Download {1, 2, 3, …, 50}. Consider the f

Probability1_Homework
Answers
1. Let the sample space consist of the integers 1 through 50. {1, 2, 3, …, 50}. Consider the following events
from that Sample Space.
• Event A: {a number is a multiple of 5| 5, 10, 15, …, 50}
• Event B: {a number is odd | 1, 3, 5, …, 49}
• Event C: {a number is a multiple of 7| 7, 14, 21, …, 49}
• Event D: {a number is a multiple of 2| 2, 4, …, 50}
Let us think of the situation as having a 50 sided fair die; any one number is equally likely to appear.
a. Are events A and B disjoint? Not disjoint as they share common values such as 5 and 15 for
example.
b. Are events B and D disjoint? Event D are multiples of 2 which are even numbers, thus they share
no common values with Event B, odd numbers. The two events are disjoint.
c. What is the probability that one observation results in event A AND B occurring? P(A AND B)
•
•
Event B: { 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49}
Event A: {5, 10, 15, 20, 25, 30, 35, 40. 45, 50}
The conjunction AND represents what is common between the two sets.
5
P(A AND B) =
50
d. What is the probability that one observation results in event B AND event D occurring? P(B AND D)
•
Event B: { 1, 3, 5, 7, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49}
•
Event D: {a number is a multiple of 2| 2, 4, …, 50}
Since the two events have nothing in common, then P(B AND D) = 0
e. What is the probability that
•
•
one observation results in event B and event C occurring? P(B AND C)?
Event B: { 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49}
Event C: {a number is a multiple of 7| 7, 14, 21, 28, 35, 42, 49}
The conjunction AND represents what is common between the two sets.
4
P(B AND C) =
50
f. What is the probability that one observation results in either event B occurring, or event D? P(B OR
D)? The two events are disjoint. Thus, it is possible to consider each scenario separately from
each other and then combine the results: P(B OR D) = P(B) + P(D)
25 25
=
+
50 50
=1
The two events are disjoint but represent the whole sample space when put together; basically event
D is the even numbers of that set.
g. What is the probability that one observation results in either event A occurring, or event C? P(A OR
C)? The two events are not disjoint. Thus, it is not possible to consider each scenario
separately from each other and then combine the results: P(A OR C) = P(A) + P(C)
10
7
=
+
50
50
=
17
This is not correct.
50
• Event A: {5, 10, 15, 20, 25, 30, 35, 40. 45, 50}
• Event C: {a number is a multiple of 7| 7, 14, 21, 28, 35, 42, 49}
Because I can just list out all the values and everything is equally likely it is possible to just count
how many values meet the criteria.
P(A OR C) =
16
I count 16 unique objects.
50
The general formula that addresses the problem of two events not being disjoint is
P(A OR C) = P(A) + P(C) – P(A AND C)
=
10
7
1
+
−
50
50 50
16
Notice I added the two probabilities for each event, but doing so has made me
50
count the value 35 twice, thus I subtract the count of the common values (the 35 value) to count the
common values once, not twice.
=
h. What is the probability that one observation results in either event B occurring, or event C? P(B OR C)
The two events are not disjoint. Thus the formula P(B OR C) = P(B) + P(C), does not work. The
fact that they share common outcomes makes the above formula invalid.
Here is then an approach; brute method.
•
•
Event B: { 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49}
Event C: {a number is a multiple of 7| 7, 14, 21, 28, 35, 42, 49}
Count only each “original” number once. So for example 7 appears in both B and C, once I
counted for one group, I cannot count it again. So how many numbers are we considering?
There are 28 numbers.
P(B OR C) =
28
50
If you understand the reason for this brute method, then you understand the meaning of OR,
and basic probability which is good.
What is a more elegant method? Well it is not much less brute, but it opens a brand new door,
that can be then exploited later.
Here is the approach: Use the formula I told you not to use, and fix it.
P(B OR C) = P(B) + P(C)
25
7
+
The problem here is that I am counting the numbers 7, 21, 35, and 49
50
50
twice if I decide to actually add the fractions. In other words, I am stating that 7, 21, 35 and 49
are more likely to occur than any other number which is not true, since the model concept is a
fair 50 –sided die. To fix it, remove those common numbers out, and thus the formula becomes.
=
P(B OR C) = P(B) + P(C) – P(A and B)
25
7
4
=
+
−
50
50 50
28
=
50
i. What is the probability that one observation results in either event A occurring, or event B?
P(A OR B)
•
•
Event B: { 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49}
Event A: {5, 10,. 15, 20, 25, 30, 35, 40. 45, 50}
We can answer the question based on counting how many things meet the criteria A OR B.
30
P(A OR B) =
50
We can also use the general formula for A OR B to guide us which still involves counting.
P(A OR B) = P(A) + P(B) – P(A AND B)
10
25
5
=
+
−
50
50
50
30
=
50
j. A number is chosen at random. It turns out that this number is odd. What is the probability that this
number belongs to set A? P(A | odd number)?
•
•
Odd number: { 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47,
49}
Event A: {5, 10, 15, 20, 25, 30, 35, 40. 45, 50}
Because the situation is simple to visualize we can just count what values meet the criteria.
P(A | odd number) =
5
25
Notice that we are only dealing with odd numbers, thus our sample space consists of 25 numbers,
and 5 of those numbers can be found in event A.
Here is the same thing but using the formula
P(A | odd number) =
P ( A AND odd number )
P ( odd number )
5
= 50
25
50
=
5
25
k. A number is chosen at random. It turns out it belongs to set B. What is the probability it also belongs
to set C? P(C | B)?
•
•
Event B: { 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49}
Event C: {a number is a multiple of 7| 7, 14, 21, 28, 35, 42, 49}
P(C | B) =
4
25
2. Heights of women. The heights of women aged 20 to 29 are approximately Normal with mean 64 inches and standard
deviation 2.7 inches. Men the same age have mean height 69.3 inches with standard deviation 2.8 inches, also normally
distributed. Let the random variable Y determine the height of a woman in inches.
a. What is the probability that a randomly chosen woman is either shorter than 60 inches, or taller than 69 inches?
Let the random variable Y measure the height of a woman.
The events are disjoints thus I can calculate each event separately and add the results.
P(Y < 60 OR Y > 69) = P(Y < 60) + P( Y > 69 )


= PZ <
60 - 64 
69 - 64 

 + PZ >

2.7 
2.7 

= P(Z < -1.48) + P(Z > 1.85)
= normsdist(-1.48) + 1 – normsdist(1.85)
=normalcdf(-10, -1.48) + normalcdf(1.85,10)
= 0.0694 + 0.0322
= 0.1016
Note: P(Z < 1.85) = 0.9678, so
P(Z > 1.85) = 1 - 0.9678 = 0.0322
b. What is the probability that a randomly chosen woman is over 64 inches tall and less than 70 inches tall?
The events are not disjoint, and I will find the region that is common to both.
P(Y > 64 AND Y < 70) = P(64 < Y < 70) Restate the question.
= P(Y < 70) - P(Y < 64) Represents what I will have to do to
answer question


= PZ <
70 - 64 
 - 0.5
2.7 
= P(Z < 2.22) - 0.5
=normsdist(2.22) – 0.5
=normalcdf(-10, 2.22) – 0.5
= 0.9868 – 0.5
= 0.4868
c. What is the probability that a randomly chosen woman is 65 inches or taller and less than 60 inches?
Let the random variable Y be the height of a randomly chosen woman.
P(Y > 65 and Y < 60) = 0, since one person cannot satisfy both criteria simultaneously.
d. Are the events mentioned in “c” disjoint? Yes, since one person cannot satisfy both criteria simultaneously.
e. Let the random variable X measure the height of a male in inches. Calculate P(X > 75.4 OR X < 69.3)
. Men the same age have mean height 69.3 inches with standard deviation 2.8 inches.
P(X > 75.4 OR X < 69.3) = P(X > 75.4) + P(X < 69.3) Events are disjoint so we can separate.


= PZ >
75.4 - 69.3 
69.3 - 69.3 

 + PZ <

2.8
2.8



= P(Z > 2.18) + 0.5
= 1 – normsdist(2.18) + 0.5
Note: P(Z < 2.18) = 0.9854, so
P(Z > 2.18) = 1 - 0.9854 =.0146
= normalcdf(2.18, 10) + 0.5
= 0.0146 + 0.5
= 0.5146
3. I set the random number generator in Excel to generate random numbers from a uniform distribution. The computer
can generate any real numbers between 5 and 30. The random variable X represents all the random numbers generated
by the computer.
σx = 7.22, µx = 17.5
1
25
a. What is the sample space of the random variable X?
5
All the real numbers between 5 and 30 including 5 and 30.
30 X
b. What is the probability that a random number generated by the computer program is either less than 12 or
greater than 22?
P(X < 12 or X > 22) = P(X < 12) + P( X > 22) The two events are disjoint
1
1
= (12 – 5) +
(30 – 22)
25
25
= 0.28 + 0.32
= 0.6
σx = 7.22, µx = 17.5
1
25
5
12
22
30 X
c. What is the probability that a random number generated by the computer is either greater than 15 or less than
20?
P(X > 15 or X < 20) = 1
σx = 7.22, µx = 17.5
1
25
Since, as you can see from the picture X> 15 or X < 20 encompasses
5
15
20
30 X
the entire sample space.
P(X > 15 or X < 20) = P(5 < X < 30)
=1
d. What is the probability that a random number generated by the computer is both less than 15 and more than 29?
P(X < 15 and X > 29) = 0
σx = 7.22, µx = 17.5
1
25
One observation cannot satisfy both events simultaneously.
5
15
29 30 X
e. Are the events mentioned in part “d” disjoint? Yes, one number
generated at random cannot satisfy both events simultaneously.
σx = 7.22, µx = 17.5
1
25
f. P(X < 24 AND X > 20) = P(20 < X < 24) Restate the question
1
= (24 – 20)
25
= 0.16
5
20
24 30 X
4. A coin is tossed 12 times into the air. The random variable X counts the number of times that a coin lands
heads. Write down the sample space of the random variable X.
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
5. A MTH 243 class has 35 students. Out of those 35 students 8, have taken the course the previous term but did
not pass. The instructor for the class will sample 6 students at random and look at their transcripts. Let the
random variable Y count the number of students out of the sample of six that did not pass the previous term.
Write down the sample space of the random variable Y.
{0, 1, 2, 3, 4, 5, 6}
6. A test is created to test if a person has been infected with HIV. The test is an over the counter exam, performed
by the individual. If a person is infected it will detect this 80% of the time. Suppose that five people with HIV
are tested using this exam. Let the random variable H count how many out of the five are correctly identified as
having HIV. Below are the probability values of the random variable H, except for one.
H
P(x)
0
1
2
3
4
0.0003 0.0064 0.0512 0.2048 0.4096
5
a. What is the probability that all of the five are correctly identified?
1 – (0.0003 + 0.0064 + 0.0512 + 0.2048 + 0.4096) = 0.3277
The sum must always equal 1.
b. What is the sample space of the random variable H? {0, 1, 2, 3, 4, 5}
c. What is the probability that out of a sample of five either 1 or 4 people have been detected as having an HIV
infection?
P(H = 1 or H = 4) = P(H = 1) + P(H = 4) Events are disjoint since a sample (one observation will only be
= 0.0064 + 0.4096
able to yield one number.
= 0.416
d. Out of a single sample of five HIV infected people, calculate P(H = 0 AND H = 3)?
P(H = 0 AND H = 3) = 0 since one observation (recall one observation consist of the result of testing five
individuals) cannot satisfy both events simultaneously.
e. Calculate P(H = 2 or H = 4 or H = 5) = P(H = 2) + P(H = 4) + P(H = 5) Events are disjoint.
= 0.0512 + 0.4096 + 0.3277
= 0.7885