ページ3:

By (c), P(A) = P(AnB³) + P(AMB) (*)
P(B) = P(BOA) + P(ANB) (**)
P(AUB) = P(An B²) + P(BAA) + P(ANB)
|| (+)
PLA)-P(ANB)
P(B)-P(AB)
=
P(A) + P(B)-P(AB)
Example 1.1-3
A fair coin is flipped successively until the same face is
observed
on
Successive flips.
Let A he the event that it will take three
(ie, A = {x1 x = 3, 4, 5, 6, ... } ). Find P(A).
Method 1
or more
flips
P(A) = P( { x | x = 3 } ) + P( { x ( x=4}) + --- + p ( {x| x = k } ) + —---
Note that p( {× 1 × = 3}) = p ( [HTH}) + p ( { THT 3)
= (±)³+ (±)³ = 2(±)³
p ( { × 1 × = ₤2 3 ) = 2. (±)
x=
Thus, P(A) =
∞
12=3
∞
=2.
P({ x 1 x = k } ) = Σ 2. ( 1 ) ^
2(4)²= 2
k
(3
k=3
=
2². (£)³
3

ページ4:

Method 2.
Note that P(A) = 1- P(A") where A`= {x1) x = 2 } = { HH, TT }
PLA²) = P( {HH₁TT})
=
(+)² + (+)² = 1/125
Hence, P(A) = 1-1/2 = ½-
Example 1.1-5.
A
Survey
events
was taken of a group's viewing habits of sporting
on TV during the last year.
A = { watched football }
B = {
"
"
basketball }
C = {
"
"
base hall }
P(A)=0.43 & P(B) = 0.4 & P(C) = 0.32
=
PLANB) = 0.29, P(Anc) : 0.22, P(BAC) = 0.2 .
& PlanBnc) = 0.15
Then PLAUBUC) = P(A) + P(B) + P(C) P(AnB) -p (Anc) -P(Bnc)
= 0.59
+ P(AOBOC)

ページ5:

$ 1.3 Conditional Probability
Rolling a die
A fair six-sided die is rolled. Defined the fulluring events
A: the outcome is even
B: the outcome is greater than 3.
Given that
it is
a
die is greater
than
3,
what is the Prob. that
condition
even? P(A|B)
The conditional probability of
S
Def 1.3-1.
an event A,
given
that
event B has occurs.
,
is defined by
P(A|B) =
PLANB)
P(B)
Let us
A
=
provided that P(B) > 0.
answer the ahove example.
{2.4.63
B = {4,5,6}
Then P(AIB)=
Example 1.3-4
& AnB = {4,6}
P(ANB)
P(B)
=
3
½
=
2/3
Two fair four-sided dice are rolled
Let A he the event that
a sum of 3
is rolled
B "
"
"
"
a tum
of
3 or
a sum of 5
is rolled

ページ6:

+
y
((1.4) (2.4) (3.4) (4.4)
A = {(1,2), (2,1)}
3
(1.3) (2.3) (3.3) (43)
B = { (1.2), (2.1),
2
(1.2)(2.2) 13,2) (42)
1
(1.1) (2.1) (3.1) (4.1)
x
1 2 3. 4
P(AIB) PLANB)
=
P(A)
(1,4), (2,5), (3,2). (4.1)}
6
3
P(A) = 1/16 = ½, P(B) = 11/1
½
8
.
16
=
8
=
=
P(B)
P(B)
3/8
Properties
Given events A & B with P(B) >O
(i) P(AIB) 0; P(BIB) = 1
(ii) P(A|B)
=
1- P(AIB)
(ii) A₁, An, disjoint events, then
PLAUUAB) = P(A₁1B) + P(A₂|B) + ... + P(AKIB)
PLA₁UUAKUIB) = P(A|B) + P(A₂IB) + P(A31B) +---
prof (i) P(AIB) PLANB)
=
>0 & P(B) =
P(BOB) PIB)
=1
PCB)
P(B)
P(B)
(ii)
P(A|B) =
P(AB)
=
=
P(B)
B
P(B)-PLANB)
P(B)
= 1- P(AIB)
PLANB)
|-
P(B)
A
Ang An

ページ7:

(iii)
P(AU AKIB) = P(LAAKJNB)
P ([AB] U... [AB])
P(B)
P(B)
=
P(A,B) + + P(AkB)
P(B)
= P(A, B) +
+ P(AIB).
property. I multiplication Rule)
PLANB) =
=
P(A) P(BIA) prided P(A)>0
= P(B) P(AIB) provided P(B)>0.
Example 1.3-7
A barn contains
8
one-
hump camels. I two-
hump camels
hump camels come out
The camels come out of the barn randomly and one at
A: 2 two-
B: Seventh camel to come out has two-
How to compute PLANB).
a time
among
the first 6 camels
5-hump.
P(A) =
(2) (4)
=
0.2937
P(BIA)
=
(15)
554
=0.5556
9
Hence, PLANB) = P(A) P(BIA) = 0.1632.

ページ8:

f P(AABIO
Remark
P(ABC) = P([AB] c) = P(AMB) P(c| AnB)
Example 1.3-9.
= P(A) P(BIA) P(c(ANB)
A School girl has 5 blue & 4 white marbles in her left pocket
& 4 blue & 5 white marbles in her night pocket
5B
4B
✗
What is the Proh. of her then
5W
4W
drawing
a blue marble from
Left
Right
her right Pocket?
BL drawing blue from left Pocket
BR:
we:
"
"
"
"
right Pocket
"
white
"
"
left Pocket
P(BR)
=
PCBROBE) + P(BRA WR)
= P(BL) P(BRIBL) + P(WL) P (BRT WL)
5 5
4 4
41
=
+
=
q 10
9 10
90.

ページ9:

§ 2. Discrete Distributions
2.1~2.3 Random Variables of Discrete type & Expectation
Def 2.1-1.
S: a set :
XSR
a
a
sample space.
function
: a random variable.
Notation
Let As R. {XEA } = { s = S | X € A }
Example 2.1-1.
A rat is selected at random. The set of possible outcomes is
female & male.
The sample space S = { female, male } = { F. M}.
The random variable X (F) = 0 & X(M) = 1.
{ X = 1 } = {M} & {X = 0} = {F}.
Example 2.1-2
A experiment in which we roll a six-sided die.
S = {1, 2, 3, 4, 5, 63
X(s)
= S.
{X=3}, {x2}-{X=1, X=2}. {2X5] = [X=2. 3. 4. 5 },
Notation
Sis finite
or countable.
A random variable X: S → R is of
the discrete type.

ページ10:

The
range
of X RIX)= {X(s) | SE S } is at most countable.
f(x) = P(X = x): the
:
Def 2.1-2.
Proliability
mass function (pmf).
1. The pmf f(x) of a discrete random variable X is
that satisfies the following conditions.
a
function
(a) f(x) >o
for all x E RIX)
(b) f(x) = 1
XERIX)
(c) P(XEA) Σ f(x)
=
A & RIX).
XEA
:
2. F(x) = P(X ≤ x) = P(X = (-00,x]) is called the cumulative
"distribution function (cdf) of X (or distribution function of XJ
"XER"
Example 2.1-2 (Continue from the above)
f(x) = P(X = x) = ½ for x = 1.2.3.4.5.6.
and F(x) = P(X ≤ x ) = P(X = 1. 2. --·, [x])
0
=
if x <1
12/
if k ≤x < fe+l
if x 26.
J
the largest integer less or equal to x
e.g. k≤ x < k+1 => [x] = k
Example 2.1-3.
Roll a fair four sided die twice.
Let X
=
{
the larger of the two outcomes if they
the common value
are different
if they
ace
the same.

ページ11:

The sample space
S = { (d₁, dz)) d₁ = 1,2,3,4 & d₂ = 1,2,3,4}
X ( (1,2)) = X ( (2,1)) = X ((2,2)) = 2
=> f(2) = P(X = 2) = P( { (1,2), (2,1), (2,2)}) = 3/16
X((1.3))X((3.1)) = X ( (2.3)) = X ( (3,2)) = X ((3.3)) = 3.
=> f(3) = P(X = 3) = 5/16.
Def 2.2-1.
X is a random variable with its pmf f.
u(X)
Then the expected value (the expectation) of UCX is
E[u(X)] = Σ u(x) · f(x).
Theorem 2.2-1.
x
C1, C2 : Constants. U₁, U₂ = functions
(a) E[c] = C₁ ;
(b) E[ c₁u, (X)] = c E [u, (X)] ;
(c) E [Cu(X) + C₂ U₂ (X)] = C₁ E [U₁(X)] + C₂ E [ U₂ (X)].
Example 2.2-3.
Let X have the pmf fixs:
x
=
x = 1.2.3.4
10
E[X]
=
二.f(a)=(六)+2(六)+3.(%) +4.(%) = 3
_E[X²] = { x²ƒ‹x) = 1² ( ½ ) + 2)² ( 21 ) + ³ ³ ( 23 ) + 4² ( 16 ) = 10.
x=1

ページ12:

and E[X (5-X)] = 5 E[X] - E[X²] = 5.3-10 = 5.
Example 2.2-5
An experiment has prohahility of success pε (0,1) and probability
of failure 9 = 1-p.
This experiment is repeated independently until the first success
this happens
Occurs; say
on trail X.
x-
f(x) = P(x = x) = 9.9.9.9.8 = 97.p
x-1 times
geometric distribution
X~ geo (p)
for all x = 1, 2, 3, ----
8
P
We observe — fix) = Σ 9x+p = P(1+9+9+9+) = 1-9 = 1)
☑
x=1
x=1
— ....
E[X] = x.fix) = 1. p + 2. qp + 3 q² p + ....
x=1
then (1-9) E[X] = p + pq + pq² + .....
=> ECX] ==
1-9
1
= 1 =1
9
Def
μ=E[X] = f(x) :
x
the
mean
of X
x
² = Van (X) = E [ (X-μ)²] = Σ (x-µ³)² f(x): the Variance of X
5: the standard derivation of X.
Remark
6²≥0 and 6²=E[X²] − (E[X])²

ページ13:

Example 2.3-2.
Auppose X has the pmf_f(x) = P(x=x)=%. x = 1,2,…, 6.
6
μ=E[X] = x ( ) = |+2+-+6
x=1
6
=
6
E[X³] = { x²³ ( ± ) = 1³½³ 2²+---+6² = %/6
x=1
6
96
then ²= E[X²] - (E[X])² = % - (½)² = 35/2
and 6 = 135/12~1.708.
Example 2.3-4
Let X have a uniform distribution on the fint m positive integers.
( That is, P(X = x) = 1/1 for each x = 1, 2, ... m )
m
u = E[X] = x (MM)
E[X²] =
m
x=1
x=1
=
m -2
m
1 m (m+1) m+1
=
2
=
(m+1) (2m+l)
6
x² ( ) = 1 m (m+1) (2m+1)
σ² = (7+) (2H)
6
property.
a, beR.
m
-
M+
m
1° E[ax+b]=aE[X] +b
=
2. Var (ax+b)= a² Van (X).
proof of 2
6
m²-1
12.
Van (ax+b) = E [{aX+b }² ] - {E[aX+b] }²

ページ14:

=
-
E[ a²x²+ 4a b X + b² ] − (a² E[X]²+ 4ab E[X] + b²)
= a² E[X³] + 4ab E[X] + b² - (a² E[X]² + 4ab E[X] + b²)
=
a² ( E[X²) - E[X]²) = a² Van (X).
Def.
E[X"] = Σx" fix) : the r-th moment
E [(X)] = E[X(X) (X-r+1)] : the r-th factorial moment.
Def.
Let X he a random variable
of
the discrete
with
type
pmf f).
Mlt) = E [ex] = Σ ex fix), - h<t<h for some h>0, is called the
moment generating function (mgf) of X.
Remark
mgf uniquely determines the distribution of that random variable.
In other words,
Suppose
that
M₁ (t) is the mgf of X, with its pmf fi
M₂ (t)
"
X2
"
fz
=
and Milt) Malt), te (-h,h)
then X, and X₂ have the same distribution; that is, f, (x) = ₤2(x) Vx.
Example 2.3-7
If X has the mgf
3t
M(t) = e* (² ²) + e²± (²) + e³ (t)
then P(X=1):
,
P(X = 2) = —, and PCX = 3) = 1.

ページ15:

Property
M'(0) = E[X], M"(0) = E[X³], -, M² (10) = E [X].
proof Note that M'(t) = Σ x e** fix then M'(0) = Σ x·1 fix) = E[X].
x
r
In general, M" (t) = Σx" etx fix) then M (0) = Σ x*·1. f(x) = E[X*].
Example 2.3-9.
x
x
Suppose X has the geometric distribution (p); that is, the pmf of X
x-|
is fox) = 9x+p, x = 1, 2, 3,...-
xx
etx
DO
Then MH+) = E[ e²x] = 2 ** q* p = — — 1, (96*) *
x=1
x=1
=
|| [ (qe³) + (get) ... ] = 1
get
Pet
↓
91-9et
get<lt<-Inq
=
1-9et
Then M'(t) = (1-e) (Pet) - Pe* (-qe²)
Pet
=
(1-9et)2
(1-98+)2
M" (t) = (1-9e*)² pe* - Pe³·2· (1-9eª) (-qe*)
(1-90*)4
Hence, u = E[X] = '(0) = ½
=
Pet (1+get)
(1-9et)3
²=E[X] - E[X]² = M"(0) - (M²(0))² =
P(1+9)
(1-913-p=797

ページ16:

§ 2.4 The Binomial Distribution
A Bemoulli trail ( experiment) is a random experiment, the
can he classified in
outcome
Success or failure
two
mutually
ex
clusive
ways:
V.S
defective)
( female v.s male; life vs death; non defective
Then, X is a random variable associated with
Bemoulli trail :
X( success) = 1 & X( failure) = 0.
q
Thus, the pmf of X is fix) = P(X = x) = p² (1-p)/x, x=0.1.
1
Bernoulli distribution
μ = E [X] = ½ x f₁x) = x px (1-p)+x = p.
x=0
x=0
1
¸²= Var(X) = E[ (X-µ³²] = ≥ cx-p)²³ f(x) = p(1-p) = pq.
x=0
In a sequence of independent Bernoulli trails, We are
interested in the total mumber of success.
Let X = number of success in n Bemoulli trails
n-x
then P(X = x) = (1) px (1-p) *-*, x = 0. 1. 2, n.
We
Say
募
often
X follows Binomial distribution (n.p) (X~ Bin (n.p) )
Dr b(n.p)

ページ17:

if it pmf flx) = ( 1 ) P* (1-P)"
where
(2)
=
n!
x! (1-x)!
Example 2.4-1 & 2.4-7
The probability of germination of
n-x
x = 0, 1, 2, ..., n.
a beet seed is 0.8.
We plan to plant 10 seeds and assume that the germination
of one seed is indep of the germination of another seed.
Let X he the member of seeds that germinate.
f(x) = (10) 10.8)* (0.2)"
Then
2
10-x
x = 0, 1, 2, ..., 10
10-x
≈0.000078
10-x
P(X ≤2) = Σ (10) 10,8)* (0.2)"
and
x=0
8
8
P(X ≤ 8) = Σ f(x) = ²² (1°) (0.8)* (0.2)
x=0
x=0
or PLX = 8) = 1- P(X = 9) - P(X = 10)
= 1 - (19) (0.8)³ (0.2) - (10) (0.8) 10
Example 2.4.3 & 2.4-5
=
0.6242
An experience archer can hit the center of the target, the bullseye,
in 20% of the
attempts.
Let X he the number of hitting the bullseye among n = 8 subsequent
Then the Probability of hitting the bullseye twice is
f(z) = ($) (0.2)² (0.8) = 0.2936.
shots..

ページ18:

Recall
n
(a+b)" = ±² ("). a*b".
Remarks
x=0
11-x
binomial expansion.
1° X ~ Bin (1. p) <=> X ~ Bernoulli cp)
2. Let X ~ Bin (n.p), then
M (t) = E [ex] = Σ ex fix) = ^^
=
x=0
11-x
tx
e
*** (*) P* (1-p)**
x=0
Σ (2) (pet)* (1-p) = [pe² + 1-p]".
x=0
n+
Thus. M'(t) = n [ Pe² + 1-p]" · (pet)
=>>
11-2
M"(t) = n (n) [ pe² + 1-p]² (pe³)² + n [Pe*+ 1-p]" (Pe*)
· M = E[X] = M'(0) = np
and σ² = E[X²] - (E[X])² = M"(0) - ( μ'(o)² = np (1-p) = npq.
Example 24-10
Suppose that observation over a long period of time has
disclosed that.
on
the
average,
one but of ten items produced
by a process is defective. Select 5 items independently from
the production line and test them.
Let X he the number of defective items among n=5.
Then X ~ Bin (5, 0.1).
Also, E[X] = 5 (0.1) = 0.5 & Var (X) = 5 (0.1) (0.9) = 0.45
In addition, PLX ≤1) = (5) 10.1)° (0.9)5 + (5) (0.1)' (0.954
= 0.9185.

ページ19:

§ 2.5 The Hypergeometric Distribution
N. objects
Nz objects
First class
Second class
N+N₂ =N
fix) = P(X = x) =
We
Say
A collection of n objects is selected
from those N objects at random
and without replacement.
Let X he the number of objects selected
that belong to the first class.
(EN) (~)
,
max (n-N₂,0) ≤ x ≤ min (n. N₁).
(N)
that X has a hypergeometric distribution with
N₁, №₂, and n. We denote it by X ~ HG (N₁, N₂, n).
Example 2.5-2.
In
a
small pond, there
are
parameters
50 fish, ten of which have been
tagged. If a fisherman's catch consists of seven fish selected
at random and without replacement.
X denotes the mumber of tagged fish, the probability that
exactly tuo tagged fish are
P(X = 2) =
(12) (40)
(500)
=0.2964.
caught is

ページ20:

Example 2.5-3
A lot consisting of 100 fuses is inspected by : 5 fuses are chosen
at random and tested; if all five blow at the correct amperage
the lot is accepted. Suppose the lot contains 20 defective fuses.
If X is the number of defective fuses in the sample of five.
the probability of accepting the lot is
P(X = 0) = (20) (80)
(150)
=
0.3193.
80
, x = 0, 1, 2, 3, 4,5.
More general, fix) = (^2)(32)
Remarks
X~ HG (NN₂, n)
1. E[X] = x
=
= n
2
Σ x
x+0
N₁!
(N)
N₂
(150)
Σx.
=
x+0
(№32)
* ! (N₁-x)! (~~) (N-1)
(x-4)1
x+0
N₂
(1-1) (n-1-(x-1))
(N-1)
(*)(M)
(N)
N₁ (N-1)!
=
(M)Σ
x+0
(x+)! (N₁-x)!
(二)
(N₂
PLY = x+), Y~HG(N1, №₂, n−1)
n(~) Σ
=n()
(N) since Σ
(M-1) (n-1-(x+1)
= 1.
x+0
(N-1)

ページ21:

2°
n
n
(DA) DA)
n-x
EX(X-1)] = x(x-1)
x=0
= Σx(x-1)
X=2
(N)
= M
x(x-1)
2
=N. (N-1)
x
N.!
N₂!
x! (N₁-x)! (1-x)! (N₂-n+x)!
(N)
(N.-2)!
N₂!
(x-2)! (N,-x)! (1-x)! (N₂-n+x)!
(N)
Now, we note that
(N)
N!
N (N-1)
=
n! (N-h?!
n(n-1)
( N-2)
n-2
Hence, EXCX-1)] = N₁ (N, −1)
h(n-1)
y
(№1-2 ) ( № 2 - y)
N₂
N(N-1)
y=0
(N-2)
1
1
y=x-2
n(n-1)
=N. (N-1)
N (N-1)
Hence,
Van (X) = E[X (X-1)] + E[X] − (E[X])²
= N. (N-1)
= n
-
( u ) - Nu
n(n-1)
+
N
N (N-1)
n (~~) (~~) (N-1)
(冊)(冊)

ページ22:

§ 2.6 The Negative Binomial Distribution
Consider a sequence of independent Bernoulli trails until exactly
Ir successes occur, where I is a fixed positive integer.
Let X he the number
of
trails needed to observe the r-th success.
In other words, X is the trail number on which the r-th success is
observed.
Then the pmf of X is
x-r
g(x) = (x-1) p² (1-p)
- *~~ ·p
Y~ Bin (x-1, p & P(Y=1-1)
binomial distribution.
r-th success
We
Say
that X has a
Remarks
(k).
10
1. how) = Σ
K=0K!
negative
wk where hows = (1-1)
-r-
observe that '(w) = r (1-w)" => hco = v
then (0)
k!
-Y-2
r
h" (w) = r (r+1) (1-w)" => "(0) = r (r+1)
(k)
h (w) =
=
-Y-K
(x)'
=> h (100) = x (V+1) ... (Y+K-1).
Y (1+1) ….. (Y+k+1) (1-1)
(1+k+)... (1+1).r
Hence, we obtain
ki
(V+k-1)!
=
=
K! (Y+)!
Y-I
DO
(1-15)=
Y+k-1
wk
K=0 Y-1
DO
{√ (11) w
x=k+r_x=Y
x-r

ページ23:

DO
where 9 = 1-P
x-r
00
x-r
We note that — glx) = — (*) prq *-* = p². ± (*) q*
X=Y
x=Y
X=Y
= pr. (1-9) = pr. pr=1.
2. If r=1, then X has
a
geometric distribution; that is,
·P = (1-p)
= (1 p)² = k
k
k
g(x) = (1-p)x+1px = 1, 2, 3, ....
DO
R) 2
x-
Hence, PIX> k) = Σ (1-p) p =
x=12+1
(1-P)*
1-(1-p)
Thus, P(X ≤ 1) = = (1-p)*p = 1- P(X>k) = 1 - (1 - p)² = 1-9
Example 2.6-1
x=1
Some biology students were checking eye color in a
large number
of fruit flies. For the individual fly, suppose the probability of white
eyes is 1/4 and the probability of red eyes is ¾/4.
The probability that at least four flies have to checked for
to observe
a
white fly is
P(X ≥4) = P(X > 3) = q³ = ( 4 )³ = −27
P(X ≤ 4) = 1-9 ² = 0.6836.
and
3
=
0.4219
64
P(X = 4) = 9 ³ p = ( 3 ) ³ ( 4 ) = 0.1055.
Color
eye

ページ24:

Example 2.6-1 ( Continue from above)
eyes.
(a) The probability that three out of eight fruit flies have white
(b) The Probability that the third white eyed fruit fly is found on the
eighth check.
Sol. (a) Note that we shall compute PCX = 3) where X ~ Bin (8,¼4)
Then P(X = 3) = ( 3 ) · (4) ³ (4) * ≈ 0.2077.
(b) We shall compute P(X = 8) where
2 x-3
g(x) = (x-1) (4)² ( 4 ) *³ (4)
Then P(X = 8) = g(8) = ( ! ) (4)² (3) ~ 0.0779.
Remark
00
tx
DO
M(t) = El ex] = — e** (x-1) p* (1-p)** = (pe³)*· Σ (**) [ (1-p) e* ]'
=
(Pet)
x=r
| [1-(1-pet]
Cl-peti
then M'(t) = r. (pet) [1-(1-p) e*] ---
-1-2
X=Y
and M" (t) = r ( Pe²)" (-r-1) [1-(1-p) e³] [-(1-p) e*]
+r (pet) (pet) [1-(1-1)*]
M'(0) = /p
x-r

ページ25:

and μ"(0) = = = (1+1 - p)
p²
Hence, we conclude μ = + and
Example 2.6-2.
2
Y (PH-P)
r(1-p)
σ =
p²
p²
p²
a
free
can make
Sequence of a
Suppose that during practice a basketball player
throw 80% of the time. Furthermore, assume that a
free-throw shooting
can he thought of as
trails.
independent Bernoulli
Let X he the minimum number of free throws that this player
must attempt to make
The pmf of X is
Then
a
total
of ten shots.
g(x)
=
(x-1) (0.8)" (0.2)*
10
x-10
x=10,11,12,-
1
& σ²=
10.(0.2)
=
= 3.125
(0.8)2
=12.5
μ = 10.
0.8
Remark
& 5 = √√3.125 = 1.768.
Let M he the mgf of a given random variable X.
Suppose Mk (0) existo for all k. then
Mit) = M (0) +
M'(0)
M"(0)
H
t +
t² +---+
M(110)
t+.....
2!
k!

ページ26:

Example 2.6-4
Let the moment of X he defined by E[X"] = 0.8, r = 1, 2, 3 ...
M(t) = M(0) + Mr) (0)
T=1
Y!
= 0.2 +0.86 tr
tr
= 1
=1+E[X]
tr
=
1+(0.8) 总点
T=1 V!
t
Y=I Y!
1.t
8 + 1 = 0.2 + 28. e² = (0.2) e°*+ (0.8) e¹t
Y=0 Y!
0.8.
Thus, PCX = 0) = 0.2 & PCX = 1) = 0.8.