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1) A corporation takes delivery of some new machinery that must be installed and checked before 
it becomes operational. The accompanying table shows a manager’s probability assessment for 
the number of days required before the machinery becomes operational. 
Number of days 3 4 5 6 7 
Probability .08 .24 .41 .20 .07 
Let, A be the event “it will be more than 4 days before the merchandise becomes operational” 
and B the event “it will be less than 6 days before the merchandise becomes available” 
Ans: 
Let, 
A denotes that, “it will be more than 4 days before the merchandise becomes operational.” 
B denotes that, “it will be less than 6 days before the merchandise becomes available.” 
A= [5, 6, 7] and B= [3, 4, 5] 
a) P (A)= P (5)+P (6)+P (7) 
=. 41 +. 20 +. 07 
= .68 
The probability of event A = .68 
So, there will be .68 chances that it will be more than 4 days before the machinery becomes 
operational. 
b) P(B) = P(3) + P(4) + P(5) 
= .08+. 24 +. 41 
=. 73 
So, there will be .73 chances that it will be less than 6 days before the machinery becomes 
available. 
c) The symbol of A complement is A . So, A denotes that, it will be at most 4 days before the 
machinery becomes operational. 
d) P ( A ) = P (S) – P (A) 
= 1-.68 
=. 32 
e) The symbol of intersection of events A and B are AÇ B denotes is the set of all the basic 
outcomes that are both in events A and B. So, 
AÇ B = [5, 6, 7] Ç [3, 4, 5] = [5] 
So, it will be 5 days before the machinery becomes available or operational. 
f) (AÇ B) = (A) Ç (B) 
=[5,6,7] Ç [3,4,5] 
= [5] 
 P (AÇ B) = P (5) 
=. 41 
g) The symbol of union of events A and B are AÈ B denotes is the set of all the basic outcomes 
will occur at least one of these two events. So, A È B = [5,6,7] È [3,4,5] = [3, 4, 5, 6, and 7] 
So, it will be at most 7 days before the machine operational or available. 
1
h) P(AÈ B) = P (3)+P(4)+P(5)+P(6)+P(7) 
= .08+.24+.41+.20+.07 
= 1 
i) No. The events A and B are not mutually exclusive. Because they have to something share 
among them. 
j) Yes, The events A and B are collectively exhaustive because, P (A È B)=P (S) = 1 
2) A fund manager is considering investment in the stock of a health care provider. The 
manager assessments of probabilities for rates of return on the stock over the next year are 
summarized in the accompanying table. Let A be the event “Rate of return will be more than 
10%” and B the event “Rate of return will be negative” 
Rate of return Less than- 
10% 
-10% to 0% 0% to 10% 10% to 
20% 
More 
than 20% 
Probability .04 .14 .28 .33 .21 
Ans: 
Let, A = Rate of return will be more than 10% 
B = Rate of return will be negative 
So, A = [10% to 20%, more than 20%] and B = [Less than -10%, -10% to 0%] 
a) P (A) = P (10% to 20%) + P (more than 20%) 
= .33+.21 = .54 
b) P (B) = P(Less than -10%) + P (-10% to 0%) 
= .04+.14 = .18 
c) The symbols of A complement is A . The element have of the event A that are must not have 
in the A . So, A = [Less than -10%, -10% to 0%, 0% to 10% ] 
d) P ( A ) = P(Less than -10%) + P (-10% to 0%) + P (0% to 10%) 
= .04 + .14 + .28 = .46 
e) The symbol of intersection of events A and B are AÇ B denotes is the set of all the basic 
outcomes that are both in events A and B. So, AÇ B = [ ], because nothing to share among 
them. 
f) P(AÇ B) = P( ) = 0 
g) The symbol of union of events A and B are A È B denotes is the set of all the basic outcomes 
will occur at least one of these two events. So, AÈ B = [Less than -10% , -10% to 0% , 10% 
to 20% , more than 20% ] 
h) P (AÈ B) = P(Less than -10%) + P (-10% to 0%) + P (10% to 20%) + P (more than 20%) 
= .04 + .14 + .33 + .21 = .72 
i) Yes. The events A and B are mutually exclusive. Because they have nothing to share among 
them, and their intersection is 0. 
j) No, the events A and B are not collectively exhaustive because, their union is not equal to the 
sample space. So, P (AÈ B) ¹ P (S) ¹ 1 
3) A manager has available a pool of eight employees who could be assigned to a project 
monitoring task. Four of the employees are women and four are men. Two of the men are 
brothers. The manager is to make the assignment at random, so that each of the eight 
employees is equally likely to be chosen. 
Let A be the event “chosen employee is a man.” 
And B the event “chosen employee is of the brothers.” 
2
Ans: 
Let, 
A denotes that, “chosen the employee is a man.” 
B denotes that, “chosen the employee is of the brother.” 
a) P (A) = 4/8 = ½ 
b) P (B) = 2/8 = ¼ 
c) P (AÇ B) = 2/8 = ¼ 
d) P (AÈ B) = P (A) + P (B) – P (A Ç B) 
= ½ + ¼ - ¼ 
= ½ 
4) In Section 3.4, we saw that if pair of events are mutually exclusive, the probability of their 
union is the sum of their individual probabilities. However, this is not the case for the events 
that are not mutually exclusive. Verify this assertion by considering the events A and B of 
exercise 1. 
Ans: The assertion by considering the events A and B of exercise 1 are not mutually exclusive 
because, they have to something share. When the two events are mutually exclusive then, they have 
nothing to share among them. So their intersection will be 0,  AÇ B = 0. And their probability P 
(AÇ B) = 0. But in exercise-1, AÇ B¹ 0, and their probability P (AÇ B) ¹ 0. In this why these two 
events are not mutually exclusive. 
5) A department store manager has monitored the number of complaints received per week 
about poor service. The probabilities for numbers of complaints in a week, established by the 
review, are shown in the table. Let A be the event “There will be at least one component in a 
work.” and B the event, “There will be less than ten components in a work.” 
NUMBER OF 
0 1-3 4-6 7-9 10-12 More than 
COMPLAINTS 
12 
PROBABILITY .14 .39 .23 .15 .06 .03 
Ans: 
Let, 
A denotes that, “at least one component in a work.” 
B denotes that, “less than ten components in a work.” 
 A= [1-3, 4-6, 7-9, 10-12, more than 12] and B= [0, 1-3, 4-6, 7-9]. 
a) P (A) = P (1-3)+P (4-6)+P (7-9)+P (10-12)+P (more than 12) 
= .39+. 23 +. 15 +. 06 +. 03 
= .86 
So, there will be 53% chances to occur the events A in a week. 
b) P (B) = P (0) + P (1-3) +P (4-6) +P (7-9) 
= .14+. 39 +. 23 +. 15 
=. 91 
So, there will be 91% chances to occur the events B in a week. 
c) P ( A ) = 1 – P (A) 
= 1 - .86 
3
= .14 
d) The probability of the union A and B is; 
P (AÈ B) = P (A) + P (B) – P (A Ç B) 
= .86 + .91 - .77 
= 1 
e) The probability of the intersection A and B is: 
P (AÇ B) = P (1-3) + P (4-6) +P (7-9) 
= .39+. 23 +. 15 
= .77 
f) The events A &B are not mutually exclusive because they have something to 
Share. 
g) The events A & B are collectively exhaustive because P (AÈ B) equal to 1. 
6) A corporation receives a particular part in shipments of 100.Research has indicated the 
probabilities shown in the accompanying table for number of defective parts in a shipments. 
NUMBER DEFECTIVE 0 1 2 3 More than 3 
PROBABILITY .29 .36 .22 .10 .03 
Ans: 
Let, A denotes that “there will be less than 3 defective part in a shipment” 
B denotes that “there will be more than 1 defective part in a shipment” 
A =[0,1,2] & B= [2, 3, More than 3] 
a) P (A)=P (0)+ P (1)+ P (2) 
=. 29 +. 36 +. 22 
=. 87 
So, there will be 87% chances to occur the event A in a shipment. 
b) P (A)=P (2)+ P (3)+ P (More than 3) 
=. 22 +. 10 +. 03 
= .35 
So, there will be 35% chances to occur the event B in a shipment. 
c) The events A & B are collectively exhaustive because P (A È B) equal to 1. 
4

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Statistics assignment 4

  • 1. 1) A corporation takes delivery of some new machinery that must be installed and checked before it becomes operational. The accompanying table shows a manager’s probability assessment for the number of days required before the machinery becomes operational. Number of days 3 4 5 6 7 Probability .08 .24 .41 .20 .07 Let, A be the event “it will be more than 4 days before the merchandise becomes operational” and B the event “it will be less than 6 days before the merchandise becomes available” Ans: Let, A denotes that, “it will be more than 4 days before the merchandise becomes operational.” B denotes that, “it will be less than 6 days before the merchandise becomes available.” A= [5, 6, 7] and B= [3, 4, 5] a) P (A)= P (5)+P (6)+P (7) =. 41 +. 20 +. 07 = .68 The probability of event A = .68 So, there will be .68 chances that it will be more than 4 days before the machinery becomes operational. b) P(B) = P(3) + P(4) + P(5) = .08+. 24 +. 41 =. 73 So, there will be .73 chances that it will be less than 6 days before the machinery becomes available. c) The symbol of A complement is A . So, A denotes that, it will be at most 4 days before the machinery becomes operational. d) P ( A ) = P (S) – P (A) = 1-.68 =. 32 e) The symbol of intersection of events A and B are AÇ B denotes is the set of all the basic outcomes that are both in events A and B. So, AÇ B = [5, 6, 7] Ç [3, 4, 5] = [5] So, it will be 5 days before the machinery becomes available or operational. f) (AÇ B) = (A) Ç (B) =[5,6,7] Ç [3,4,5] = [5] P (AÇ B) = P (5) =. 41 g) The symbol of union of events A and B are AÈ B denotes is the set of all the basic outcomes will occur at least one of these two events. So, A È B = [5,6,7] È [3,4,5] = [3, 4, 5, 6, and 7] So, it will be at most 7 days before the machine operational or available. 1
  • 2. h) P(AÈ B) = P (3)+P(4)+P(5)+P(6)+P(7) = .08+.24+.41+.20+.07 = 1 i) No. The events A and B are not mutually exclusive. Because they have to something share among them. j) Yes, The events A and B are collectively exhaustive because, P (A È B)=P (S) = 1 2) A fund manager is considering investment in the stock of a health care provider. The manager assessments of probabilities for rates of return on the stock over the next year are summarized in the accompanying table. Let A be the event “Rate of return will be more than 10%” and B the event “Rate of return will be negative” Rate of return Less than- 10% -10% to 0% 0% to 10% 10% to 20% More than 20% Probability .04 .14 .28 .33 .21 Ans: Let, A = Rate of return will be more than 10% B = Rate of return will be negative So, A = [10% to 20%, more than 20%] and B = [Less than -10%, -10% to 0%] a) P (A) = P (10% to 20%) + P (more than 20%) = .33+.21 = .54 b) P (B) = P(Less than -10%) + P (-10% to 0%) = .04+.14 = .18 c) The symbols of A complement is A . The element have of the event A that are must not have in the A . So, A = [Less than -10%, -10% to 0%, 0% to 10% ] d) P ( A ) = P(Less than -10%) + P (-10% to 0%) + P (0% to 10%) = .04 + .14 + .28 = .46 e) The symbol of intersection of events A and B are AÇ B denotes is the set of all the basic outcomes that are both in events A and B. So, AÇ B = [ ], because nothing to share among them. f) P(AÇ B) = P( ) = 0 g) The symbol of union of events A and B are A È B denotes is the set of all the basic outcomes will occur at least one of these two events. So, AÈ B = [Less than -10% , -10% to 0% , 10% to 20% , more than 20% ] h) P (AÈ B) = P(Less than -10%) + P (-10% to 0%) + P (10% to 20%) + P (more than 20%) = .04 + .14 + .33 + .21 = .72 i) Yes. The events A and B are mutually exclusive. Because they have nothing to share among them, and their intersection is 0. j) No, the events A and B are not collectively exhaustive because, their union is not equal to the sample space. So, P (AÈ B) ¹ P (S) ¹ 1 3) A manager has available a pool of eight employees who could be assigned to a project monitoring task. Four of the employees are women and four are men. Two of the men are brothers. The manager is to make the assignment at random, so that each of the eight employees is equally likely to be chosen. Let A be the event “chosen employee is a man.” And B the event “chosen employee is of the brothers.” 2
  • 3. Ans: Let, A denotes that, “chosen the employee is a man.” B denotes that, “chosen the employee is of the brother.” a) P (A) = 4/8 = ½ b) P (B) = 2/8 = ¼ c) P (AÇ B) = 2/8 = ¼ d) P (AÈ B) = P (A) + P (B) – P (A Ç B) = ½ + ¼ - ¼ = ½ 4) In Section 3.4, we saw that if pair of events are mutually exclusive, the probability of their union is the sum of their individual probabilities. However, this is not the case for the events that are not mutually exclusive. Verify this assertion by considering the events A and B of exercise 1. Ans: The assertion by considering the events A and B of exercise 1 are not mutually exclusive because, they have to something share. When the two events are mutually exclusive then, they have nothing to share among them. So their intersection will be 0, AÇ B = 0. And their probability P (AÇ B) = 0. But in exercise-1, AÇ B¹ 0, and their probability P (AÇ B) ¹ 0. In this why these two events are not mutually exclusive. 5) A department store manager has monitored the number of complaints received per week about poor service. The probabilities for numbers of complaints in a week, established by the review, are shown in the table. Let A be the event “There will be at least one component in a work.” and B the event, “There will be less than ten components in a work.” NUMBER OF 0 1-3 4-6 7-9 10-12 More than COMPLAINTS 12 PROBABILITY .14 .39 .23 .15 .06 .03 Ans: Let, A denotes that, “at least one component in a work.” B denotes that, “less than ten components in a work.” A= [1-3, 4-6, 7-9, 10-12, more than 12] and B= [0, 1-3, 4-6, 7-9]. a) P (A) = P (1-3)+P (4-6)+P (7-9)+P (10-12)+P (more than 12) = .39+. 23 +. 15 +. 06 +. 03 = .86 So, there will be 53% chances to occur the events A in a week. b) P (B) = P (0) + P (1-3) +P (4-6) +P (7-9) = .14+. 39 +. 23 +. 15 =. 91 So, there will be 91% chances to occur the events B in a week. c) P ( A ) = 1 – P (A) = 1 - .86 3
  • 4. = .14 d) The probability of the union A and B is; P (AÈ B) = P (A) + P (B) – P (A Ç B) = .86 + .91 - .77 = 1 e) The probability of the intersection A and B is: P (AÇ B) = P (1-3) + P (4-6) +P (7-9) = .39+. 23 +. 15 = .77 f) The events A &B are not mutually exclusive because they have something to Share. g) The events A & B are collectively exhaustive because P (AÈ B) equal to 1. 6) A corporation receives a particular part in shipments of 100.Research has indicated the probabilities shown in the accompanying table for number of defective parts in a shipments. NUMBER DEFECTIVE 0 1 2 3 More than 3 PROBABILITY .29 .36 .22 .10 .03 Ans: Let, A denotes that “there will be less than 3 defective part in a shipment” B denotes that “there will be more than 1 defective part in a shipment” A =[0,1,2] & B= [2, 3, More than 3] a) P (A)=P (0)+ P (1)+ P (2) =. 29 +. 36 +. 22 =. 87 So, there will be 87% chances to occur the event A in a shipment. b) P (A)=P (2)+ P (3)+ P (More than 3) =. 22 +. 10 +. 03 = .35 So, there will be 35% chances to occur the event B in a shipment. c) The events A & B are collectively exhaustive because P (A È B) equal to 1. 4