AMS 311 Joe Mitchell
PROBABILITY THEORY: Practice Exam 1 Solution Notes
Statistics: n= 41, µ= 70.0, median 76, σ= 22.0; score range: 3–100
1. The sample space, S={(1,1,1),(1,1,2),...}, consists of 63= 216 equally likely outcomes. Let Ebe the
event that all three dice show the same number. Then E={(1,1,1),(2,2,2),...,(6,6,6)}consists of six
outcomes. We want the probability of the complement: P(Ec) = 1 −P(E) = 1 −6/216 = 35
36 .
2. First, note that the information about the windows is superfluous – it is not needed at all in solving the
problem. Let Edenote the event that a randomly selected house needs a paint job; let Fdenote that it
needs a new roof. We know that P(E) = 0.30, P(E∩F) = 0.15.
(a). By definition, p=P(E|F) = P(E∩F)
P(F)=.15
P(F). What bounds can we place on the number
P(F)? Draw a Venn diagram! Since the portion of Ethat is NOT in Fhas probability 0.15, we know that
P(F)≤0.85. (Formally, F⊆(E∩Fc)c, which implies that P(F)≤1−P(E∩Fc) = 1 −0.15 = 0.85)
Also, we know that P(F)≥P(E∩F) = 0.15, but we do not need this here. Since P(F)≤0.85, we get that
p≥0.15
0.85 =3
17 . (This lower bound is achievable, by putting F= (E∩Fc)c.)
(b). We now know that P(F|Ec) = 0.5, so we get
0.5 = P(F∩Ec)
P(Ec)=P(F∩Ec)
0.70 ,
implying that P(F∩Ec) = 0.35, so P(F) = P(F∩E) + P(F∩Ec) = 0.15 + 0.35 = 0.5. (You can also argue
it straight from the Venn diagram.)
3. (This is a modified version of problem 6, section 3.2, page 90, which was given on HW2.) Let Ebe the
event that a random patient has had skin cancer; let Fbe the event that the patient is a redhead. We know
that P(E|F) = 0.30¡ P(E|Fc) = 0.20, and P(F) = 0.40.
(a). We want P(E), which we obtain by conditioning: P(E) = P(E|F)P(F) + P(E|Fc)P(Fc) =
(.3)(.4) + (.2)(.6) = 0.24.
(b). We want to compute P(F|E), which we get straight from the definition, and utilize the P(E) we
calculated in (a). (This is Bayes formula, but I like to do it from scratch every time.)
P(F|E) = P(E∩F)
P(E)=P(E|F)P(F)
P(E)=(.3)(.4)
.24 = 0.5
4. (This is problem 5, page 90, which is very similar to problem 4, page 90, which was on the practice exam.)
Let Eibe the event that there are ispades lost, for i= 0,1,2. Let Fbe the event that the drawn card (from
the 50-card deck) is a spade. We compute P(F) by conditioning on how many spades are missing:
P(F) = X
i
P(F|Ei)P(Ei) = 13
50 ·13
039
2
52
2+12
50 ·13
139
1
52
2+11
50 ·13
239
0
52
2
5. We know that P(A) = 4/5, P(B|A) = 1/2, and that Aand Bare NOT independent, which means that
P(B|A)6=P(B) (i.e., that P(B)6= 1/2 – but we do not know if P(B) is larger or smaller than 1/2). Since
P(B|A) = 1/2, we know that P(A∩B)
P(A)= 1/2, so P(A∩B) = (4/5)(1/2) = 2/5.
(i). Aand Bare mutually exclusive: FALSE. If they were mutually exclusive, then by definition, A∩B=
∅, so P(A∩B) = 0, but we know that P(A∩B) = 2/5.
(ii). Aand A∩Bare independent: FALSE. Note that P(A|A∩B) = 1 6=P(A) = 4/5. (Knowing that
A∩Boccurs implies that Aoccurs.)
