Stat 312 - Dr. Uebersax
October 21, 2013 – Review Notes for Midterm 1
1. Correction to Standard Normal Curve Formula
Incorrect version:
1 z
f(z;0,1) =
e2  < x < 
2
2
Correct version:
2. Rules of Probability
Stat 312 - Dr. Uebersax
October 21, 2013 – Review Notes for Midterm 1
Stat 312 - Dr. Uebersax
October 21, 2013 – Review Notes for Midterm 1
Problem 1
4.3 Diagnostic tests are available for a variety of medical conditions. Although these tests are
extremely reliable, they sometimes provide incorrect results.
Suppose that:



the probability that the test will give a positive result is 0.04.
Given that the test result is positive, the probability that the person has the disease is 0.95.
Given that the test result is negative, the probability that the person does not have the
disease is 0.99.
(a) Give an example of a simple event.
Answer: disease present, disease absent, positive test result, or negative test result are all
simple events.
(b) Give an example of a joint event.
Answer: (disease present, positive test result), (disease present, negative test result),
(disease absent, positive test result), (disease absent, negative test result) are all joint
events.
(c) For a diagnostic test result selected at random what is the probability that
(1) the test result is negative?
Answer: P(negative test result) = 1 – P(positive test result) = 1 – .04 = .96.
(2) the test diagnosis is negative and the patient does not have the disease?
Answer: P(negative test result and no disease)
= P(negative test result) P(no disease | negative test result) = .96 × .99
(3) the test result is positive and the patient has the disease?
Answer: P(positive test result and disease)
= P(positive test result) P(disease | positive test result) = .04 × .95
Stat 312 - Dr. Uebersax
October 21, 2013 – Review Notes for Midterm 1
Problem 2
An engineer has a box filled with 10 identical batteries. Unknown to her, two of the batteries are
dead. She will take batteries out of the box until she finds one that works. Let X = the number
of batteries taken out of the box by the engineer. Find the probability distribution of X.
Answer:
The question asks you to find the probability distribution. A probability distribution lists each
possible outcome, and the probability of each.
What are the possible outcomes? We know that the most batteries she will draw will be three:
– if the first battery is dead, she will draw a second
– if the second batter is also dead, she will draw a third
– there are only two dead batteries; if the first two are dead, the third must be 'live'
Therefore the only possible values of X are 1, 2 or 3 batteries drawn from the box. To determine
our probability distribution, we need to calculate the probability of each. Let Di and Li denote a
dead or live battery, respectively, at draw i.
x
1
2
3
Formula
P(L1)
P(D1) P(L2|D1)
P(D1) P(D2|D1) P(L3|D1 and D2)
P(x)
8/10
(2/10)(8/9)
(2/10)(1/9)(1)
The probabilities in column three sum to 1, showing that our probability distribution is complete
(every possible outcome is accounted for).
Note how with x = 3 we extend the multiplication rule to the case of three joint events. That is,
the multiplication rule for two joint events is:
P(A and B) = P(A) P(B|A)
and by extension the rule for three joint events is:
P(A and B and C) = P(A) P(B|A) P(C|A,B)
The multiplication rule can be extended like this to joint events involving any number of
individual events.
Also note that if the question had been, what is probability of drawing two batteries
simultaneously and both being dead", you would use the hypergeometric distribution.