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1) Researchers are interested in the relationship between cigarette
smoking and lung cancer. Suppose and adult male is randomly
selected from a particular population. Assume that the following
table shows some probabilities involving the compound event that
the individual does or does not smoke and the person is or is not
diagnosed with cancer.
Event
Probability
Smokes and gets cancer
.05
Smokes and does not get cancer .20
Does not smoke and gets cancer .03
Does not smoke and does not get cancer
.72
Suppose further that the probability that the randomly selected
individual is a smoker is .25.
a) Find the probability that the individual gets cancer, given that he
is a smoker.
b) Find the probability that the individual does not get cancer,
given that he is a smoker.
c) Find the probability that the individual gets cancer, given that he
does not smoke.
d) Find the probability that the individual does not get cancer,
given that he does not smoke.
a) P(cancer given smoker) = P(cancer and smoker)/P(smoker) =
.05/0.25=.2
b) P(not cancer given smoker) = P(cancer and
smoker)/P(smoker) = .20/.25=.8
c) P(cancer given not smoker) = P(cancer and not
smoker)/P(not smoker) = .03/.75
d) P(not cancer given not smoker)=P(not cancer and not
smoker)/P(not smoker) = .72/.75=.96
2. The probability that a car will skid on a bridge on a rainy
day is .75. Today the weather station announced that
there is a 20% chance of rain. What is the probability
that it will rain today and that a car will skid on the
bridge.
a) .0300 b) .0375 c) .1500 d) .3000 e) .9500
(.75)(.20) = .1500
3. The probability that Ted will enroll in an English class in
1/3. If he does enroll in an English class, the probability
that he would enroll in a math class is 1/5. What is the
probability that he enrolls in both classes?
a) 1/15
b) 2/15
c) 7/15
d) 3/5
e) 13/15
 1  1 
  
 3  5 
4. The local Chamber of Commerce conducted a survey of one
thousand randomly selected shoppers at a mall. For all
shoppers, “gender of shopper” and “items shopping for” was
recorded.
Items shopping for
Gender Clothing Shoes
Other
Total
Male
75 25 150 250
Female
350
230
170
750
Total
425
255
320
1000
a) What is the probability that the shopper is a female?
75/1000=.75
b) What is the probability that the shopper is shopping for
shoes?
255/1000=.255
c) What is the probability that the shopper is a female
shopping for shoes?
230/1000=.23
d) What is the probability that the shopper is shopping for
shoes given that the shopper is a female?
Shopping for shoes and female = .23 and female = .75, so
.23/.75 = .3067
e) Are the events “female” and “shopping for shoes” disjoint?
Shopping for shoes and female = .23 which is not 0 so they are
not disjoint.
f) Are the events “female” and “shopping for shoes”
independent?
Shopping for shoes given female = .3067 and shopping for
shoes = .255 which means that those two probabilities are not
equal. Therefore “female” and shopping for shoes” are not
independent.
5) Two companies offer overnight shipping that is supposed to
arrive by 10:00 a.m. Your company ships 30% of its packages
with shipping service 1 and 70% with shipping service 2.
Shipping service 1 fails to meet the 10:00 a.m. deadline 10% of
the time and shipping service 2 fails to meet it 8% of the time.
If a customer is expecting a package and it is late, which of the
shipping services is the more likely to have been used? (Hint:
use a tree diagram).
Let S1 = shipping service 1, S2 = shipping service 2. Let S =
package arrives by 10am, F = package doesn’t arrive by 10am.
Find P(S1 given F) = P(S1 and F)/P(F) = .03/.086=.349
Find P(S2 given F) = P(S2 and F)/P(F) = .056/.086