HSA 523 Homework #4
Answer Key
Dr. Robert Jantzen
Economics Department
1. Given the contingency table, and let the two risk factors of low-birth weight and
illegal drugs be denoted as LB & ID:
Had Illegal Drugs:
Yes
No
Total
Had Low-Birth Weights:
Yes
No
30
20
50
200
80
220
Total
50
250
300
What’s the probability that an infant will be born w/ low-birth weight?
P(LB) = 80/300 = .267
What’s the probability that an infant will be born w/ illegal drugs in their blood?
P(ID) = 50/300 = .167
B.
What’s the joint probability that an infant will have both low birth weight and
blood w/ drugs in it?
P(LB&ID) = 30/300 = .1. This is the chance of being born w/ both risk
factors.
What’s the joint probability that an infant will have neither?
P(LB’ & ID’) = 200/300 = .667. This is the chance of being born w/
neither risk factor.
C.
What’s the conditional probability that a low birth weight infant will have drugs
in their blood?
P(ID|LB) = 30/80 = .375. This is the chance a low birth weight infant
will be born w/ illegal drugs.
What’s the conditional probability that an infant w/ drugs in their blood will have
low birth weight?
P(LB|ID) = 30/50 = .6. This is the chance an infant w/ drugs in their
blood will be low birth weight.
A.
2. Given P(HIV) = .02, P(HIV & Male) = .015, and P(Male) = P(Female) = .5:
A.
What’s the joint probability of being infected w/ HIV and being a women?
(HINT: simple probability is = to the sum of the joints)
Since P(HIV) = P(HIV&Male) + P(HIV&Female), then .02 = .015 +
P(HIV&Female), so the joint probability of being infected w/ HIV and being a women
infant, i.e., P(HIV&Female) is .005.
B.
What’s the conditional probability of being infected w/ HIV for men? (HINT:
conditional is = to joint / simple)
P(HIV|Male) = P(HIV&Male) / P(Male) = .015 / .5 = .03. Men have a 3%
chance of being HIV infected.
What’s the conditional probability of being infected w/ HIV for women?
P(HIV|Female) = P(HIV&Female) / P(Female = .005 / .5 = .01. Women
have a 1% chance of being HIV infected.
What’s the conditional probability of an HIV infected person being a man?
P(Male|HIV) = P(HIV&Male) / P(HIV) = .015 / .02 = .75. The chance
that you’re a male if you’re HIV+ is 75%.
What’s the conditional probability of an HIV infected person being a woman?
P(Female|HIV) = P(HIV&Female) / P(HIV) = .005 / .02 = .25. The
chance that you’re a female if you’re HIV+ is 25%.
D.
HIV infection and gender are not statistically independent because the simple
probability of infection is not equal to the conditionals, i.e., P(HIV) is not equal to
P(HIV|Male) or P(HIV|Female). Men are more likely (.03) to be infected than women
(.01).
C.
3. Given that 10% of transplant patients die after surgery, 30% survive w/ marginal
improvement, and 60% completely recover, and alcohol abusers comprise 30% of deaths,
20% of those w/ marginal improvements, and 10% of complete recoveries. We can form
a Bayes Table to answer the questions because the first three percentages are the simple
probabilities after surgery and the latter three are conditional probabilities of alcohol
abuse for each of the surgical outcomes.
Let P(SOi) indicate the surgical outcome probabilities, and P(AB|SOi) the probabilities of
alcohol abuse for each surgical outcome.
Surgical Outcome
Death
Marginal
Improvement
Complete Recovery
Prior prob. of each
Conditional prob. of
alcohol abuse for
surgical outcome
P(SOi)
0.1
each surgical outcome
P(AB|SOi)
0.3
0.3
0.6
0.2
0.1
Joint prob.
of alcohol
abuse for each
surgical
outcome
P(AB&SOi)
0.03
Revised conditional
prob. for each
surgical outcome given
0.06
0.06
0.15
.4
.4
alcohol abuse
P(SOi|AB)
.2
Given the above:
A. What fraction of all transplant patients abuse alcohol?
0.15 which is the sum of all of the joint probabilities of abusing alcohol
and the surgical outcomes.
B. Given a patient abuses alcohol, what are the probabilities that they’ll die, show
marginal improvement or make complete recoveries?
The probabilities are .2, .4 & .4. There’s a 20% chance of death for an
alcohol abuser, a 40% chance of marginal improvement and only a 40% chance of
complete recovery.
4. Given that a hospital knows that 5% of their inpatients will suffer a “hospital caused”
complication and 95% won’t, and weekend admissions comprise 3% of those w/
complications and 2% of those w/o complications. We can form a Bayes Table to answer
the questions because the first two percentages are the simple probabilities of
complications & no complications, and the latter two are the conditional probabilities of
being admitted on the weekend for each of these initial outcomes.
Let P(COMPL?i) indicate the two initial outcome (complication & no complication)
probabilities, and P(WKEND|COMPL?i) the probabilities of weekend admission for each
outcome.
inpatient outcome
Conditional prob. of
weekend admission
for
each inpatient
outcome
P(COMPL?i)
0.05
0.95
P(WKEND|COMPL?i)
0.03
0.02
Prior prob. of
each
Inpatient
Outcome
Complication
NoComplication
Joint prob.
of weekend
admit for each
Revised conditional
prob. for each
inpatient outcome
given
inpatient outcome
weekend admit
P(WKEND&COMPL?i)
0.0015
0.019
0.0205
P(COMPL?i|WKEND)
.073
.927
A. What’s the simple probability of suffering a complication? .05 Not suffering a
complication? .95
B. What’s the simple probability of being admitted on the weekend, i.e., what fraction
of all inpatients were admitted on the weekend?
0.0205 which is the sum of all of the joint probabilities of being admitted on the
weekend and the two inpatient outcomes, i.e., had complications or didn’t..
C. Given an inpatient was admitted on the weekend, what’s the probability that
they’ll suffer a complication? Won’t suffer a complication?
The probabilities are .073 & .927. There’s a 7.3% chance of complication for a
weekend admission (& 92.7% chance of no complication), which is 50% higher than the
general 5% complication rate.