Homework 3 Solutions
1. Ross 3.19 p. 113
a) Choose a random member of the class as a whole. Let A be the event that this person
attends the success party, let W be the event that this person is a woman and let M be the
event that this person is a man. Then, using Bayes’s formula, we have
P( A | W ) P(W )
(0.48)(0.38)
P(W | A) 

 0.443 ,
P( A | W ) P(W )  P( A | M ) P( M ) (0.48)(0.38)  (0.37)(0.62)
so about 44.3% of those attending the party were women.
b) The percentage of the original class that attended the party is just the denominator from
above, so (0.48)(0.38)  (0.37)(0.62)  0.4118 , or 41.18%, of the original class attended
the party.
2. Ross 3.26 p. 114
a) If we assume there are an equal number of males and females, then the probability of a
randomly chosen colorblind person being male is just (using M to denote the event of a
person being male, F to denote the event that the person is a female, C to denote the
event of a person being colorblind)
P(C | M ) P( M )
(0.05)(0.5)
P( M | C ) 

 0.952 .
P(C | M ) P( M )  P(C | F ) P( F ) (0.05)(0.5)  (0.0025)(0.5)
b) If we instead assume that there are twice as many males as females, then this probability
becomes
(0.05)( 2 / 3)
P( M | C ) 
 0.976 .
(0.05)( 2 / 3)  (0.0025)(1 / 3)
3.
a) Ross 3.42 p. 116
Let A, B, and C denote the events that a cake is baked by cook A, B, and C, respectively,
and let F denote the event that a cake fails to rise. Then, the proportion of failures caused
by A is just
(0.02)(0.5)
10
P( A | F ) 

 0.345
(0.02)(0.5)  (0.03)(0.3)  (0.05)(0.2) 29
b) Ross 3.55 p. 118
Let x be the number of sophomore girls that are present. We want sex and class to be
independent when a student is selected at random; hence, we want
P(sophomore girl )  P(sophomore ) P(girl ) . Using the numbers that we are given and
then solving for x, this means that
x
 x  6  x  6 



x  16  x  16  x  16 
x( x  16)  ( x  6) 2
x 2  16 x  x 2  12 x  36
4 x  36  x  9
So, we need 9 sophomore girls to be present in order for sex and class to be independent
in this scenario.
4. Ross 3.57 p. 118
a) The probability that after 2 days the stock will be at its original price is just the
probability of (move up day 1, move down day 2) plus the probability of (move down
day 1, move up day 2). Since the changes on different days are assumed to be
independent, each of these events has probability p(1  p) . Hence, the probability of the
stock being at its original price after 2 days is just 2 p(1  p) .
b) The probability that after 3 days the stock’s price will have increased by 1 unit is
obtained by considering the probability of seeing the stock move up on 2 different days
and move down on another, multiplied by the number of ways you can arrange 2 moves
up among 3 total moves; in other words, the probability of the stock’s price having
 3
increased by 1 unit after 3 days is just   p 2 (1  p)  3 p 2 (1  p) .
 2
c) Out of the 3 ways to arrange 2 moves up and 1 move down, only 1 of the 3 ways does
NOT have a move up on the first day. Hence, the probability of the stock’s price moving
up the first day given that it increased by 1 unit after 3 days is just 2/3.
5. Ross 3.60 p. 118-19
a) Based on the information from the problem, we can reasonably deduce that Smith’s
parents both have one brown-eyed gene and one blue-eyed gene (else, the combination of
Smith having brown eyes and Smith’s sister having blue eyes could not occur). Then the
possible gene pairings for Smith would be {(Br, Br), (Br, Bl), (Bl, Br)}. Let X denote the
number of blue-eyed genes that Smith received. The values of X corresponding to the
above set of outcomes are just {0, 1, 1}. Then the probability of him having a blue-eyed
gene is 2/3.
b) Since Smith’s wife has blue eyes, then the probability that their first child has blue eyes
depends entirely on Smith contributing a blue-eyed gene, and as such, this probability is
2
1 1
P(child blue )  P(Bl | 1 Bl ) P(1 Bl )  P(Bl | 0 Bl ) P(0 Bl )  (0.5)   (0)   .
3
3 3
c) First, we need to compute the probability that Smith has a blue-eyed gene given that his
first child has brown eyes. Then, using this updated probability, we can compute the
probability that his second child also will have brown eyes.
P(child brown | 1 Bl ) P(1 Bl ) (1 / 2)( 2 / 3) 1
P(1 Bl | child brown ) 


P(child brown )
1  1/ 3
2
 1  1   1  3
P(2nd child brown)  P(brown | 1 Bl ) P(1 Bl )  P(brown | 0 Bl ) P(0 Bl )      1  
 2  2   2  4
6. Ross 2.26 p. 58
Let E i denote the event the initial outcome is i and the player wins. Then the desired probability
12
is
 P( E ) .
i
i2
Before we go any further, note that P( E2 )  P( E3 )  P( E12 )  0 , P ( E 7 ) 
1
, and
6
1
. For any other value of i, define the event E i , n to be the event that the initial
18
outcome is i and the player wins on roll n. It is clear that the E i , n are mutually exclusive for any
P( E11 ) 
value of i (since one cannot win on roll n if one has already won on roll n  1 , and so on). Hence,
the probability of event E i should just be the sum over n of the events E i , n ; in other words,

P( Ei )   P( Ei ,n ) . Then, all that is needed is to determine P ( E i ,n ) for i = 4, 5, 6, 8, 9, 10; but
n2
since the pairs (4, 10), (5, 9), and (6, 8) are identical in terms of probability, we can simplify
matters by looking just at values i = 4, 5, 6 (taking care to double the probability for 4, 5, and 6
to reflect the fact that they’re pulling double-duty for 10, 9, and 8, respectively). Leaning back
toward our original problem, this means that the probability of winning at craps can be expressed
as
1 1
2
P( win)  2 P( E 4 )  2 P( E5 )  2 P( E6 )  
 2[ P( E 4 )  P( E5 )  P( E6 )]  .
6 18
9
Now, if our initial outcome is i (where we restrict our attention to i being one of {4, 5, 6}), then
in order for us to win on roll n, we would need to have rolled something other than i or 7 on all
rolls between 1 and n (non-inclusive); i.e., we would need n  2 rolls that are not i or 7
sandwiched between our 2 rolls of i. This means that P( Ei ,n )  P(i) 2 (5 / 6  P(i)) n2 (where
P (i ) is just the probability of rolling i and since P (7)  1 / 6 ). Substituting this expression into
the sum for P( Ei ) gives us


n2
k 0
P( Ei )  P(i ) 2  (5 / 6  P(i )) n  2  P(i ) 2  (5 / 6  P(i )) k ,
where we have chosen to re-index the sum using k  n  2 . Note that the sum on the right-hand
side is a geometric series with 0  r  (5 / 6  P(i ))  1 ; hence, the expression for P( Ei ) should
simplify to
P(i) 2
P(i) 2
.

1  (5 / 6  P(i)) 1 / 6  P(i)
Since P(4)  1 / 12 , P(5)  1 / 9 , and P(6)  5 / 36 , this means that P( E4 )  1 / 36 , P( E5 )  2 / 45 ,
and P( E6 )  25 / 396 ; plugging these values back into our expression for P (win ) gives us
P( Ei ) 
2
25  2
1
 67  2 244
P( win)  2 

  2
 0.493 .
 

 36 45 396  9
 495  9 495
7. Ross 3.89 p. 123
a) Let J1 , J 2 , and J 3 represent judges 1, 2, and 3, respectively. Then the probability of
judge 3 voting guilty given that judges 1 and 2 voted guilty should just be the probability
of all 3 voting guilty divided by the probability of the first 2 judges voting guilty (using
the definition of conditional probability); the probability of all 3 voting guilty (as well as
judges 1 and 2 voting guilty) can be decomposed into 2 components: the probability of
all 3 (or judges 1 and 2) voting guilty and the defendant being guilty and the probability
of all 3 (or judges 1 and 2) voting guilty and the defendant being innocent. In other
words,
P(all 3 vote guilty )
P( J 3 votes guilty | J 1 and J 2 vote guilty ) 
P( J 1 and J 2 vote guilty )

(0.7)(0.7) 3  (0.3)(0.2) 3
97

 0.683
2
2
142
(0.7)(0.7)  (0.3)(0.2)
b) Now, we want the probability of judge 3 voting guilty given that one of judges 1 and 2
voted guilty and the other not guilty. Again, we can decompose into the case where the
defendant is guilty and the case where the defendant is innocent; this gives us the
probability
P( J 3 votes guilty | one of J 1 , J 2 votes guilty )

(0.7)(0.7)[ 2(0.7)(0.3)]  (0.3)(0.2)[ 2(0.2)(0.8)] 15

 0.577
(0.7)[ 2(0.7)(0.3)]  (0.3)[ 2(0.2)(0.8)]
26
c) Now, we want the probability of judge 3 voting guilty given that both judges 1 and 2
voted not guilty; as before, we decompose into a probability for when the defendant is
guilty and another for when the defendant is innocent. So, the probability in this case is
P( J 3 votes guilty | J 1 and J 2 vote innocent )
(0.7)(0.7)(0.3) 2  (0.3)(0.2)(0.8) 2 11


 0.324
34
(0.7)(0.3) 2  (0.3)(0.8) 2
The events E i are conditionally independent given the guilt or innocence of the defendant.
8.
a) We start off with probability 0.5 of the original fish being a piranha (or a goldfish). Once
our sushi lover throws a piranha into the bowl, we have probability 0.5 of there being 2
piranhas (or 1 goldfish and 1 piranha). The probability of him removing a piranha when
there’s 2 piranhas is naturally 1, and the probability of him removing a piranha when
there’s 1 piranha and 1 goldfish is 0.5. Let go o , pio , and pi g represent the events that a
goldfish was originally in the bowl, a piranha was originally in the bowl, and a piranha
was grabbed from the bowl respectively; then, the probability of the original fish being a
piranha given that the fish he grabbed was a piranha should be
P( pi g | pio ) P( pio )
(1)(0.5)
2
P( pio | pi g ) 

 .
P( pi g | pio ) P( pio )  P( pi g | goo ) P( goo ) (1)(0.5)  (0.5)(0.5) 3
b) Our sushi lover now throws a piranha and a goldfish into the bowl, so we have
probability 0.5 of there being 2 piranhas and 1 goldfish (or conversely, 2 goldfish and 1
piranha). The probability of him removing a piranha when the original fish was a piranha
is 2/3, and the probability of him removing a piranha when the original fish was a
goldfish is 1/3. Using the same notation as before, the probability of the original fish
being a piranha given that he removed a piranha is just
(2 / 3)(0.5)
2
P( pio | pi g ) 
 .
(2 / 3)(0.5)  (1 / 3)(0.5) 3
(Side note: This suggests that, using the fact that our original bowl is equally likely to have a
piranha or a goldfish, if our sushi lover tosses 1 piranha and n goldfish into the bowl, the
probability of the original fish being a piranha given that he removes a piranha will be 2/3 for all
values of n.)