P F ( ) = . 0 10 0 20 ( ) . + . 0 005 0 80 ( ) . = . 0 024
which can be interpreted as just the weighted average of the two probabilities of failure.
Example 2-27
Section 2-5/Multiplication and Total Probability Rules 47
FIGURE 2-16 Partitioning an event into several mutually exclusive subsets.
E1 B > E1
E2
E3 E4 B > E2
B > E3 B > E4 B = (B > E1) < (B > E2) < (B > E3) < (B > E4)
A A'
B B > A
B > A'
FIGURE 2-15 Partitioning an event into two mutually exclusive subsets.
Semiconductor Failures Continuing with semiconductor manufacturing, assume the following probabilities for product failure subject to levels of contamination in manufacturing:
Probability of Failure Level of Contamination
0.10 High
0.01 Medium
0.001 Low
In a particular production run, 20% of the chips are subjected to high levels of contamination, 30% to medium levels of contamination, and 50% to low levels of contamination. What is the probability that a product using one of these chips fails? Let
r
H denote the event that a chip is exposed to high levels of contamination
rM denote the event that a chip is exposed to medium levels of contamination
rL denote the event that a chip is exposed to low levels of contamination Then,
P F ( ) = P F H P H ( ) ( ) + P F M P M ( ) ( ) + P F L P L ( ) ( )
= . ( ) . + . ( .
| | |
0 10 0 20 0 01 0 30 )) + . 0 001 0 50 ( ) . = . 0 0235
The calculations are conveniently organized with the tree diagram in Fig. 2-17.
0.10(0.20)
5 0.02 0.90(0.20)
5 0.18 0.01(0.30)
5 0.003 0.99(0.30)
5 0.297 0.001(0.50)
5 0.0005 0.999(0.50) 5 0.4995 P(Fail) 5 0.02 + 0.003 + 0.0005 5 0.0235
Contamination
0.20 0.50
0.30
High Medium Low
P(FailuHigh) 5 0.10
P(Not FailuHigh) 5 0.90
P(FailuMedium) 5 0.01
P(Not FailuMedium) 5 0.99
P(FailuLow) 5 0.001
P(Not FailuLow) 5 0.999
FIGURE 2-17 Tree diagram for Example 2-28.
Example 2-28
2-121. Suppose that P A B
(
|)
= .0 4 and P B( )
= . .0 5 Deter- mine the following:(a) P A
(
∩B)
(b) P A(
′ ∩B)
2-122. Suppose that P A B
(
|)
= . 0 2, P A B(
| ′)
= .0 3, and P B( )
= . .0 8 What is P A( )
?2-123. The probability is 1% that an electrical connector that is kept dry fails during the warranty period of a portable computer. If the connector is ever wet, the probability of a fail- ure during the warranty period is 5%. If 90% of the connectors are kept dry and 10% are wet, what proportion of connectors fail during the warranty period?
2-124. Suppose 2% of cotton fabric rolls and 3% of nylon fabric rolls contain flaws. Of the rolls used by a manufacturer, 70% are cotton and 30% are nylon. What is the probability that a randomly selected roll used by the manufacturer contains flaws?
2-125. The edge roughness of slit paper products increases as knife blades wear. Only 1% of products slit with new blades have rough edges, 3% of products slit with blades of average sharpness exhibit roughness, and 5% of products slit with worn blades exhibit roughness. If 25% of the blades in manufacturing are new, 60% are of average sharpness, and 15% are worn, what is the proportion of products that exhibit edge roughness?
2-126. In the 2012 presidential election, exit polls from the critical state of Ohio provided the following results:
Total Obama Romney
No college degree (60%) 52% 45%
College graduate (40%) 47% 51%
What is the probability a randomly selected respondent voted for Obama?
2-127. Computer keyboard failures are due to faulty electrical connects (12%) or mechanical defects (88%). Mechanical defects are related to loose keys (27%) or improper assembly (73%).
Electrical connect defects are caused by defective wires (35%), improper connections (13%), or poorly welded wires (52%).
(a) Find the probability that a failure is due to loose keys.
(b) Find the probability that a failure is due to improperly connected or poorly welded wires.
2-128. Heart failures are due to either natural occurrences (87%) or outside factors (13%). Outside factors are related to induced substances (73%) or foreign objects (27%). Natural occurrences are caused by arterial blockage (56%), disease (27%), and infection (e.g., staph infection) (17%).
(a) Determine the probability that a failure is due to an induced substance.
(b) Determine the probability that a failure is due to disease or infection.
2-129. A batch of 25 injection-molded parts contains 5 parts that have suffered excessive shrinkage.
(a) If two parts are selected at random, and without replace- ment, what is the probability that the second part selected is one with excessive shrinkage?
(b) If three parts are selected at random, and without replace- ment, what is the probability that the third part selected is one with excessive shrinkage?
2-130. A lot of 100 semiconductor chips contains 20 that are defective.
(a) Two are selected, at random, without replacement, from the lot. Determine the probability that the second chip selected is defective.
(b) Three are selected, at random, without replacement, from the lot. Determine the probability that all are defective.
2-131. An article in the British Medical Journal [“Comparison of treatment of renal calculi by operative surgery, percutaneous nephrolithotomy, and extracorporeal shock wave lithotripsy”
(1986, Vol. 82, pp. 879–892)] provided the following discussion of success rates in kidney stone removals. Open surgery had a success rate of 78% (273/350) and a newer method, percutane- ous nephrolithotomy (PN), had a success rate of 83% (289/350).
This newer method looked better, but the results changed when stone diameter was considered. For stones with diameters less than 2 centimeters, 93% (81/87) of cases of open surgery were successful compared with only 83% (234/270) of cases of PN.
For stones greater than or equal to 2 centimeters, the success rates were 73% (192/263) and 69% (55/80) for open surgery and PN, respectively. Open surgery is better for both stone sizes, but less successful in total. In 1951, E. H. Simpson pointed out this apparent contradiction (known as Simpson’s paradox), and the hazard still persists today. Explain how open surgery can be bet- ter for both stone sizes but worse in total.
2-132. Consider the endothermic reactions in Exercise 2-50.
Let A denote the event that a reaction's final temperature is 271 K or less. Let B denote the event that the heat absorbed is above target. Determine the following probabilities.
(a) P A
(
∩B)
(b) P A(
∪B)
(c) P A(
′∪B′)
(d) Use the total probability rule to determine P A( )
2-133. Consider the hospital emergency room data in Exam- ple 2-8. Let A denote the event that a visit is to hospital 4 and let B denote the event that a visit results in LWBS (at any hos- pital). Determine the following probabilities.
(a) P A
(
∩B)
(b) P A(
∪B)
(c) P A′ ′(
∪B′)
(d) Use the total probability rule to determine P A( )
2-134. Consider the hospital emergency room data in Example 2-8. Suppose that three visits that resulted in LWBS are selected randomly (without replacement) for a follow-up interview.
(a) What is the probability that all three are selected from hospital 2?
(b) What is the probability that all three are from the same hospital?
FOR SECTION 2-5
Exercises
Problem available in WileyPLUS at instructor’s discretion.
Tutoring problem available in WileyPLUS at instructor’s discretion
Section 2-6/Independence 49 2-135. Consider the well failure data in Exercise 2-53. Let A
denote the event that the geological formation has more than 1000 wells, and let B denote the event that a well failed. Deter- mine the following probabilities.
(a) P A
(
∩B)
(b) P A(
∪B)
(c) P A(
′∪B′)
(d) Use the total probability rule to determine P A( )
2-136. Consider the well failure data in Exercise 2-53. Sup- pose that two failed wells are selected randomly (without replacement) for a follow-up review.
(a) What is the probability that both are from the gneiss geo- logical formation group?
(b) What is the probability that both are from the same geo- logical formation group?
2-137. A Web ad can be designed from four different colors, three font types, fi ve font sizes, three images, and fi ve text phrases. A specifi c design is randomly generated by the Web server when you visit the site. Determine the probability that the ad color is red and the font size is not the smallest one.
2-138. Consider the code in Example 2-12. Suppose that all 40 codes are equally likely (none is held back as a delimiter).
Determine the probability for each of the following:
(a) The code starts and ends with a wide bar.
(b) Two wide bars occur consecutively.
(c) Two consecutive wide bars occur at the start or end.
(d) The middle bar is wide.
2-139. Similar to the hospital schedule in Example 2-11, sup- pose that an oper ating room needs to schedule three knee, four
hip, and fi ve shoulder surgeries. Assume that all schedules are equally likely. Determine the following probabil ities:
(a) All hip surgeries are completed fi rst given that all knee surgeries are last.
(b) The schedule begins with a hip surgery given that all knee surgeries are last.
(c) The fi rst and last surgeries are hip surgeries given that knee surgeries are scheduled in time periods 2 through 4.
(d) The fi rst two surgeries are hip surgeries given that all knee surgeries are last.
2-140. Suppose that a patient is selected randomly from those described in Exercise 2-98. Let A denote the event that the patient is in group 1, and let B denote the event for which there is no progression. Determine the following probabilities:
(a) P A( ∩B) (b) P B( )
(c) P A( ′ ∩B) (d) P A( ∪B) (e) P A( ′ ∪B) 2-141. A computer system uses passwords that contain exactly eight characters, and each character is one of the 26 lowercase letters (a–z) or 26 uppercase letters (A–Z) or 10 integers (0–9).
Let Ω denote the set of all possible password, and let A and B denote the events that consist of passwords with only letters or only integers, respectively. Suppose that all passwords in Ω are equally likely. Determine the following robabilities:
(a) P(A|B′) (b) P A( ′ ∩B)
(c) P (password contains exactly 2 integers given that it con- tains at least 1 integer)