
1
6
1
)(  DPCPAP
=> Events A, C, and D are pairwise independent because      CPAPCAP  and
     DPAPDAP  and      DPCPDCP  .
However, the events are not all independent because        DPCPAPDCAP 
1.5.10
Urn #1 - 5 red and 7 blue balls
Urn #2 - 6 red and 12 blue balls
3 balls picked uniformly at random from each of the two urns.
Let R be the event that all six balls are red, then
  0011.0
16
4
17
5
18
6
10
3
11
4
12
5 




 




 RP
Let B be the event that all six balls are blue, then
  0429.0
16
10
17
11
18
12
10
5
11
6
12
7 




 




 BP
Therefore, the probability that all six ball are the same color, P(S) = P(R) + P(B) = 0.044
The conditional probability is calculated as:
   
 
 
  025.0
044.0
0011.0
| 

 SP
RP
SP
SRP
SRP

4
1.5.13
Let C1 be the card with red on both sides; P(C1) = 1/3 and P(R1) = 1
Let C2 be the card with black on both sides; P(C2) = 1/3) and P(B2) = 1
Let C3 be the card with red on one side and black on the other side; P(C3) = 1/3, P(R3) = 1/2 and P(B3)
= 1/2
(a) If P® is the probability that one side on the table is red, then
            2
1
6
1
3
1
2
1
3
1
1
3
1
3311 




 




  RPCPRPCPRP
(b)    
 
 
  3
2
2
1
3
1
| 11
1 

 RP
CP
RP
RCP
RCP
(c) For this experiment, S = {(C1,Red), (C1,Red), (C2,Black), (C2,Black), (C3,Red)(C3,Black)}
|S| = 6
One side being red, R = {(C1,Red), (C1,Red),(C3,Red)}; |R| = 3
Case when one side red for card one, E = {(C1,Red), (C1,Red)}; |E| = 2
Therefore,    
  3
2
6
3
6
2
|1  RP
EP
RCP
1.5.15
   
 
   
     
 
   
   APBAPThen
AP
BP
ABP
APBPABP
BP
AP
ABP
BPABPSuppose









|,
|,
1.5.18
The probability that the car is behind door A, B, or C is given by:       3
1
 CPBPAP
The probability that the host opens door B, given that the car is behind A,   2
1
| AOP B

5
The probability that the host opens door B, given the car is behind B,   0| BOP B
The probability that the host opens door B, given that the car is behind C,   1| COP B
The probability door B is opened,        BBBB OCPOBPOAPOP 
           COPCPBOPBPAOPAP BBB ||| 
2
1
1
3
1
0
3
1
2
1
3
1 




 




 




 
(a)    
 
   
  3
1
2
1
6
1|
| 


B
B
B
B
B OP
AOPAP
OP
OAP
OAP
(b)    
 
   
  3
2
2
1
3
1|
| 


B
B
B
B
B OP
COPCP
OP
OCP
OCP
(c) The result of parts (a) and (b) indicates that the probability of winning will be double when we
switch our choice. First, the probability that the car is behind A, B, or C is equal to 1/3. And we can
choose A, B, or C (there are three outcomes: i, ii, and iii below).
Suppose the car is behind door A.
i) Suppose we choose A.
Then, host will open B or C.
If we switch to B or C, whichever closed door, we will lose.
ii) Suppose we choose B.
Then, host must open C.
If we switch to A, we will win.
iii) Suppose we choose C.
Then, host must open B.
If we switch to A, we will win.
Thus, switching our choice will double the probability of winning.

6
Problem A:
S = { AAA, AAa, AaA, aAA, aaa, aAa, aaA, Aaa} ; |S| = 8
Event B = { AAa, AAA, aaA, aaa}; |B| = 4, thus P(B) = 4/8 = 1/2
Event C = { aAA, AAA, Aaa, aaa}; |C| = 4, thus P(C) = 4/8 = 1/2
Event D = { AaA, AAA, aAa, aaa}; |D| = 4, thus P(D) = 4/8 = 1/2
Now,
1.      4
1
8
2
,  aaaAAAPCBP ;     4
1
2
1
2
1 CPBP
      4
1
 CPBPCBP
2.      4
1
8
2
,  aaaAAAPDBP ;     4
1
2
1
2
1 DPBP
      4
1
 DPBPDBP
3.      4
1
8
2
,  aaaAAAPDCP ;     4
1
2
1
2
1 DPCP
      4
1
 DPCPDCP
Therefore, events B, C, and D are pairwise independent.
However,
4.      4
1
8
2
,  aaaAAAPDCBP ;       8
1
2
1
2
1
2
1 DPCPBP
       DPCPBPDCBP 
The Events B, C and D are not mutually independent
Problem B:
Let   pXP i  be the probability that component i fails
System A represents the path with component 1 and 2, System B represents the path with component
3, and System C represents the path with component 4 and 5.

7
a) The probability that system A fails,
            22
212121 21111 pppXPXPXXPXXPAP cccc 
b) The probability that system B fails,
    pXPBP  3
c) The probability that system C fails,
            22
545454 21111 pppXPXPXXPXXPCP cccc 
d) The probability that the entire system fails,
              34522 4422 ppppppppCPBPAPCBAPSP 
Chapter 2 Problems
2.1.5 (a)
X is an indicator function of BA  .
    










 Bsif
Bsif
sIand
Asif
Asif
sI BA 0
1
0
1
BABA IIIX 
2.1.5 (b)
From the problem, BAifABA  and ABifBBA  . Thus,  BABA ,min
 BABA III ,min 
2.1.5 (c)
We have that, BAifBBA  and ABifABA  . Therefore,  BABA ,max
 BABA III ,max 
2.1.5 (d)
    










 cA
c
A Asif
Asif
sIand
Asif
Asif
sI c
0
1
0
1
If 0,  C
A
IAs and 1AI ; and If 1,  C
A
c IAs and 0AI

8
Thus, for any s, AA II C 1
2.1.8
(a)         10011111  ZYXW
(b)         00112222  ZYXW
(c)         11005555  ZYXW
(d) To determine whether ZW  , we find   13 W and   14 W
So,           15,14,13,02,11  WWWWW and
          15,14,13,02,01  ZZZZZ
Then, for all s,    sZsW 
ZW 
2.2.4
Z = {1,2,3,4,5,6}
Let 43 ZW and ZV 
(a)  wWP 
 220,129,68,31,12,5W
  6
1
 wWP for all w
(b)  6,5,4,3,2,1V
  vallforvVP ,
6
1

(c)       6
1
5,55111  ZWPthenWZ
      6
1
24,2412222  ZWPthenWZ
  xallforxZWPSimilarly ,
6
1
, 

9
(d)       6
1
15,5111  VWPthenWV
      6
1
212,12222  VWPthenWV
  yallforyVWPSimilarly ,
6
1
, 
(e)       6
1
15,5111  WVPthenWV
      6
1
212,12222  WVPthenWV
  rallforrWVPSimilarly ,
6
1
, 
2.3.3
 1,0X ;  5,1Y ;  5,1,0Z
Thus,
  2
1
0 ZP ;   4
1
1 ZP ;   4
1
5 ZP
   5,1,0;
,
4
1
0,
2
1






 
 z
otherwise
zif
zZP
2.3.4
 1,0X ;  1,0Y
(a) Probability function of Z is given by:
YXZ  =>  2,1,0Z
  2
1
0 ZP ;   4
1
1 ZP ;   4
1
2 ZP
   2,1,0;
,
4
1
0,
2
1






 
 z
otherwise
zif
zZP

10
(b) Probability function of W is given by:
XYW  =>  1,0W
  4
3
0 WP ;   4
1
1 WP
   1,0;
,
4
1
0,
4
3






 
 w
otherwise
wif
wWP
2.3.8
Let W ∼ Poisson(λ). For what value of λ is P(W = 11) maximized?
   
 eWP !11
11
11
  0
!11!11
.1111
1110
   
 eeWP
d
d
11 
2.3.10
Let X ∼ Geometric(1/5). Compute P(X2 ≤ 15).
   3152  XPXP
  5904.0
5
4
1
5
1
5
4
3
43
0


















 
x
x
XP
2.3.13
Let X ∼ Hypergeometric(20, 7, 8). What is the probability that X = 3? What is the probability that X
= 8?
N=20; M=7; n=8
=>   3576.0
125970
45045
8
20
38
720
3
7
3 



















XP

11
=>   0
8
20
88
720
8
7
8 



















XP because X>M(7)
2.3.14
(a) We consider each roll to be a success if one of the {2,3,5,6} shows up. Therefore, the rolls are
independent Bernoulli trials with probability of success, 4/6 = 2/3.
)32,20(~ bX
(b)   000142.0
3
1
3
2
5
20
5
155


















XP
2.3.15
(a) Let X be the number of basket sunk (successes) in 10 independent throws.
)35.0,20(~ bX
Thus,           2522.065.035.0
3
10
35.0135.0
3
10
3 733103 











 
XP
(b) Let Y be the number of trows until the first basket is sunk
)35.0(~ GeometircY
Thus,      0047.065.035.011 1010  pqYP
2.3.18
Let tY be the random number of calls in the next t minutes.
(a) If tY is Poisson process with rate, per minute, where 2t ; 2 ; y =5 then
      1562.0
!5
4
!5
5
5
4
5
  e
t
eYP t 
(b) The probability that 5 calls arrive between 0t and 2t is 0.1562, which is equal to the
probability that 5 calls arrive between 2t and 4t
     0244.01562.05 22  YP
(c) We have 10t ; 2 ; y = 0 then       20
0
20
0
!0
20
!0
0   ee
t
eYP t 