D ESCRIPTION OF D ATA
3.6 Conditional Probability
As we have defined probability, it is meaningful to ask for the probability of an event only if we refer to a given sample spaceS. To ask for the probability that an engineer earns at least $90,000 a year is meaningless unless we specify whether we are referring to all engineers in the western hemisphere, all engineers in the United States, all those in a particular industry, all those affiliated with a university, and so forth. Thus, when we use the symbol P(A) for the probability of A, we really mean the probability of A given some sample spaceS. Since the choice of S is not always evident, or we are interested in the probabilities of A with respect to more than one
Sec 3.6 Conditional Probability 79
sample space, the notation P(A|S) makes it clear that we are referring to a particular sample spaceS. We read P(A|S) as the conditional probability of A relative to S, and every probability is thus a conditional probability. Of course, we use the simplified notation P(A) whenever the choice ofS is clearly understood.
20 D ù I 10 I
Figure 3.14
Reduced sample space
To illustrate some of the ideas connected with conditional probabilities, let us consider again the 500 machine parts of which some are improperly assembled and some contain one or more defective components as shown in Figure 3.8. Assum-ing equal probabilities in the selection of one of the machine parts for inspection, it can be seen that the probability of getting a part with one or more defective compo-nents is
P(D)= 10+ 5
500 = 3
100
To check whether the probability is the same if the choice is restricted to the machine parts that are improperly assembled, we have only to look at the reduced sample space of Figure 3.14 and assume that each of the 30 improperly assembled parts has the same chance of being selected. We thus get
P(D| I) =N(D∩ I) N(I) = 10
30 = 1 3
and it can be seen that the probability of getting a machine part with one or more defective components has increased from 3
100 to 1
3. Note that if we divide the nu-merator and denominator of the preceding formula for P(D| I) by N(S), we get
P(D| I) =
N(D∩ I) N(S)
N(I) N(S)
= P(D∩ I) P(I)
where P(D| I) is given by the ratio of P(D ∩ I) to P(I).
Looking at this example in another way, note that with respect to the whole sample spaceS we have
P( D∩ I ) = 10 500 = 1
50 and P( D∩ I) = 20 500 = 2
50
assuming, as before, that each of the 500 machine parts has the same chance of being selected. Thus, the probabilities that the machine part selected will or will not contain one or more defective components, given that it is improperly assembled, should be in the ratio 1 to 2. Since the probabilities of D and D in the reduced sample space must add up to 1, it follows that
P(D| I) = 1
3 and P( D| I ) = 2 3
which agrees with the result obtained before. This explains why, in the last step, we had to divide by P(I) to
P(D| I) = P( D∩ I ) P( I )
Division by P(I), or multiplication by 1/P(I), takes care of the proportionality factor, which makes the sum of the probabilities over the reduced sample space equal to 1.
Following these observations, let us now make the following general definition:
If A and B are any events inS and P(B) = 0, the conditional probability of A given B is
P( A| B ) = P( A∩ B ) P( B ) Conditional probability
EXAMPLE 20 Calculating a conditional probability
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
P( A| B ) = 0.18 0.81 = 2
9 j
EXAMPLE 21 The conditional probability that a used car has low mileage given that it is expensive to operate
Referring to the used car example, for which the probabilities of the individual out-comes are given in Figure 3.9, use the results on page 73 to find P( M1|C3).
Solution Since we had P( M1∩ C3)= 0.08 and P(C3)= 0.40, substitution into the formula for conditional probability yields
P( M1|C3)= P( M1∩ C3)
P( C3) = 0.08 0.40 = 0.20
It is of interest to note that the value of the conditional probability obtained here, P(M1|C3)= 0.20, equals the value for P(M1) obtained on page 73. This means that the probability a used car has low mileage is the same whether or not it is expensive to operate. We say that M1is independent of C3. As the reader is asked to verify in Exercise 3.59, it also follows from the results on page 73 that M1is not independent of P1, namely, that low mileage is related to the car’s price. j In general, if A and B are any two events in a sample spaceS, we say that A is independent of B if and only if P(A| B ) = P( A ), but as it can be shown that B is independent of A whenever A is independent of B, it is customary to say simply that A and B are independent events.
Theorem 3.8 If A and B are any events inS, then
P( A∩ B ) = P( A ) · P( B | A ) if P( A )= 0
= P( B ) · P( A | B ) if P( B )= 0 General multiplication
rule of probability
The second of these rules is obtained directly from the definition of conditional probability by multiplying both sides by P(B); the first is obtained from the second by interchanging the letters A and B.
Sec 3.6 Conditional Probability 81
EXAMPLE 22 Using the general multiplication rule of probability
The supervisor of a group of 20 construction workers wants to get the opinion of 2 of them (to be selected at random) about certain new safety regulations. If 12 workers favor the new regulations and the other 8 are against them, what is the probability that both of the workers chosen by the supervisor will be against the new safety regulations?
Solution Assuming equal probabilities for each selection (which is what we mean by the selections being random), the probability that the first worker selected will be against the new safety regulations is 8
20, and the probability that the second worker selected will be against the new safety regulations given that the first one is against them is
7
19. Thus, the desired probability is 8 20· 7
19 = 14
95 j
In the special case where A and B are independent so P(A| B) = P(A), Theo-rem 3.8 leads to the following result:
Theorem 3.9 Two events A and B are independent events if and only if P( A∩ B ) = P( A ) · P( B )
Special product rule of probability
Thus, the probability that two independent events will both occur is simply the prod-uct of their probabilities. This rule is sometimes used as the definition of indepen-dence. It applies even when P(A) or P(B) or both equal 0. In any case, it may be used to determine whether two given events are independent.
EXAMPLE 23 The outcomes to unrelated parts of an experiment can be treated as independent
What is the probability of getting two heads in two flips of a balanced coin?
Solution Since the probability of heads is1
2 for each flip and the two flips are not physically connected, we treat them as independent. The probability is
1 2 ·1
2 = 1
4 j
EXAMPLE 24 Independence and selection with and without replacement
Two cards are drawn at random from an ordinary deck of 52 playing cards. What is the probability of getting two aces if
(a) the first card is replaced before the second card is drawn;
(b) the first card is not replaced before the second card is drawn?
Solution (a) Since there are four aces among the 52 cards, we get 4
52· 4 52 = 1
169
(b) Since there are only three aces among the 51 cards that remain after one ace has been removed from the deck, we get
4 52· 3
51 = 1 221 Note that
1 221 = 4
52· 4 52
so independence is violated when the sampling is without replacement. j EXAMPLE 25 Checking if two events are independent under an assigned
probability
If P(C) = 0.65, P(D) = 0.40, and P(C ∩ D) = 0.24, are the events C and D independent?
Solution Since P(C)· P(D) = (0.65)(0.40) = 0.26 and not 0.24, the two events are not
independent. j
In the preceding examples we have used the assigned probabilities to check if two events are independent. The concept of independence can be—and frequently is—employed when probabilities are assigned to events that concern unrelated parts of an experiment.
EXAMPLE 26 Assigning probability by the special product rule
Let A be the event that raw material is available when needed and B be the event that the machining time is less than 1 hour. If P(A) = 0.8 and P(B) = 0.7, assign probability to the event A∩ B.
Solution Since the events A and B concern unrelated steps in the manufacturing process, we invoke independence and make the assignment
P(A∩ B) = P(A)P(B) = 0.8 × 0.7 = 0.56 j
The special product rule can easily be extended so that it applies to more than two independent events—again, we multiply together all the individual probabilities.
EXAMPLE 27 The extended special product rule of probability
What is the probability of not rolling any 6’s in four rolls of a balanced die?
Solution The probability is 5 6· 5
6· 5 6· 5
6 = 625
1,296 j
For three or more dependent events the multiplication rule becomes more com-plicated, as is illustrated in Exercise 3.70.
EXAMPLE 28 The probability of falsely signaling a pollution problem
Many companies must monitor the effluent that is discharged from their plants into rivers and waterways. In some states, it is the law that some substances have water-quality limits that are below the limit of detection, L, for the current method of measurement. The effluent is judged to satisfy the quality limit if every test spec-imen is below the limit of detection L. Otherwise it will be declared to fail compli-ance with the quality limit. Suppose the water does not contain the contaminant of
Sec 3.6 Conditional Probability 83
interest but that the variability in the chemical analysis still gives a 1% chance that a measurement on a test specimen will exceed L.
(a) Find the probability that neither of two test specimens, both free of the contaminant, will fail to be in compliance.
(b) If one test specimen is taken each week for two years, and they are all free of the contaminant, find the probability that none of the test specimens will fail to be in compliance.
(c) Comment on the incorrect reasoning of having a fixed limit of detection no matter how many tests are conducted.
Solution (a) If the two samples are not taken too closely in time or space, we treat them as independent. We use the special product rule to obtain the probability that both are in compliance:
0.99 × 0.99 = 0.9801 (b) Treating the results for different weeks as independent,
(0.99)104= 0.35
so, even with excellent water quality, there is almost a two-thirds chance that at least once the water quality will be declared to fail to be in compliance with the law.
(c) With this type of law, no company would want to collect test specimens more than maybe once a year. This is in direct opposition to the scientific idea that more information is better than less information on water quality. Some effort should be made to allow for higher limits when the testing is more frequent. j EXAMPLE 29 Using probability to compare the accuracy of two schemes for
sending messages
Electrical engineers are considering two alternative systems for sending messages.
A message consists of a word that is either a 0 or a 1. However, because of random noise in the channel, a 1 that is transmitted could be received as a 0 and vice versa.
That is, there is a small probability, p, that
P[A transmitted 1 is received as 0]= p P[A transmitted 0 is received as 1]= p
One scheme is to send a single digit. The message is short but may be unreliable.
A second scheme is to repeat the selected digit three times in succession. At the receiving end, the majority rule will be used to decode. That is, when any of 101, 110, 011, or 111 are received, it is interpreted to mean a 1 was sent.
(a) Evaluate the probability that a transmitted 1 will be received as a 1 under the three-digit scheme when p= 0.01, 0.02, or 0.05. Compare this with the scheme where a single digit is transmitted as a word. Treat the results for different digits as independent.
(b) Suppose a message, consisting of the two words, a 1 followed by 0, is to be transmitted using the three-digit scheme. What is the probability that the total message will be correctly decoded under the majority rule with p= 0.05?
Compare with the scheme where a single digit is transmitted as a word.
Solution (a) The three digits 111 are transmitted. By independence, the sequence 111 has probability (1− p)(1 − p)(1 − p) of being received as 111. Also the
probability of receiving 011 is p(1− p)(1 − p), so the probability of exactly one 0 among the three received is 3p(1− p)2. Using the majority rule,
P[Correct]= P[transmitted 1 received as 1] = (1 − p)3+ 3p(1 − p)2
p 0.01 0.02 0.05
P[Correct] 0.9997 0.9988 0.9928
All three probabilities are considerably above the corresponding single-digit scheme probabilities 0.99, 0.98, and 0.95, respectively.
(b) Both of the words 1 and 0 must be received correctly. As in part (a), the probability that a 0 is received correctly is also 0.9928. Consequently, using independence, the probability that the total message is correctly received is (0.9928)2= 0.986. This improves over the scheme where single digits are sent for each word since that scheme has only the probability (0.95)2= 0.903 of correctly receiving the total message.
Redundancy helps improve accuracy, but more digits need to be transmitted,
which results in a significantly lower throughput. j