Hints and answers

30.17 HINTS AND ANSWERS
constraint ni=1 ci Xi = 0, where the ci are constants (and not all equal to zero),
show that the variable
χ2n = (x − µ)T V−1 (x − µ)
follows a chi-squared distribution of order n − 1.
30.17 Hints and answers
30.1
30.3
30.5
30.7
(a) Yes, (b) no, (c) no, (d) no, (e) yes.
Show that, if px /16 is the probability that the total will be x, then the corrsponding
gain is [px (x2 + x) − 16x]/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.
P1 = α(α + β − αβ)−1 ; P2 = α(1 − β)(α + β − αβ)−1 ; P3 = P2 .
If pr is the probability that before the rth round both players are still in the
tournament (and therefore have not met each other), show that
r−1 n+1−r
1 2n+1−r − 2
2
−1
1
pr+1 =
and
hence
that
p
=
p
.
r
r
4 2n+1−r − 1
2
2n − 1
(a) The probability that they meet in the final is pn = 2−(n−1) (2n − 1)−1 .
(b) The probability
that they meet at some stage in the tournament is given by
the sum nr=1 pr (2n+1−r − 1)−1 = 2−(n−1) .
30.9
30.11
30.13
30.15
The relative probabilities are X : Y : Z = 50 : 36 : 8 (in units of 10−4 ); 25/47.
Take Aj as the event that a family consists of j boys and n − j girls, and B as
the event that the boy has at least two sisters. Apply Bayes’ theorem.
(i) For a even, the number of ways is 1 + 3 + 5 + · · · + (a − 3), and (ii) for a odd
it is 2 + 4 + 6 + · · · + (a − 3). Combine the results for a = 2m and a = 2m + 1,
with m running from 2 to N, to show that the total number of non-degenerate
triangles is given by N(4N + 1)(N − 1)/6. The number of possible selections of a
set of three rods is (2N + 1)(2N)(2N − 1)/6.
Show that k = e2 and that the average duration of a call is 1 minute. Let pn
be the probability that the call ends during the interval 0.5(n − 1) ≤ t < 0.5n
and cn = 20n be the corresponding cost. Prove that p1 = p2 = 14 and that
pn = 12 e2 (e − 1)e−n , for n ≥ 3. It follows that the average cost is
30
e2 (e − 1) −n
ne .
+ 20
2
2
n=3
∞
E[C] =
30.17
30.19
30.21
The arithmetico-geometric series has sum (3e−1 − 2e−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.
(a) The scores must be equal, at r each, after five attempts each.
(b) M can only be even if team 2 gets too far ahead (or drops too far behind) to
be caught (or catch up), with conditional probability p2 (or q2 ). Conversely, M
can only be odd as a result of a final action by team 1.
(c) Pr(i : x, y) = y Cx pxi qiy−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.
Show that dY /dX = f and use g(y) = f(x)|dx/dy|.
(a) Use result (30.84) to show that the PGF for S is Q/(1 − P q − P pt). Then use
equations (30.74) and (30.76).
(b) The PGF for the score is 6/(21 − 10t − 5t2 ) and the average score is 10/3.
The variance is 145/9 and the standard deviation is 4.01.
1219
PROBABILITY
30.23
30.25
30.27
30.29
30.31
30.33
30.35
30.37
30.39
Mean = 4/π. Variance = 2 − (16/π 2 ). Probability that X exceeds its mean
= 1 − (2/π) sin−1 (2/π) = 0.561.
Consider, separately, 0, 1 and ≥ 2 errors on a page.
Show that the maximum occurs at x = (r − 1)/λ, and then use Stirling’s approximation to find the maximum
value.
Pr(k chicks hatching) = ∞
n=k Po(n, λ) Bin(n, p).
There is not much to choose between the schemes. In (a) the critical value of
the standard variable is −2.5 and the average fine would be 15.5 euros. For
(b) the corresponding figures are −1.0 and 15.9 euros. Scheme (c) is governed
by a geometric distribution with p = q = 12 , and leads to an expected fine
∞
4n(n − 1)( 12 )n . The sum can be evaluated by differentiating the result
of
∞ n=1
n
n=1 p = p/(1 − p) with respect to p, and gives the expected fine as 16 euros.
(a) [12!(0.5)6 (0.3)3 (0.2)3 ]/(6! 3! 3!) = 0.0624.
You will need to establish the normalisation constant for the distribution (36),
the common mean value (3/5) and the common standard deviation (3/10). The
marginal distributions are f(x) = 3x(6x2 − 8x + 3), and the same function of y.
The covariance has the value −3/50, yielding a correlation of −2/3.
A = 3/(24a4 ); µX = µY = 5a/8; σX2 = σY2 = 73a2 /960; E[XY ] = 3a2 /8;
Cov[X, Y ] = −a2 /64.
(b) With the continuity correction Pr(xi ≥ 15) = 0.0334. The probability that at
least three are 15 or greater is 7.5 × 10−4 .
1220