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Probability Calculations based on Events and Conditional Probability, Assignments of Statistics

University of Illinois - Urbana-ChampaignStatistics

Calculations of probabilities of events based on given conditional probabilities. It includes the use of total probability theorem and bayes' rule to find the probabilities of certain events. The document also discusses the concept of independent events and their impact on probability calculations.

Typology: Assignments

Pre 2010

Uploaded on 03/10/2009

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University of Illinois Fall 1998
ECE 313: Solutions to Problem Set #4
1. Let
B
be the event that the target is hit, and let
A
be the eventthat there is a gust of wind. Then, by the theorem
of total probability, we have
(a)
P
(
B
)=
P
(
B
j
A
)
P
(
A
)+
P
(
B
j
A
c
)
P
(
A
c
)=(0
:
4)(0
:
3) + (0
:
7)(0
:
7) = 0
:
61
Part (b) is asking for the conditionalprobability of the event
A
c
given that the event
B
c
occurred. Using Bayes
inversion formula, we obtain
P
(
A
c
j
B
c
)=
P
(
B
c
j
A
c
)
P
(
A
c
)
=P
(
B
c
)
.
(b) ¿From part (a), we have
P
(
B
c
)=0
:
39
.Furthermore
P
(
B
c
j
A
c
)=1
,
P
(
B
j
A
c
)=0
:
3
. Therefore
P
(
A
c
j
B
c
)=(0
:
3)(0
:
7)
=
0
:
39 = 0
:
54
.
2. (a) Define the followingevents:
T
=
f
Exactly 2 white balls were chosen
g
W
F
=
f
Ball from urn F was white
g
W
E
=
f
Ball from urn E was white
g
W
G
=
f
Ball from urn G was white
g
The probability we seek is
P
(
W
E
j
T
)=
P
(
W
E
T
)
=P
(
T
)
. Since the picks from the three baskets are
all independent, we can write
P
(
T
)=
P
(
W
E
W
F
W
c
G
)+
P
(
W
E
W
c
F
W
G
)+
P
(
W
c
E
W
F
W
G
)
a
=
3
6
8
12
3
4
+
3
6
4
12
1
4
+
3
6
8
12
1
4
=
9
24
P
(
W
E
T
)=
P
(
W
E
W
F
W
c
G
)+
P
(
W
E
W
c
F
W
G
)
b
=
3
6
8
12
3
4
+
3
6
4
12
1
4
=
7
24
;
where
(
a
)
and
(
b
)
follow from independence. Therefore
P
(
W
E
j
T
)=
7
=
24
9
=
24
=
7
9
.
(b) Define the followingevents:
W
=
f
Ball picked was white
g
P
F
=
f
Ball picked was from urn F
g
P
E
=
f
Ball picked was from urn E
g
P
G
=
f
Ball picked was from urn G
g
The probability we seek is
P
(
P
E
j
W
)=
P
(
P
E
W
)
=P
(
W
)
, which, by Bayes’ Rule, can be written as
P
(
P
E
j
W
)=
P
(
W
j
P
E
)
P
(
P
E
)
P
(
W
)
. From the theorem of total probability, we can write
P
(
W
)=
P
(
W
j
P
E
)
P
(
P
E
)+
P
(
W
j
P
F
)
P
(
P
F
)+
P
(
W
j
P
G
)
P
(
P
G
)
:
Now
P
(
P
E
)=
P
(
P
F
)=
P
(
P
G
)=1
=
3
, since the ball was chosen with equal probability from any of
the baskets; also
P
(
W
j
P
E
)=3
=
6
;P
(
W
j
P
F
)=8
=
12
;P
(
W
j
P
G
)=1
=
4
.Thisgives
P
(
W
)=
3
6
1
3
+
8
12
1
3
+
1
4
1
3
=
17
36
:
Therefore,
P
(
P
E
j
W
)=
3
=
6
1
=
3
17
=
36
=
6
17
.
pf3
pf4

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Download Probability Calculations based on Events and Conditional Probability and more Assignments Statistics in PDF only on Docsity!

University of Illinois Fall 1998

ECE 313: Solutions to Problem Set #

1. Let B be the event that the target is hit, and let A be the event that there is a gust of wind. Then, by the theorem

of total probability, we have

( a) P (B ) = P (B jA)P (A) + P (B jAc^ )P (Ac^ ) = (0:4)(0:3) + (0:7)(0:7) = 0 : 61

Part ( b ) is asking for the conditional probability of the event Ac^ given that the event B c^ occurred. Using Bayes

inversion formula, we obtain P (Ac^ jB c^ ) = P (B c^ jAc^ )P (Ac^ )=P (B c^ ).

( b) ¿From part ( a ), we have P (B c^ ) = 0 : 39. Furthermore P (B c^ jAc^ ) = 1 P (B jAc^ ) = 0 : 3. Therefore

P (Ac^ jB c^ ) = (0:3)(0:7)= 0 : 39 = 0 : 54.

2. ( a) Define the following events:

T = fExactly 2 white balls were choseng WF = fBall from urn F was whiteg

WE = fBall from urn E was whiteg WG = fBall from urn G was whiteg

The probability we seek is P (WE j T ) = P (WE T )=P (T ). Since the picks from the three baskets are

all independent, we can write

P (T ) = P (WE WF W Gc ) + P (WE W Fc WG ) + P (W Ec WF WG )

=^ a 3

6 ^

12 ^

4 +^

6 ^

12 ^

4 +^

6 ^

12 ^

4 =^

P (WE T ) = P (WE WF W Gc ) + P (WE W Fc WG )

=^ b 3 6

where (a) and (b) follow from independence. Therefore P (WE j T ) = 7 =^24

( b) Define the following events:

W = fBall picked was whiteg PF = fBall picked was from urn Fg

PE = fBall picked was from urn Eg PG = fBall picked was from urn Gg

The probability we seek is P (PE j W ) = P (PE W )=P (W ), which, by Bayes’ Rule, can be written as

P (PE j W ) = P^ (W^ j^ PE^ )P^ (PE^ )

P (W )

. From the theorem of total probability, we can write

P (W ) = P (W j PE )P (PE ) + P (W j PF )P (PF ) + P (W j PG )P (PG ):

Now P (PE ) = P (PF ) = P (PG ) = 1 = 3 , since the ball was chosen with equal probability from any of

the baskets; also P (W j PE ) = 3 = 6 ; P (W j PF ) = 8 = 12 ; P (W j PG ) = 1 = 4. This gives

P (W ) = 3

Therefore, P (PE j W ) = 3 =^6 ^1 =^3

3. Ternary Channel: The numbers given in the table are all the conditional probabilities of receiving sj ; j =

1 ; 2 ; 3 given that you send si ; i = 1 ; 2 ; 3 , i.e. P [sj received j si sent]. Therefore, the probability that the signal

sj is received can be calculated using the theorem of total probability as

P [sj received] =

X^3

i=

P [sj received j si sent]  P [si sent]:

Since P [si sent] = 1 = 3 for all i, the required probabilities above are found to be

s 1 s 2 s 3

P [sj received] 0 : 29 0 : 36 0 : 35

Now, we compute P [si sent j sj received]. For example,

P [s 1 sent j s 1 received] = P^ [s^1 received^ j^ s^1 sent]^ ^ P^ [s^1 sent]

P [s 1 received]

= 0 :^8 ^1 =^3

In this fashion, we can form a table similar to the one given in the problem, but consisting of the inverse

conditional probabilities P [si sent j sj received]:

Send, i

s 1 s 2 s 3

s 1 0 : 9195 0 : 0575 0 : 0230

Receive, j s 2 0 : 0926 0 : 8333 0 : 0741

s 3 0 : 0952 0 : 0476 0 : 8571

4. ( a) Define the following events:

A = fan airb orne narb ee b eing involved in an accidentg

S = fnarb ee driver is a sob er Nargiang

D = fnarb ee driver is a drunk Nargiang

X = fnarb ee driver is an alien g

Let us denote the probability that a driver is a sober Nargian by P (S ) = p. Then from the problem,

P (D ) = 2 P (S ) = 2 p and since P (S ) + P (D ) + P (X ) = 1 , we have P (X ) = 1 3 P (S ) = 1 3 p.

Now using the theorem of total probability, we have

P (A) = P (A j S )P (S ) + P (A j D )P (D ) + P (A j X )P (X )

) 0 : 59 = 0 : 2 p + 0 :8(2p) + 0 :5(1 3 p) =) p = 3

Therefore, the probability that a randomly chosen driver is an alien is 1 3  103 = 1 = 10. And also,

P (D ) = 6 = 10.

( b) We want to find the probability P (D j Ac^ ). We can use Bayes’ rule for this

P (D j Ac^ ) = P^ (A

c j D ) P (D )

P (Ac^ )

= 0 :^2 ^0 :^6

5. Let Ai denote the event that the student passes the i-th exam. Then we are given the following (conditional)

probabilities:

P (A 1 ) = 0 : 9 P (A 2 jA 1 ) = 0 : 8 P (A 3 jA 1 A 2 ) = 0 : 7

Based on this information, we can answer the questions in this problem as follows.

7. ( a) Let A 1 and A 2 respectively denote the events that the first (Dilbert) and second (Wally) players initially

pick the correct curtain. A 1 and A 2 are disjoint events. Let A = A 1 [ A 2 be the event that one of

the players picked the correct curtain, and let B denote the event that the final choice of the player not

eliminated is the correct curtain. Clearly P (A) = 2 = 3 and hence, P (Ac^ ) = 1 = 3.

Stick to pick strategy: If the player not eliminated sticks to the original choice, then P (B j A) = 1 ,

while P (B j Ac^ ) = 0. Hence P (B ) = P (B j A)  P (A) + P (B j Ac^ )  P (Ac^ ) = 2 = 3.

Ditch and switch strategy: If the player switches, then P (B j A) = 0 , while P (B j Ac^ ) = 1. Hence,

P (B ) = 1 = 3 now.

Thus it is better to stay with the original pick in this game. Switching at random gives a probability of

1 = 2 of winning.

( b) Assuming that both players stick to their original choices, the ith^ player, i = 1 ; 2 , will win if and

only if event Ai occurs, i.e., P (ith^ player wins prize) = P (Ai ). Obviously, P (A 1 ) = 1 = 3 , and hence,

P (A 2 ) = P (A) P (A 1 ) = 1 = 3. Both have equal probabilities of winning. The remaining 1 = 3 of the

time Ac^ occurs and neither wins.

( c) Extra credit: The method for solving this problem is similar to the “3-door Monty Hall” problem that

we solved in class. Except we now have two decisions to take into account, and hence 4 possibilities —

Stay-Stay (St-St), Stay-Switch (St-Sw), Switch-Stay (Sw-St), and Switch-Switch (Sw-Sw). Define the

following events:

R = fDilbert^0 s final choice is correctg

R 1 = fDilbert^0 s first choice is rightg

R 2 = fDilbert^0 s second choice is rightg

We will partition the sample space into 4 mutually exhaustive events: R 1 R 2 ; Rc 1 R 2 ; R 1 Rc 2 , and Rc 1 Rc 2.

Now use the theorem of total probability to get

P (R) = P (R j R 1 R 2 )P (R 1 R 2 ) + P (R j Rc 1 R 2 )P (Rc 1 R 2 ) + P (R j R 1 Rc 2 )P (R 1 Rc 2 )

+ P (R j Rc 1 Rc 2 )P (Rc 1 Rc 2 )

The above probabilities are tabulated in the table below under each of the 4 scenarios possible.

St-St St-Sw Sw-St Sw-Sw

P (R 1 R 2 ) 14 14 0 0

P (R j R 1 R 2 ) 1 0 x x

P (Rc 1 R 2 ) 0 0

P (R j Rc 1 R 2 ) x x 1 0

P (R 1 Rc 2 ) 0 0

P (R j R 1 Rc 2 ) x x 0 1

P (Rc 1 Rc 2 ) 34343838

P (R j Rc 1 Rc 2 ) 0 1 0 1

P (R) 14343858

The “x” in a row indicates that the conditioning event is the null event. For example, if Dilbert switches

after his first pick (no matter what he does on his second choice) it is impossible that both his first and

second choices are right, i.e., P (R 1 R 2 ) = 0 , and so it makes no sense to condition the event R on the

event R 1 R 2.

Thus we see that Dilbert’s best policy is to Stay-Switch, i.e., pick a curtain, stay with his pick after Monty

opens the first curtain, then switch after Monty opens a curtain for the second time.

If Monty had 5 curtains on stage and repeated the game, except now Dilbert had 3 stages at which he

could make a choice, you can work out and see that his best policy is to Stay-Stay-Switch. And this

generalizes to n curtains too — Dilbert’s best policy is to Stay-Stay-   (n 1) times    -Switch.