Download Homework 4 Solutions

Homework 4 Solutions
1.
Let p1 be the probability that Andre wins a set against his father, and let p 2 be the probability
that Andre wins a set against the club champion. Clearly, because the club champion is a better
player than Andre’s father, we have p1  p2 .
For sequence (a) father-champion-father, the probability of Andre winning at least 2 tennis sets
in a row is just the probability of him winning the first 2 sets plus the probability of him losing
the 1st set and winning the 2nd and 3rd sets:
P( Andre wins sets 1  2)  P( Andre loses set 1 and wins sets 2  3)
 p1 p 2  (1  p1 ) p 2 p1  (2  p1 ) p1 p 2
For sequence (b) champion-father-champion, the probability of Andre winning at least 2 tennis
sets in a row is instead p2 p1  (1  p2 ) p1 p2  (2  p2 ) p1 p2 .
Comparing the probabilities for (a) and (b), note that because p1  p2 , it follows that
2  p1  2  p2 ; this means that Andre would have a higher probability of winning at least 2 sets
in a row if he chose to play the champion-father-champion sequence.
2. Ross 3.80 p. 122
a) Let Ai be the event that A plays in exactly i contests, where i can be any integer value
between 1 nd n, inclusive. In order for A to play in exactly i contests, A must lose the i-th
contests for values of i less than n ( i  n  1 ). Why does this not hold when i  n ?
Because once A reaches the n-th contest, there are none after, so A will play in exactly n
contests regardless of whether he wins or loses the n-th contest. In other words, the
probability that A plays in exactly n contests is just the probability that A wins all contests
up to and including contest n  1 . These statements give us P( Ai ) as
(1 / 2) i i  n  1
.
P( Ai )  
n 1
in
(1 / 2)
b) Let E denote the event that contestants A and B ever play each other. We can obtain this
probability by conditioning on the events Ai . In other words, we know that
n
P( E )   P( E | Ai ) P( Ai ) .
i 1
Note that Ai means A faced i different contestants, and each of the 2 n  1 other
contestants have an equal chance of ever playing with A; hence, P( E | Ai )  i /( 2 n  1) .
Using what we determined for P( Ai ) from (a), we have
n 1
 i 
 n 
i
n 1
P( E )    n
(1 / 2)   n
(1 / 2)
 2 1
i 1  2  1 
 1  n 1
 n 
i
n 1
 n
 i (1 / 2)   n
(1 / 2)
 2  1  i 1
 2 1
n 1
n
 1  1  1  n(1 / 2)  (n  1)(1 / 2)
 n
 
(1  1 / 2) 2
 2  1  2 
  1 
n 1
   n
 n(1 / 2)
2

1

 


 1 
n 1
n
n 1
 n
 2(1  n(1 / 2)  (n  1)(1 / 2) )  n(1 / 2)
 2 1
 1 
n 1
n
n
 n
 2  n(1 / 2)  2n(1 / 2)  2(1 / 2)
2

1






2
1
 2 
n
 n
[1  (1 / 2) ]  n  n 1
2
2
 2 1
c) Let Pn  P(E ) be the probability that A and B meet in a contest with 2 n players. Now,
let F be the event that A and B play in the first contest; then
P( E )  Pn  P( E | F ) P( F )  P( E | F C ) P( F C ) . But since we know that the probability
that A and B meet in the first contest is just 1 /( 2 n  1) , we get that
n
 1 
C 2 2
 .
Pn  (1) n
  P( E | F ) n
 2 1
 2 1 
Now, let G be the event that both A and B advance to the second round of this contest (i.e.,
the event that both win their first match). Then we have that
P( E | F C )  P( E | F C  G) P(G | F C )  P( E | F C  G C ) P(G C | F C )
2
2
1
1
 P( E | F C )  Pn 1    0    Pn1
2
2
In other words, the probability that they meet given that they both advance to the second
round and don’t meet in the first round is just the probability that out of 2 n 1 possible
contestants, we see A and B meet ( 2 n 1 because the number of contestants left after 1
round has been completed is half the original number). Hence, we have that
2
 2 n  2  1 
1
  Pn 1 .
P( E )  Pn  n
  n
2  1  2  1  2 
To verify that this holds, let us plug in P( E )  Pn  1 /( 2 n 1 ) (and hence Pn 1  1 /( 2 n  2 ) )
to verify that this identity holds:
 2 n  2  1 
 2 n  2  1   1 
1
1



  

P





n 1
2 n  1  2 n  1  2 
2 n  1  2 n  1  2   2 n  2 
2
2
 2 n  2  1 
1
1
    
 n
  n
2  1  2  1  2 
2
n
1
 
2
(Checks out)
n
n
 2n  2n  2 


n
 2 1 
 2 n 1  2   1   2(2 n  1)  2
1
 n
     n
  n  n 1  Pn
2
 2 1   2   2 1  2
n
d) Note that every game played yields 1 loser. Also, with 2 n contestants and only 1 winner,
there must be 2 n  1 losers; hence, we require 2 n  1 games in order to obtain 2 n  1
losers.
e) Let Bi denote the event that A and B play each other in game i, where i  1,,2 n  1 .
 2n 
Note that there are   possible pairings that can occur. Also, since players are equally
2
 2n 
skilled, the 2 players in game i are equally likely to be any of the   pairs. Hence, it
2
n
2 
2
follows that P( Bi )  1    n n
.
 2  2 (2  1)
f) Since the events Bi are mutually exclusive, then the event E is just the union of the
events Bi ; hence,

 2
2
1
  n  n1 .
P( E )   P( Bi )  (2 n  1) n n
2
 2 (2  1)  2
3.
Let a k denote the probability of having no two consecutive heads in a sequence of k tosses of a
fair coin. Then let bk (i ) be the probability of having at least 2 heads in a row if there are k
tosses remaining, i consecutive heads have already appeared, and there has not already been a
run of at least 2 heads prior to the current run of i heads in a row. A recursive formula using
bk (i ) ’s will give the complement to a k .
Adapting the FT shooting recursive formula from the addendum to the notes for a coin-tossing
scenario, we have that
1
1
bk (i )  bk 1 (i  1)  bk 1 (0) for i  0,1
2
2
with boundary conditions
b0 (i)  0 for i  0,1 and bk (2)  1 .
Then the probability that we have no runs of at least 2 consecutive heads in a sequence of k
tosses of a fair coin is just ak  1  bk (0) .
The following R function will be used to calculate bk (0) for k = 5, 10, and 25 by calculating
b0 (0), b0 (1), b0 (2), b1 (0), b1 (1), , bk 1 (1), bk 1 (2), bk (0) .
bmat <- matrix(rep(0, 26*3), nrow = 26)
# bmat will store the b_k(i)’s. The 1, ..., 26 rows of bmat correspond to
# k=0,...,25 and the 1,...3 columns of bmat correspond to i=0,1,2.
bmat[,3] <- 1 # set b_k(2)=1
for(k in 2:26){
bmat[k,1:2] <- .5*bmat[(k-1),2:3]+.5*bmat[(k-1),1]
formula
}
# Apply recursive
To get a 5 , a10 , and a 25 from the matrix, we just need to input 1 – bmat[c(5, 10, 25)+1, 1]:
> 1 - bmat[c(5,10,25)+1, 1] # gives us a_5, a_10, and a_25, respectively
[1] 0.406250000 0.140625000 0.005853713
So, as we can see, the probability of not seeing at least 1 run of 2 heads or longer in a sequence
of k tosses decreases fairly rapidly as k increases.
4. Ross 4.4 p. 187
We have 5 men and 5 women who are ranked according to their scores on an exam, where no
two scores are alike and all 10! possible rankings are equally likely. Let X denote the highest
ranking achieved by a woman. Clearly, since 6 is the lowest possible rank attainable by the
highest-scoring female, we must have P( X  7)  P( X  8)  P( X  9)  P( X  10)  0 .
For X  1 (female is highest-ranking scorer), we have 5 possible choices out of 10 for the top
spot that satisfy this requirement; hence P( X  1)  1 / 2 .
For X  2 (female is 2nd-highest scorer), we have 5 possible choices for the top male, then 5
possible choices for the female who ranked 2nd overall, and then any arrangement of the
remaining 8 individuals is acceptable (out of 10! possible arrangements of 10 individuals); hence,
P( X  2)  (5  5  8!) / 10! 5 / 18  0.278 .
For X  3 (female is 3rd-highest scorer), acceptable configurations yield (5)(4)=20 possible
choices for the top 2 males, 5 possible choices for the female who ranked 3rd overall, and 7!
different arrangement of the remaining 7 individuals (out of a total of 10! possible arrangements
of 10 individuals); hence, P( X  3)  (5  4  5  7!) / 10! 5 / 36  0.139 .
For X  4 (female is 4th-highest scorer), acceptable configurations yield (5)(4)(3)=60 possible
choices for the top 3 males, 5 choices for the female who ranked 4th overall, and 6! different
arrangements of the remaining 6 individuals (out of a total of 10! possible arrangements of 10
individuals); this gives us P( X  4)  (5  4  3  5  6!) / 10! 5 / 84  0.060 .
For X  5 (female is 5th-highest scorer), satisfactory arrangements have (5)(4)(3)(2)=120
possible choices for the top 4 males, 5 choices for the female who ranked 5th overall, and 5!
different configurations of the remaining 5 individuals, thus giving us a probability of
P( X  5)  (5  4  3  2  5  5!) / 10! 5 / 252  0.020 .
For X  6 (female is 6th-highest scorer), we basically have 5! arrangements of the males in the
top 5 positions, and another 5! arrangements of the females in the bottom 5 positions; hence, the
probability of all 5 males outperforming the top female is P( X  6)  (5!5!) / 10! 1 / 252  0.004 .
5. Ross 4.14 p. 188
So we have 5 numbers randomly distributed to players numbered 1 through 5; players compare
numbers in pairs, with the higher number winning. Let X denote the number of times player 1 is
a winner. Then the probabilities P ( X  k ) are as follows:
P( X  0)  P(1 loses to 2)  1 / 2
P( X  1)  P(of 1, 2, 3 : 3 has largest, then 1, then 2)  (1 / 3)(1 / 2)  1 / 6
P( X  2)  P(of 1, 2, 3, 4 : 4 has largest and 1 has next largest )  (1 / 4)(1 / 3)  1 / 12
P( X  3)  P(of 1, 2, 3, 4, 5 : 5 has largest then 1)  (1 / 5)(1 / 4)  1 / 20
P( X  4)  P(1 has largest)  1 / 5
6.
i) Ross 4.21 p. 190
a) It would seem reasonable to believe that E ( X ) should be larger than E (Y ) . Note that
each of the bus drivers is equally likely to be selected, whereas the student selected is
more likely to have come from a bus with larger capacity. In other words, X has greater
weight placed on the larger busses, and Y does not.
b) For X, we know that P( X  k )  k / 148 , where k can take on values of 40, 33, 25, and 50.
Hence, E ( X )  (40)( 40 / 148)  (33)(33 / 148)  (25)( 25 / 148)  (50)(50 / 148)  39.284 .
On the other hand, E (Y )  (40)(1 / 4)  (33)(1 / 4)  (25)(1 / 4)  (50)(1 / 4)  37 .
ii) Ross 4.37 p. 192
We now want Var ( X ) and Var (Y ) for the random variables from (i). To determine their values,
let us first calculate E ( X 2 ) an E (Y 2 ) :
E ( X 2 )  (40) 2 (40 / 148)  (33) 2 (33 / 148)  (25) 2 (25 / 148)  (50) 2 (50 / 148)  1625.419
E (Y 2 )  (40) 2 (1 / 4)  (33) 2 (1 / 4)  (25) 2 (1 / 4)  (50) 2 (1 / 4)  1453.5
This gives us the variances of X and Y:
Var ( X )  E ( X 2 )  E ( X ) 2  1625.419  (39.284) 2  82.186
Var (Y )  E (Y 2 )  E (Y ) 2  1453.5  (37) 2  84.5
.
7. Ross 4.32 p. 192
a) Let T be a random variable denoting the number of tests that are required for a group of
10 people. Then we have that T  1 if all 10 people turn up negative T  11 if at least 1
people turns up positive. Hence, P(T  1)  (1  0.9)10  1  P(T  11) , and this gives us
as the expected number of tests necessary for each group
E (T )  (1)(0.9)10  (11)[1  (0.9)10 ]  11  (10)(0.9)10  7.513
b) Let p be the probability that an individual has the disease. Then the expected number of
tests required for each group of 10 people under the group testing scheme is
E (T )  11  (10)(1  p)10 compared to 10 tests required for each group of 10 people under
a “test each individual” scheme. Hence, let us find that value of p for which E (T )  10 :
11  (10)(1  p )10  10
1  (10)(1  p )10
0.1  (1  p )10
0.794  1  p
p  0.206
Noting from (a) that p  0.1 yielded an expected number of tests less than 10, we can
conclude that the group testing scheme is better when p  0.206 and consequently, it is
better to test each individual separately when p  0.206 .
8. Ross 4.35 p. 192
a) Since we are drawing without replacement, we can see that (letting R denote drawing a
red marble and B denote drawing a blue marble) P( RR )  P( BB )  2 / 9 and
P( RB )  P( BR )  5 / 18 . Hence, if we let X be the random variable representing your
winnings playing this marble game, then P( X  1.1)  4 / 9  1  P( X  1) . This gives
us an expected value for X of
E ( X )  (1.10)( 4 / 9)  (1.00)(5 / 9)  0.6 / 9  0.067 .
In other words, you would expect to lose about 7 cents every time you play this game.
b) To compute the variance, we can use the fact that Var ( X )  E ( X 2 )  E ( X ) 2 , which only
requires to determine what E ( X 2 ) is, since we have E ( X ) from part (a):
E ( X 2 )  (1.1) 2 (4 / 9)  (1.0) 2 (5 / 9)  (1.21)(4 / 9)  (1.0)(5 / 9)  1.093 .
Hence, the variance of the amount you win is just Var ( X )  1.093  (0.067) 2  1.089 .