Engineering 323
Kuszmar
Beautiful Homework Set 3
Problem 2.80
2.80 Before we get started, we might as well define some terms and/or equations that we
might expect to see in this problem. Some of this will be a restatement of what was
already mentioned in Beautiful Homework Solution 2.51.
TERMS
“complement”- The complement of an event A is the set of all outcomes in the sample
space that are not contained in A. Complementary events are mutually exclusive, and the
sum of their probabilities is always equal to 1. (p. 46)
“mutually exclusive”- When events A and B have no outcomes in common, they are said
to be mutually exclusive or disjoint events. In other words, it is impossible for both A and
B to occur at the same time. (p. 47)
“exhaustive”- Events A1, … , An are exhaustive if one Ai must occur, so that
A1 ∪ ⋅⋅⋅∪ An = S, where S is the sample space. A set of exhaustive events completely
defines its own sample space (p. 73)
“beautiful”- that which best describes the quality of the work presented here
EQUATIONS
The Definition of Probability
The probability that an event A will occur, given N number of possible outcomes is
P ( A) =
N ( A)
N
where N(A) is the number of outcomes contained in A. (p. 55)
The Definition of Conditional ProbabilityFor any two events A and B with P(B) > 0, the conditional probability of A given that B
has occurred is defined (p. 69) as
P( A | B )=
P ( A ∩ B)
P ( B)
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Engineering 323
Kuszmar
Beautiful Homework Set 3
Problem 2.80
The Multiplication RuleThe definition of conditional probability leads us directly to
P ( A ∩ B ) = P ( A | B) ⋅P ( B)
The multiplication rule is an algebraic manipulation of the definition of conditional
probability. It may be used to calculate the probability of several stages of individual
events occurring in sequence. (p. 70)
Total Law of ProbabilityIf A1, … , An are mutually exclusive and exhaustive events, then for any other event B,
P ( B) = P ( B | A1 ) P ( A1 ) + ⋅⋅⋅+ P ( B | An ) P ( An )
P ( B) = P ( A1 ∩ B ) ∪ ⋅⋅⋅∪ P ( An ∩ B)
The total law of probability may be used to calculate the probability of a particular event
occurring within a sequence of a number of events. (p.73)
The Definition of Independent EventsTwo events are independent if the occurrence or nonoccurrence of one event has no
bearing on the chance that the other will occur. Therefore, A and B are independent iff
P(A|B) = P(A)
or
P(A∩ B)=P(A)⋅P(B)
If two events are mutually exclusive, they cannot be independent events.
GLOSSARY OF SYMBOLS FOR EXPRESSIONS IN PROBABILITY
When confronted with mathematical expressions in the upcoming solutions, simply
substitute the following words for the symbols indicated, and everything will be just fine!
P(A)
is to be replaced with
“the probability that (event A) occurs”
=
is to be replaced with
“is equal to”
∩
is to be replaced with
“and”
∪
is to be replaced with
“or”
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Engineering 323
Kuszmar
Beautiful Homework Set 3
Problem 2.80
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a. A lumber company has just taken delivery on a lot of 10,000 2x4 boards. Suppose that
20% of these boards (2000) are actually too green to be used in first-quality
construction. Two boards are selected at random, one after the other. Let A = {the
first board is green} and B = {the second board is green}. Compute P(A), P(B), and
P(A ∩ B) (a tree diagram might help). Are A and B independent?
Here’s a recap of what we’ve been given…
A = {the first board is green}
B = {the second board is green}
Using the definition of probability, we may calculate the probability that the first board
selected is green as
P ( A) =
2000
N ( A) # of green boards
=
=
= 0.2
total # of boards
10,000
N
Because there is no replacement in this experiment, the probability that the second board is
green depends on whether or not the first board selected was green. Intuitively, we may
already infer that A and B are not independent! Let’s define A′as the complement of A.
A′= {the first board is not green}
According to the definition of complementary events,
P(A′
) = 1 – P(A) = 1 – 0.2 = 0.8
The definition of probability provides us with information pertaining to the probability that
the second board selected is green. The two possible ways that event B may occur are
(cont.)
Engineering 323
Kuszmar
Beautiful Homework Set 3
Problem 2.80
P ( B | A) =
N ( B | A)
total # of green boards after A
1999
=
=
= 0.19992
N−1
total # of boards after one selection 9999
P ( B | A′
)=
N ( B | A′
)
total # of green boards after A′ 2000
=
=
= 0.20002
N−1
total # of boards after one selection 9999
and
The probability that the second board selected is green may be found by applying the total
law of probability and using everything we’ve defined so far.
P ( B) = P ( B | A) ⋅P ( A) + P ( B | A′
) ⋅P ( A′
)=
1999
2000
⋅0.2 +
⋅0.8 = 0.2
9999
9999
Whew! Now, we can finally calculate the intersection of events A and B. Actually, we
just did that in the solution for P(B)! The probability that the first two boards selected are
green may be found using the multiplication rule.
P ( A ∩ B) = P ( B | A) ⋅B =
1999
⋅0.2 = 0.039984
9999
It was previously mentioned that the events A and B are not independent. We may verify
this by establishing the following inequality, which is a violation of the definition of
independence.
P ( A ∩ B) = 0.039984 ≠ 0.04 = P ( A) ⋅P ( B)
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Engineering 323
Kuszmar
Beautiful Homework Set 3
Problem 2.80
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b. With A and B independent and P(A) = P(B) = 0.2, what is P(A ∩ B)? How much
difference is there between this answer and P(A ∩ B) in part (a)? For purposes of
calculating P(A ∩ B), can we assume that A and B of part (a) are independent to
obtain essentially the correct probability?
Assuming the independence of A and B, we may calculate the intersection of A and B as
P(A ∩ B) = P(A)⋅P(B) = 0.2 ⋅0.2 = 0.04
There is little difference between this answer and that computed in part (a). The difference
is 0.000016. Because this number is so small, it is quite reasonable to assume that A and B
are independent for purposes of simplifying future calculations.
c. Suppose the lot consists of ten boards, of which two are green. Does the assumption of
independence now yield approximately the correct answer for P(A ∩ B)? What is the
critical difference between the situation here and that of part (a)? When do you think
that an independence assumption would be valid in obtaining approximately the
correct answer to P(A ∩ B)?
Now the total number of boards, N = 10. Without assuming independence, we may
calculate the intersection of A and B in the same fashion as part (a). Note that the
probability that A occurs is still equal to 0.2.
P ( A ∩ B) = P ( B | A) ⋅P ( A) =
1
⋅0.2 = 0.02222
9
Whoa! Now, making the assumption of independence would be a big mistake! The
difference between the solutions in parts (c) and (b) is 0.01778. Let’s figure out why the
approximation became a bad one once the population size decreased…
(cont.)
Engineering 323
Kuszmar
Beautiful Homework Set 3
Problem 2.80
By comparing the experiments described in parts (a) and (c), we see that the validity of the
assumption of independence improves as the population size increases relative to the
sample size.
If we were to read ahead to Chapter 3 in our textbook, we would find a discussion on
binomial experiments. The analysis of a binomial experiment is not computationally
difficult. However, such an analysis requires that independence exist between events in
the population. It would make our lives easier if we were able to assume that an
experiment has binomial properties whenever possible. A rule of thumb exists for the
appropriate application of such an assumption.
The rule states that, in an experiment such as the one described in this problem, as long as
the sampling size is less than 5% of the sample population, the assumption of
independence is appropriate. So there’s our answer!
When given an experiment for which the assumption of independence would be useful, we
may assume independence as long as the sampling size is much smaller (i.e. <5%) of the
sample population.
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