Page 71 - Applied statistics and probability for engineers

Section 2-6/Independence 49

2-135. Consider the well failure data in Exercise 2-53. Let A hip, and i ve shoulder surgeries. Assume that all schedules are
denote the event that the geological formation has more than equally likely. Determine the following probabil ities:
1000 wells, and let B denote the event that a well failed. Deter- (a) All hip surgeries are completed i rst given that all knee
mine the following probabilities. surgeries are last.
(
(
(
(a) P A∩ B) (b) P A∪ B) (c) P A′ ∪ B′) (b) The schedule begins with a hip surgery given that all knee
(d) Use the total probability rule to determine P A( ) surgeries are last.
(c) The irst and last surgeries are hip surgeries given that knee

2-136. Consider the well failure data in Exercise 2-53. Sup-
pose that two failed wells are selected randomly (without surgeries are scheduled in time periods 2 through 4.

replacement) for a follow-up review. (d) The irst two surgeries are hip surgeries given that all knee
(a) What is the probability that both are from the gneiss geo- surgeries are last.
logical formation group? 2-140. Suppose that a patient is selected randomly from those
(b) What is the probability that both are from the same geo- described in Exercise 2-98. Let A denote the event that the
logical formation group? patient is in group 1, and let B denote the event for which there
is no progression. Determine the following probabilities:
2-137. A Web ad can be designed from four different colors, (a) P A( ∩
three font types, i ve font sizes, three images, and i ve text B) (b) P B( )
phrases. A speciic design is randomly generated by the Web (c) P A( ′ ′ B) (d) P A( ∪ B) (e) P A( ′ B)

server when you visit the site. Determine the probability that 2-141. A computer system uses passwords that contain exactly
the ad color is red and the font size is not the smallest one. eight characters, and each character is one of the 26 lowercase
2-138. Consider the code in Example 2-12. Suppose that all letters (a–z) or 26 uppercase letters (A–Z) or 10 integers (0–9).
40 codes are equally likely (none is held back as a delimiter). Let Ω denote the set of all possible password, and let A and B
Determine the probability for each of the following: denote the events that consist of passwords with only letters or
(a) The code starts and ends with a wide bar. only integers, respectively. Suppose that all passwords in Ω are
(b) Two wide bars occur consecutively. equally likely. Determine the following robabilities:
(c) Two consecutive wide bars occur at the start or end. (a) P(A|B′)
(d) The middle bar is wide. (b) P A′ ′ B)
(
2-139. Similar to the hospital schedule in Example 2-11, sup- (c) P (password contains exactly 2 integers given that it con-
pose that an oper ating room needs to schedule three knee, four tains at least 1 integer)

2-6 Independence
In some cases, the conditional probability of P B A| ( ) might equal P B( ). In this special case,
knowledge that the outcome of the experiment is in event A does not affect the probability that
the outcome is in event B.

Example 2-29 Sampling with Replacement Consider the inspection described in Example 2-14. Six parts are
selected ran domly from a bin of 50 parts, but assume that the selected part is replaced before the next
one is selected. The bin contains 3 defective parts and 47 nondefective parts. What is the probability that the second part is
defective given that the irst part is defective?

In shorthand notation, the requested probability is P(B | A), where A and B denote the events that the irst and second parts

are defective, respectively. Because the irst part is replaced prior to selecting the second part, the bin still contains 50 parts,

of which 3 are defective. Therefore, the probability of B does not depend on whether or not the irst part is defective. That is,
P B A) = 3
|
(
50
Also, the probability that both parts are defective is
P A∩ B) = P B A P A) = 3 ⋅ 3 = 9
|
(
)
(
(
50 50 2500
Example 2-30 Flaws and Functions The information in Table 2-3 related surface laws to functionally defec-

/
0
tive parts. In that case, we determined that P D F( | ) = 10 40 = .25 and P D) = 28 400 = .07 .
/
0
(
Suppose that the situation is different and follows Table 2-4. Then,
(
0
0
P D F) = 2 40 = .05 and P D ( ) = 20 400 = .05
/
|
/
That is, the probability that the part is defective does not depend on whether it has surface l aws. Also,
(
0
/
|
/
P F D) = 2 20 = .10 and P F ( ) = 40 400 = .10
0