Page 70 - Applied statistics and probability for engineers

48 Chapter 2/Probability

Exercises FOR SECTION 2-5

Problem available in WileyPLUS at instructor’s discretion.
Tutoring problem available in WileyPLUS at instructor’s discretion
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2-121. Suppose that P A B) = .4 and P B ( ) = . . Deter- (a) If two parts are selected at random, and without replace-
5
0
0
mine the following: ment, what is the probability that the second part selected
(
(
(a) P A∩ B) (b) P A′ ′ B) is one with excessive shrinkage?
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2-122. Suppose that P A B| ( ) = .0 2 , P A B ) = .0 3′ , and (b) If three parts are selected at random, and without replace-
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ment, what is the probability that the third part selected is
P B ( ) = . .0 8 What is P A ( ) ?
one with excessive shrinkage?
2-123. The probability is 1% that an electrical connector 2-130. A lot of 100 semiconductor chips contains 20 that
that is kept dry fails during the warranty period of a portable are defective.
computer. If the connector is ever wet, the probability of a fail- (a) Two are selected, at random, without replacement, from the
ure during the warranty period is 5%. If 90% of the connectors lot. Determine the probability that the second chip selected
are kept dry and 10% are wet, what proportion of connectors is defective.
fail during the warranty period? (b) Three are selected, at random, without replacement, from
2-124. Suppose 2% of cotton fabric rolls and 3% of nylon the lot. Determine the probability that all are defective.
fabric rolls contain laws. Of the rolls used by a manufacturer,
2-131. An article in the British Medical Journal [“Comparison
70% are cotton and 30% are nylon. What is the probability that a
of treatment of renal calculi by operative surgery, percutaneous
randomly selected roll used by the manufacturer contains laws?
2-125. The edge roughness of slit paper products increases nephrolithotomy, and extracorporeal shock wave lithotripsy”
(1986, Vol. 82, pp. 879–892)] provided the following discussion
as knife blades wear. Only 1% of products slit with new blades
of success rates in kidney stone removals. Open surgery had a
have rough edges, 3% of products slit with blades of average
success rate of 78% (273/350) and a newer method, percutane-
sharpness exhibit roughness, and 5% of products slit with worn
ous nephrolithotomy (PN), had a success rate of 83% (289/350).
blades exhibit roughness. If 25% of the blades in manufacturing
This newer method looked better, but the results changed when
are new, 60% are of average sharpness, and 15% are worn, what
stone diameter was considered. For stones with diameters less
is the proportion of products that exhibit edge roughness?
2-126. In the 2012 presidential election, exit polls from than 2 centimeters, 93% (81/87) of cases of open surgery were
successful compared with only 83% (234/270) of cases of PN.
the critical state of Ohio provided the following results:
For stones greater than or equal to 2 centimeters, the success
rates were 73% (192/263) and 69% (55/80) for open surgery and
Total Obama Romney PN, respectively. Open surgery is better for both stone sizes, but
No college degree (60%) 52% 45% less successful in total. In 1951, E. H. Simpson pointed out this
College graduate (40%) 47% 51% apparent contradiction (known as Simpson’s paradox), and the
hazard still persists today. Explain how open surgery can be bet-
What is the probability a randomly selected respondent voted ter for both stone sizes but worse in total.
for Obama? 2-132. Consider the endothermic reactions in Exercise 2-50.
2-127. Computer keyboard failures are due to faulty electrical Let A denote the event that a reaction's inal temperature is 271
connects (12%) or mechanical defects (88%). Mechanical defects K or less. Let B denote the event that the heat absorbed is above
are related to loose keys (27%) or improper assembly (73%). target. Determine the following probabilities.
(
Electrical connect defects are caused by defective wires (35%), (a) P A∩ B) (b) P A∪ B) (c) P A′ ∪ B′)
(
(
improper connections (13%), or poorly welded wires (52%). (d) Use the total probability rule to determine P A( )
(a) Find the probability that a failure is due to loose keys.
(b) Find the probability that a failure is due to improperly 2-133. Consider the hospital emergency room data in Exam-
connected or poorly welded wires. ple 2-8. Let A denote the event that a visit is to hospital 4 and
2-128. Heart failures are due to either natural occurrences let B denote the event that a visit results in LWBS (at any hos-
(87%) or outside factors (13%). Outside factors are related to pital). Determine the following probabilities.
′(
(
(
induced substances (73%) or foreign objects (27%). Natural (a) P A∩ B) (b) P A∪ B) (c) P A′ ∪ B′)
occurrences are caused by arterial blockage (56%), disease (d) Use the total probability rule to determine P A( )
(27%), and infection (e.g., staph infection) (17%). 2-134. Consider the hospital emergency room data in Example
(a) Determine the probability that a failure is due to an induced 2-8. Suppose that three visits that resulted in LWBS are selected
substance. randomly (without replacement) for a follow-up interview.
(b) Determine the probability that a failure is due to disease or (a) What is the probability that all three are selected from
infection. hospital 2?
2-129. A batch of 25 injection-molded parts contains 5 (b) What is the probability that all three are from the same
parts that have suffered excessive shrinkage. hospital?