HW 5_solutions
HW problems
Problem 1
Let’s modify Problem 7 (page 23) to a more general test.
Assume a test has a probability p of indicating the disease among
patients actually having a disease. Assume also that the test indicates
the presence of the diseases with probability 1-p among the patients
not having the disease (false positive). Finally, suppose the incident rate
of the disease is r.
(a) Find the probability that the person tested positive actually has the
disease.
Hint: Consider the events: T+ = “tested positive” and A = “Has a
disease”. You have to find P(A|T+).
(b) If p = 0.95 and r =0.005, what is the probability of having a disease being
tested positive?
(c)
Leaving p = 0.95, find how the result depends on r (use Mathematica to
make a plot). At what r the test reliably indicates the disease?
A= “tested positive”. P(T+) = p*r +(1-r) (1-p).
P(A|T+) = p*r/ [p*r + (1-r)(1-p)] = 0.087.
Problem 2
Formula P(AUB) = P(A) + P(B) – P(AB) can be generalized for any
number of events. Thus for the union of three events the result is
P(AUBUCU) = P(A) + P(B) +P( C) – P(AB) – P(AC) – P(BC) +P (ABC)
Suppose that A and B are independent. B and C are mutually exclusive , and
A and C are independent.
If P(AUBUCU) = 0.9, P(B) = 0.5 and P© = 0.3, find P(A).
Solution:
0.9 = 0.8 + x – 0.3 x – 0.5 x;
x = 0.5.
Problem 3
A loaded coin with p(H) = ¾ is tossed twice.
Let the events A, B and C be respectively “first toss is heads”, “second
toss is heads” and “tosses show the same face”.
(a) Are A and B independent?
(b) Are A and BUC independent?
( c) are A,B and C independent?
Solution
Tosses
Probability
HH
9/16
TH
3/16
HT
TT
A = HH + HT; B = HH + TH; C = HH + TT.
P(A) = ¾; P(B) = ¾; P(C) = 5/8.
(a)
P(AB) = P(HH) = 9/16 = ¾ ¾ = P(A) P(B)
=> A and B are independent
3/16
(b)
P(BUC) = 13/16. P(A(BUC)) = P(HH) = 9/16  3 / 4 * 13/16
1/16
=> A and BUC are not independent
Both are heads
(c) P(ABC)= P(HH) =9/16  P(A) P(B) P(C) => the events are not
independent
Problem 4 : Your friend flips three coins and tells you that there is
at least two heads. What is the probability that the first coins is heads?
Solution:
A = “At least two heads”; B = “The first is heads”.
P(A) = 1 – P(at least two tails ) = 1/2. (due to symmetry).
P(AB) = P(HHT or HTH or HHH) = 3/8.
P(B|A) = ¾.
Problem 5: Suppose that the probability that married man votes is 0.45, the
probability a married woman votes is 0.4 and the probability a woman votes
given her husband votes is 0.6.
What is the probability (a) that both vote (b) a man votes given that his wife
does?
A = man votes; B = woman votes; P(A) = 0.45; P(B) = 0,4; P(B|A) = 0.6
P(AB) = P(B|A)P(A) = 0.6*0.45 = 0.27
P(A|B) = P(AB)/P(B)= 0.27/0.4=0.675
A
B
AB