Solution - Classes

ECE353: Probability and Random Processes
Winter 2011 - Dr. Thinh Nguyen
Homework 2 Solution
1. Genetics.
(a) The basic rules of genetics were discovered in mid-1800s by Mendel, who found that
each characteristics of a pea plant, such as whether the seeds were green or yellow, is
determined by two genes, one from each parent. Each gene is either dominant d or
recessive r. Mendel’s experiment is to select a plant and observe whether the genes are
both dominant d, both recessive r, or one of each (hybrid) h. In this pea plants, Mendel
found that yellow seeds were a dominant trait over green seeds. A yy pea with two yellow
genes has yellow seeds; a gg pea with two recessive genes has green seeds; a hybrid gy or
yg pea has yellow seeds. In one of Mendel’s experiments, he started with a parental
generation in which half the pea plants were yy and half the plants were gg. The two
groups were crossbred so that each pea plant in the first generation was gy. In the second
generation, each plant was equally likely to inherit a y or a g gene from each first
generation parent. What is the probability P(Y ) that a randomly chosen pea plant in the
second generation has yellow seeds?
Solution:
P(Y) = 1 – P(G) = 1 – P(gg) = 1 – 1/4 = 3/4
(b) One of Mendel’s most significant results was the conclusion that genes determining
different characteristics are transmitted independently. In pea plants, Mendel found that
round peas are a dominant trait over wrinkled peas. Mendel crossbred a group of (rr, yy)
peas with a group of (ww, gg) peas. In this notation, rr denotes a pea with two “round”
genes and ww denotes a pea with two “wrinkled” genes. The first generation were either
(rw, yg), (rw, gy), (wr, yg), or (wr, gy). plants with both hybrid shape and hybrid color.
Breeding among the first generation yielded second-generation plants in which genes for
each characteristics were equally likely to be either dominant or recessive. What is the
probability P(Y ) that a second-generation pea plant has yellow seeds? What is the
probability P(R) that a second-generation plant has round peas? Are R and Y independent
events? How many visibly different kinds of pea plants would Mendel observe in the
second generation? What are the probabilities of each of these kinds?
Solution:

Similar to part (a), P(Y) = 3/4 and P(R) = 3/4
R and Y are independent events.
P(G) = 1/4 and P(W) = 1/4
There are 4 kind of visibly different pea plants: (R, Y) (R,G) (W,Y) (W,G) with
probability as follow
P
P
P
P
3 3
, = × =
4 4
( R Y )
3 1
, = × =
4 4
( R G)
1 3
, = × =
4 4
( W Y )
1 1
, = × =
4 4
( W G)
9
16
3
16
3
16
1
16
2. Medical Test. Suppose that for the general population, 1 in 5000 people carries the
human immunodeficiency virus (HIV). A test for the presence of HIV yields either a
positive (+) or negative (-) response. Suppose the test gives the correct answer 99% of the
time.
(a) What is P(−|H), the conditional probability that a person tests negative given that the
person does have the HIV virus?
P
( − | H ) = P( incorrect test) = 0. 01
(b) What is P(H|+), the conditional probability that a randomly chosen person has the
HIV virus given that the person tests positive?
P
( H | + )
=
P
( H,
+ )
P( + )
=
P
( + | H ) P( H )
( + | H ) P( H ) + P( + | H ) P( H )
0.99×
0.0002
=
= 0.019
0.99×
0.0002 + 0.01×
0.9998
P
3. Cards. Shuffle a deck of cards and pick two cards at random. Observe the sequence of
the two cards in the order in which they were chosen.
(a) How many outcomes are in the sample space?
52
P
2
= 52 × 51 = 2652

(b) How many outcomes are in the event that two cards are the same type but different
suits?
4
13× P
2
= 13×
12 = 156
(c) What is the probability that the two cards are the same type but different suits?
P = 156 / 2652 = 0.059
(d) Suppose the experiment specifies observing the set of two cards without considering
the order in which they are selected, and redo parts (a)-(c)
Without considering card order, we have:
52
52 2
i. C
2
= P = 1326
2
4
4 P2
ii. 13×
C
2
= 13×
= 78
2
78
iii. P = = 0. 059
1326
4. Basketball. A basketball team has three pure centers, four pure forwards, four pure
guards, and one swingman who can play either guard or forward. A pure position player
can play only the designated position. If the coach must start a lineup with one center,
two forwards, and two guards, how many possible lines can the coach choose?
⎛3⎞⎡⎛4⎞⎛
4⎞
⎛4⎞⎛4⎞
⎛4⎞⎛
4⎞⎤
⎜ ⎟⎢⎜
⎟⎜
⎟ + ⎜ ⎟⎜
⎟ + ⎜ ⎟⎜
⎟⎥
= 3 36
⎝1
⎠⎣⎝2⎠⎝
2⎠
⎝1
⎠⎝2⎠
⎝2⎠⎝1
⎠⎦
( + 24 + 24) = 252
5. Football. There is a collection of field goal kickers, which can be divided into two
groups 1 and 2. Group i has 3i kickers. On any kick, a kicker from group i will kick a
field goal with probability 1/(i+1), independent of the outcome of any other kicks by that
kicker or any other kicker.
(a) A kicker is selected at random from among all the kickers and attempts one field goal.
Let K be the event that a field goal is kicked. Find P(K).
Let G i is the event that a selected kicker belongs to group i, then
P(
K ) = P( K | G1
) P( G1
) + P( K | G2
) P( G2
)
1 3 1 6 7
= × + × =
2 3 + 6 3 3 + 6 18

(b) Two kickers are selected at random. For j = 1, 2, let Kj be the event that kicker j kicks
a field goal. Find P(K 1 ∩ K 2 ). Are K 1 and K 2 independent events?
P
Now P ( )
Let
( K1
∩ K
2
) = P( K1
∩ K
2
| G1,
G1
) P( G1,
G1
) + P( K1
∩ K
2
| G1,
G2
) P( G1,
G2
)
+ P( K ∩ K | G , G ) P( G , G ) + P( K ∩ K | G , G ) P( G , G )
K 1 =
7
18
1
2
2
1 1 3 2 1 1 3 6 1
= × × × + × × × + ×
2 2 9 8 2 3 9 8 3
195
= = 0.226
864
'
G
i
is the event that the second kicker belongs to group i, then
P
Therefore,
P
' '
' '
( K2
) = P( K2
| G1
) P( G ) + P( K2
| G ) P( G )
1
2
2
' '
'
= P( K2
| G1
)( P( G | G1
) P( G1
) + P( G | G2
) P( G2
))
1
1
' '
'
+ P( K | G )( P( G | G ) P( G ) + P( G | G ) P( G ))
2
2
2
1
1
1
2
1
1
2
1
2
2
6 3 1 1 6 5
× × + × × ×
9 8 3 3 9 8
1 ⎛ 2 3 3 6 ⎞ 1 ⎛ 6 3 5 6 ⎞
= ⎜ × + × ⎟ + ⎜ × + × ⎟
2 ⎝ 2 + 6 3 + 6 3 + 5 3 + 6 ⎠ 3⎝
2 + 6 3 + 6 3 + 5 3 + 6 ⎠
1 ⎛ 2 3 3 6 ⎞ 1 ⎛ 6 3 5 6 ⎞ 1
= ⎜ × + × ⎟ + ⎜ × + × ⎟ =
2 ⎝ 8 9 8 9 ⎠ 3 ⎝ 8 9 8 9 ⎠ 3
( K ) P( K ) × = ≠ P( K ∩ K )
1
2
2
7 1 7
=
1 2
, K 1 and K 2 are NOT independent.
18 3 48
2
2
2
2
2
(c) A kicker is selected at random and attempts 10 field goals. Let M be the number of
misses. Find P(M = 5).
P
( M = 5) = P( M = 5 | G ) P( G ) + P( M = 5 | G ) P( G )
5
1
1
5
⎛10⎞⎛
1 ⎞ ⎛ 1 ⎞ 3 ⎛10⎞⎛
1 ⎞
= ⎜ ⎟⎜
⎟ ⎜1
− ⎟ + ⎜ ⎟⎜
⎟
⎝ 5 ⎠⎝
2 ⎠ ⎝ 2 ⎠ 9 ⎝ 5 ⎠⎝
3 ⎠
10
10 6
⎛10⎞⎛
1 ⎞ 1 ⎛10⎞⎛
1 ⎞ 2
= ⎜ ⎟⎜
⎟ + ⎜ ⎟⎜
⎟
⎝ 5 ⎠⎝
2 ⎠ 3 ⎝ 5 ⎠⎝
3 ⎠ 3
6
1 ⎛10⎞⎡
1 2 ⎤
= ⎜ ⎟ = 0.173
10 10
3 5
⎢ + ⎥
⎝ ⎠⎣2
3 ⎦
5
2
⎛ 1 ⎞
⎜1
− ⎟
⎝ 3 ⎠
2
5
6
9

6. System reliability. A particular operation has six components. Each component has a
failure probability q, independent of any other component. The operation is successful if
both 1) Components 1, 2, and 3 all work, or component 4 works and 2) Either component
5 or component 6 works.
(a) What is the probability P(W) that the operation is successful?
Let C i is the event that component i works, then
And
W
=
( C C C ∪ C ) ∩ ( C ∪ )
1 2 3 4 5
C6
( ) = ( P( C1C2C3
) + P( C4
) − P( C1C2C3C4
))( P( C5
) + P( C6
) − P( C5C6
))
3
4
2
= ( 1−
q) + ( 1−
q) − ( 1−
q)
)( ( 1−
q) + ( 1−
q) − ( 1−
q)
)
2
2
3
= ( 1+
q)( 1−
q) 1+
( 1−
q) − ( 1−
q)
P W
( )
(b) Suppose we can replace any one of the components with an ultrareliable component
that has a failure probability of q/2. Which component should we replace?
Since C 1 , C 2 , and C 3 are symmetric; C 5 , and C 6 are symmetric. Let consider 3 following
cases:
- Replace C 1 :
P
1
P
- Replace C 4 :
P
P
'
'
( W ) = ( P( C1C2C3
) + P( C4
) − P( C1C2C3C4
)( P( C5
) + P( C6
) − P( C5C6
))
⎛⎛ q ⎞ 2 ⎛ q ⎞ 3 ⎞
2
= ⎜⎜1 − ( 1 − q) + ( 1 − q) − ⎜1
− ⎟( 1−
q) ⎟( 1−
q) + ( 1−
q) − ( 1−
q)
)
⎝⎝
⎟
2 ⎠
⎝
2 ⎠
⎛ q 2 q 3 ⎞ 2 q 2
3 2
( W ) − P( W ) = ⎜ ( 1−
q) − ( 1−
q) ⎟( 1−
q ) = ( 1−
q) − ( 1−
q)
)( − q )
1
1
2
P
⎝ 2
2
'
'
( W ) = ( P( C1
C2C3
) + P( C4
) − P( C1
C2C3C4
) P( C5
) + P( C6
) − P( C5C6
)
⎛ 3 ⎛ q ⎞ ⎛ q ⎞ 3 ⎞
2
= ⎜( 1 − q) + ⎜1
− − ⎜1
− ⎟( 1−
q) ⎟( 1−
q) + ( 1−
q) − ( 1−
q)
)
⎝
⎝
⎟
2 ⎠
⎝
2 ⎠
⎠
⎠
⎠
2
( )
⎛ q q 3 ⎞ 2 q
3 2
( W ) − P( W ) = ⎜ − ( 1−
q) ⎟( 1−
q ) = ( 1−
( 1−
q)
)( − q )
2
1
⎝ 2
q
2
( W ) − P ( W ) = 1−
( 1−
q)
2
2 2
( )( 1−
q ) 0
2 1
>
⎠
2

- Replace C 5 :
P
P
3
'
'
( W ) = ( P( C1
C2C3
) + P( C4
) − P( C1
C2C3C4
)( P( C5
) + P( C6
) − P( C5C6
)
3
4 ⎛⎛ q ⎞ ⎛ q ⎞
= ( 1 − q) + ( 1 − q) − ( 1 − q)
) ⎜⎜1
− + ( 1 − q) − ⎜1
− ⎟( 1−
q)
⎞
⎟ ⎟
⎝⎝
2 ⎠ ⎝ 2 ⎠ ⎠
=
2
3
4 ⎛ q
( ) ( ) ( ) ) ⎟ ⎞
1 − q + 1 − q − 1 − q
⎜1
−
⎝ 2 ⎠
2
q
2
( W ) − P ( W ) = ( 1−
q)( 2 − q) 0
2 3
>
So we should replace C 4 .
7. King of Hearts (Bonus Problem)
Two identical packs of 52 cards A and B are shuffled thoroughly. One card is picked
from pack A and shuffled with pack B. The top card from packet A is turned up. If this is
the Queen of Hearts, what is the chance that the top card in B will be the King of Hearts?
Ans:
C 1A = card picked first from pack A
C 2A = card picked second from pack A
C B = card picked from pack B
We need P{ C KH | C QH}
B
2 A
{ C
B
= ,
2 A
= }
P{ C = QH}
P KH C QH
= = =
2 A
{ C
B
= ,
2A = } = { C
B
= ,
2A = |
1A = } { 1A
= }
+ P{ C
B
= KH, C2A = QH | C1A = QH} P{ C1A
= QH}
+ P{ C = KH, C = QH | C ≠ KHorQH} P{ C ≠ KHorQH}
P KH C QH P KH C QH C KH P C KH
B 2A 1A 1A
Each of the joint probability in the above equation can be expressed as a product since
they are independent conditional on the value of C 1A .

{ C
B
= ,
2A = } = { C
B
= |
1A = } { 2A = |
1A = } { 1A
= }
+ P{ C
B
= KH | C1A = QH} P{ C2A = QH | C1A = QH} P{ C1A
= QH}
+ P{ C = KH | C ≠ KHor QH} P{ C = QH | C ≠ KHorQH} P{ C ≠ KHorQH}
P KH C QH P KH C KH P C QH C KH P C KH
B 1A 2A 1A 1A
2 1 1 1 1 1 1 5 0
= ⋅ ⋅ + ⋅0⋅ + ⋅ ⋅
53 51 52 53 52 53 51 52
52
=
53⋅52⋅51
P C = QH = P C = QH | C = QH P C = QH
{ 2A } { 2A 1A } { 1A
}
+ P{ C = QH | C ≠QH} P{ C ≠QH}
2A 1A 1A
1 1 51
= 0⋅ + ⋅ 52 51 52
=
1
52
{ C | }
B
2 A
{ C
B
= ,
2 A
= }
P{ C = QH}
P KH C QH
P = KH C = QH =
2 A
52 52
= =
53⋅51
2703