Download HW #3 Example 4.8 – Applying Probability Rule #1 Two dice are

HW #3
Example 4.8 – Applying Probability Rule #1
Two dice are tossed, and the numbers of dots on the upper faces are observed. Find the
probability that the sum of the two numbers is equal to 7.
SOLUTION: Using Rule #1 it is understood that, the probability of any event is the sum of the
probabilities of the outcomes that compose that event. Use two dice (marked die #1 and die #2)
that each contains six different sides with corresponding dots that represent one through six. The
probability is determined by formula: P(7) = 1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 6/36
STEPS:
 Identify the process that generates the outcomes
o Count the possible outcomes associated with rolling two dice: 2 dice, 6 sides per
die equals 36 possible outcomes.
 Recognize a person can roll a 1 on the first die and 2 on the second and vice versa; one
outcome will be the number 2, two outcomes will be the number 3, and so on and so forth
 The probability of rolling any number is the ratio of the number of ways of rolling that
number to the total number of possible outcomes
o 36 possible outcomes from rolling the two dice one time. Therefore, the
probability of rolling a: 3 is 2/36, 4 is 3/36, and so on
 The probability of rolling a 7 is 6/36 = 1/6 = 0.167
This example outlines how rule #1 can be modified when an experiment calls for the use of only
one likely mutually exclusive event.
Example 4.13 – Making an Inference Based on Probability
Experience has shown that a manufacturing operation produces, on the average, only one
defective unit in 10. These are removed from the production line, repaired, and returned to the
warehouse. Suppose that during a given period of time you observe five defective units
emerging in sequence from the production line.
a) Assume Randomness – What is the probability of observing five consecutive defective
units?
b) If the event “a)” occurred, what could be concluded about the process
SOLUTION:
a) The events observed are considered unconditional probabilities and the defectives occur at
random. Therefore, the units are independent of each other and cannot be treated any other
way.
P(all 5 are defective) = (1/10) (1/10) (1/10) (1/10) (1/10) = 1/100,000 = 0.000001
STEPS:
 Define the events:
o D1: first unit is defective/D 2: second unit is defective /D 3: third unit defective/D 4:
fourth unit defective / D 5: fifth unit defective

Determine probability of defective units: 1 defect per 10 units = 1/10
o P(D1) = P(D 2) = P(D 3) = P(D 4) = P(D 5) = 1/10
b) By observing historical data it is easy to understand observing five consecutive defective
units is highly unlikely.
STEPS:
 Examine historical data to find out if anything similar has happened in the past.
Example 4.15
SOLUTION:
Suppose an experiment consists of tossing a pair of coins and observing the upper faces. Define
the following events A: Observe at least one head
B: Observe at least one tail
Use the additive law of probability
P( A or B or both)=P(A U B)= P(A)+P(B) –P(A and B)
1.
P(A)= P(head)= 3/4
2.
P(B)=P(tail)= 3/4
P(A) + P(B)- P(A and B)
The chance of tossing either a head or tail on each coin is ½
(3/4)+(3/4)-(1/2)= 1
Computer Implementation:
The JavaScript supported the above calculations
Example 4.16
SOLUTION:
A) P (A and B) = .14
B) PA = .24 and PB = .39
C)
a. PA + PAbar = 1
b. 1 - .24 = .76
D) The prob that A or B occur can be using the previous solutions:
a. .24 + .39 - .14 = 49
Computer Implementation:
The use of the JavaScript re-enforced the outcomes of the calculations
Example 5.2
SOLUTION:
Construct a relative frequency distribution for the 100,000 values of X
1. To determine the values of relative frequency of sample- this value can be obtained by
dividing the frequency of each x value by the total number of observations
X
Frequency
Relative frequency
0
32891
.32891
1
40929
.40929
2
20473
.20473
3
5104
.05104
4
599
.00599
5
4
.00004
0.45
0.4
0.35
0.3
0
1
0.25
2
0.2
3
0.15
0.1
4
5
0.05
0
2. Show that the properties of a probability distribution are satisfied
To determine sum up all values of X and determine whether distribution is between 0 and 1.
Computer Implementation: The above charts and graphs were created in Excel to illustrate the
findings for part one and two.
Example 5.3
SOLUTION:
Find the mean for example 5.2
Using the formula- find the expected value of X- You find the expected value of X by referring
to table 5.1. X is the observations of X in the sample size of N=5
μ = Sum of each x times P( X = x) = Σ x p(x)
a.
1. Substitute where p(x) is given in table – these values can be obtained by dividing the
frequency by the total number of observations ie x=0 frequency 32,891/100,000- total
number of observations- relative frequency
2. Sum up all the values of x corresponding to the sample
b. μ = 1.0.
i. Interpretation of μ indicates that over a period of time that number of
consumers who favor Brand A will be equal to 1.o. The mean of a random
variable provides the long run average of the variable or expected average
outcome over many observations
c. Find the standard deviation
1. The variance is equal to the [sum of x^2p(x)]- μ^2 .
Substitute the values of the unknown and compute to get the standard deviation
Compute the mean using the formula from part A , square the value of X corresponding
to the sample and solve
μ =sumxp(x), μ = 0(p)0+ 1(p)1+2(p)2
Next square the values of X ie 0(p)0+ 1(p)+ 4(p)2+ 9(p)3 to (1)^2
Now plug in values of p(x) ie. 0(.32768)+ 1(.40960)+4(.20480) to - 1
This gives the variance,
The square root of the value gives the standard deviation.
Example 5.4
SOLUTION:
Empirical rule application μ ±2(standard deviation)
Compute both the mean and standard deviation,
substitute the unknown values for the known values.
Compute and complete the problem.
The interval for the problem show a range of -.788 to 2.788
includes the values=0, x=1,x=2.
Add the values of the relative frequency or probability to determine the interval
P(0)+p(1)+p(2)= .32768+.40960+.20480= .94208.
The value illustrates the empirical rule that 95% will lie within the 2(standard deviation) of the
mean
Example 5.8
SOLUTION:
Summing binomial probabilitiesthat three or more persons in the sample prefer Brand A.
Find the probability of 3 or more observations - when applying probability rule #1
probability of 3 mutually exclusive events is equal to the sum of their probabilities.
Computer Implementation: N/A
Example 5.9
SOLUTION:
Determine if the statement is true:
p0+p1+p2+p3+p4+p5+p6 = .0010
Because the probability is very small the event is likely a rare event, not likely to happen if all
other factors remain constant