Chapter 1 Problems: Probability Calculations and Independence Analysis

In this assignment we will discuss about probabilities and below are the summaries point:-  Problems 1.5.3: Calculated probabilities for events A and B, and their conditional probabilities.  Problems 1.5.7: Examined the probability of hitting a home run and conditional probabilities.  Problems 1.5.9: Analyzed the independence of events A, B, C, and D.

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) = = =
Where,
(c) Let C be the event that the number of heads is even, then C = {HHT,HTH, THH)
Then, P(C) = 3/8
But,
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% .
(a) The probability that the batter will hit a home run on the pitchers next pitch is given by:
P(H) =
P(home run) = 7.4% of the time
(b) P(curve ball | home run ) =
(c) P(Curve ball | No home run) =
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) ;
;
=> Events A and B are dependent ( not independent) because
(b) ; ;
=> Events A and C are independent because

(c) ;
;
=> Events A and D are independent because
(d) ;
;
=> Events C and D are independent because
(e) ; ;
;
=> Events A, C, and D are pairwise independent because and
and .
However, the events are not all independent because
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
Let B be the event that all six balls are blue, then
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:
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

(b)
(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,
1.5.15
1.5.18
The probability that the car is behind door A, B, or C is given by:
The probability that the host opens door B, given that the car is behind A,
The probability that the host opens door B, given the car is behind B,
The probability that the host opens door B, given that the car is behind C,
The probability door B is opened,
(a)
(b)
(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.
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. ;
2. ;
3. ;
Therefore, events B, C, and D are pairwise independent.
However,
4. ;
The Events B, C and D are not mutually independent
Problem B:
Let be the probability that component 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.
a) The probability that system A fails,
b) The probability that system B fails,
c) The probability that system C fails,

d) The probability that the entire system fails,