Engineering Statistical Methods
(IND203)
Lecture #6
Elements of probability
(Continue)
Additive Rule of Probability
The probability of the union of events A and B is
the sum of the probabilities of events A and B
minus the probability of the intersection of events
A and B:
Additive Rule for Mutually Exclusive Events
Events A and B are mutually exclusive if A∩ B contains no
simple events.
If two events A and B are mutually exclusive, the probability
of the union of A and B equals the sum of the probabilities of
A and B:
Three or More Events:
P(A∪ B∪C) = P(A)+ P(B)+ P(C) − P(A∩ B) −P(A∩C) −
P(B∩C)+ P(A∩ B∩C)
Independent Events
Events A and B are independent if the occurrence of B does not alter the
probability that A has occurred, i.e., events A and B are independent if
When events A and B are independent, it will also be true that
Or
P(A ∩ B) = P(A)P(B)
Events that are not independent are said to be dependent.
Multiplicative Rule of Probability:
Multiplicative Rule for Independent Events
If events A and B are independent, the probability of the intersection of A and
B equals the product of the probabilities of A and B, i.e.,
Similarly, if A, B, and C are mutually independent events (all pairs of events are
independent), then the probability that A, B, and C all occur is
P(A ∩ B ∩ C) = P(A)P(B)P(C)
Multiplicative Rule for Independent Events
Conditional Probability
A conditional probability of an event is a probability obtained with the
additional information that some other event has already occurred. P(B | A)
denotes the conditional probability of event B occurring, given that event A
has already occurred, and it can be found by dividing the probability of
events A and B both occurring by the probability of event A:
To find the conditional probability that event A occurs given that event B
occurs, divide the probability that both A and B occur by the probability that B
occurs, that is,
The conditional probability of event B, given that event A has occurred is:
Notice that, in this form, you need to know P(A ∩ B)!
Conditional Probability
If there are n total outcomes,
P(A) = (number of outcomes in A) / n
Also,
P(A∩ B) = (number of outcomes in A∩ B) / n
Consequently,
Therefore, P(B|A) can be interpreted as the
relative frequency of event B among the trials that
produce an outcome in event A.
Example 9:
If the probability that a communication system will have
high fidelity is 0.81 and the probability that it will have high
fidelity and high selectivity is 0.18, what is probability that a
system with high fidelity will also have high selectivity?
Solution
If A is the event that a communication system has high
selectivity and B is the event that it has high fidelity, we
have P(B) = 0.81 and P(A∩B) = 0.18, and substitution into the
formula yields
Example 10:
A system contains two components, A and B, connected in series as
shown in the following diagram.
The system will function only if both
probability that A functions is given
probability that B functions is given by
and B function independently. Find the
functions.
components function. The
by P(A) = 0.98, and the
P(B) = 0.95. Assume that A
probability that the system
Solution Since the system will function only if both components function, it
follows that
P(system functions) = P(A ∩ B)
= P(A)P(B) by the assumption of independence
= (0.98)(0.95) = 0.931
Example 11:
A system contains two components, C and D, connected in
parallel as shown in the following diagram.
The system will function if either C or D functions. The probability that C
functions is 0.90, and the probability that D functions is 0.85. Assume C
and D function independently. Find the probability that the system
functions.
Solution Since the system will function so long as either of the two components
functions, it follows that
P(system functions) = P(C ∪ D)
= P(C) + P(D) − P(C ∩ D) = P(C) + P(D) − P(C)P(D)
by the assumption of independence
= 0.90 + 0.85 − (0.90)(0.85) = 0.985
Example 12:
Consider the data on wafer
contamination and location
in the sputtering tool shown
in Table 3-1. Assume that
one wafer is selected at
random from this set.
Let A denote the event that
a wafer contains four or
more particles, and
let B denote the event that
a wafer is from the center of
the sputtering tool
Determine:
Solution
3. Random Variables
Probability Distributions
Example 1:
Construct a probability distribution for rolling a
single die.
Solution
Since the sample space is 1, 2, 3, 4, 5, 6 and each
outcome has a probability of, the distribution is as
shown.
Example 2:
Represent graphically the probability distribution
for the sample space for tossing three coins.
Solution
The values that X assumes
are located on the x axis,
and the values for P(X)
are located on the y axis.
The graph is shown in Figure 1.
Example 3:
The number of games played in each series is
represented by the variable X. Find the probability
P(X) for each X, construct a probability
distribution, and draw a graph for the data.
Solution
The probability P(X) can be computed for each X by
dividing the number of games X by the total.
The probability distribution is
The graph is shown in Figure 4–2.
Probability Distributions
Example 4:
Determine whether each
probability distribution.
distribution
is
a
The mean, variance, and standard
deviation for a probability distribution
Mean
Example 5:
Find the mean of the number of spots that appear
when a die is tossed.
Solution
In the toss of a die, the mean can be computed
thus.
Variance and Standard Deviation
Example 6:
A box contains 5 balls. Two are numbered 3, one is
numbered 4, and two are numbered 5. The balls
are mixed and one is selected at random. After a
ball is selected, its number is recorded. Then it is
replaced. If the experiment is repeated many
times, find the variance and standard deviation of
the numbers on the balls.
Solution
Let X be the number on each ball. The probability
distribution is
Solution
The mean is
The variance is
The standard deviation is
Solution
The mean, variance, and standard deviation can also be
found by using vertical columns, as shown.
mean
variance
standard deviation