Mathematical Methods for Physics and Engineering : A Comprehensive Guide

30.17 HINTS AND ANSWERS

constraint

∑n

i=1ciXi=0,wheretheciare constants (and not all equal to zero),

show that the variable

χ^2 n=(x−μ)TV−^1 (x−μ)

follows a chi-squared distribution of ordern−1.

30.17 Hints and answers

30.1 (a) Yes, (b) no, (c) no, (d) no, (e) yes.
30.3 Show that, ifpx/16 is the probability that the total will bex, then the corrsponding
gain is [px(x^2 +x)− 16 x]/16. (a) A loss of 0.36 euros; (b) a gain of 27/64 euros;
(c) a gain of 2.5 euros, provided he takes your advice and guesses ‘5’ each time.
30.5 P 1 =α(α+β−αβ)−^1 ;P 2 =α(1−β)(α+β−αβ)−^1 ;P 3 =P 2.
30.7 Ifpris the probability that before therth round both players are still in the
tournament (and therefore have not met each other), show that

pr+1=

1

4

2 n+1−r− 2

2 n+1−r− 1

pr and hence that pr=

(

1

2

)r− 1

2 n+1−r− 1

2 n− 1

.

(a) The probability that they meet in the final ispn=2−(n−1)(2n−1)−^1.

(b) The probability that they meet at some stage in the tournament is given by

the sum

∑n

r=1pr(2

n+1−r−1)− (^1) =2−(n−1).
30.9 The relative probabilities areX:Y:Z= 50 : 36 : 8 (in units of 10−^4 ); 25/47.
30.11 TakeAjas the event that a family consists ofjboys andn−jgirls, andBas
the event that the boy has at least two sisters. Apply Bayes’ theorem.
30.13 (i) Foraeven, the number of ways is 1 + 3 + 5 +···+(a−3), and (ii) foraodd
it is 2+4+6+···+(a−3). Combine the results fora=2manda=2m+1,
withmrunning from 2 toN, to show that the total number of non-degenerate
triangles is given byN(4N+1)(N−1)/6. The number of possible selections of a
set of three rods is (2N+ 1)(2N)(2N−1)/6.
30.15 Show thatk=e^2 and that the average duration of a call is 1 minute. Letpn
be the probability that the call ends during the interval 0.5(n−1)≤t< 0. 5 n
andcn=20nbe the corresponding cost. Prove thatp 1 =p 2 =^14 and that
pn=^12 e^2 (e−1)e−n,forn≥3. It follows that the average cost is

E[C]=

30

2

+20

e^2 (e−1)

2

∑∞

n=3

ne−n.

The arithmetico-geometric series has sum (3e−^1 − 2 e−^2 )/(e−1)^2 and the total
charge is 5(e+1)/(e−1) = 10.82 pence more than the 40 pence a uniform rate
would cost.
30.17 (a) The scores must be equal, atreach, after five attempts each.
(b)Mcan only be even if team 2 gets too far ahead (or drops too far behind) to
be caught (or catch up), with conditional probabilityp 2 (orq 2 ). Conversely,M
can only be odd as a result of a final action by team 1.
(c) Pr(i:x, y)=yCxpxiqiy−x.
(d) If the match is still alive at the tenth kick, team 2 is just as likely to lose it
as to take it into sudden death.
30.19 Show thatdY /dX=fand useg(y)=f(x)|dx/dy|.
30.21 (a) Use result (30.84) to show that the PGF forSisQ/(1−Pq−Ppt). Then use
equations (30.74) and (30.76).
(b) The PGF for the score is 6/(21− 10 t− 5 t^2 ) and the average score is 10/3.
The variance is 145/9 and the standard deviation is 4.01.