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STAT 241/541 Homework 8 Solution
Section 5.1 : 10, 24, 32, 44
10 (a) The probability of X = k is:
Pr(X = k) =
(N
k
)(N −k n 1 −k
)(N −n 1 n 2 −k
(N
n 1
)(N
n 2
= (N^ −^ n^1 )!(N^ −^ n^2 )!n^1 !n^2! N !(N − n 1 − n 2 + k)!(n 1 − k)!(n 2 − k)!k!
(b) Let pN denote the probability X = n 12 , given that the total population is N. Then pN equals the expression in the answer to part (a), with k replaced by n 12. According to the hint, we consider the ratio: pN pN − 1 =^
(N − n 1 )(N − n 2 ) N (N − n 1 − n 2 + n 12 )
= N^
(^2) − (n 1 + n 2 )N + n 1 n 2 N 2 + (n 12 − n 1 − n 2 )N By setting the above ratio greater than or equal to 1, we obtain
N ≤ n^1 n^2 n 12
Therefore, N =
[ (^) n 1 n 2 n 12
]
will give us the maximum. In case n n^112 n^2 is integral, n 1 n 2 n 12 and^
n 1 n 2 n 12 −^ 1 both achieve the maximum.
24 The probability that John receives no mail on a given weekday can be ob- tained as follows:
p = Pr(N = 0) =
e−^10100 0! =^ e
− 10 ≈ 4. 54 × 10 − 5
So in 3000 days, the average number of days that John has no mail is 3000 ×
- 54 × 10 −^5. Let D be the number of days that bring no mail for John in a ten- year period, then D has a Poisson distribution with mean 3000× 4. 54 × 10 −^5. Therefore, the probability John wants to find is
Pr(D ≥ 1) = 1 − Pr(D = 0) = 1 − e−^3000 ×^4.^54 ×^10 −^5 ≈ 0. 127.
32 Let’s look at the probability mass function of X + Y :
Pr(X + Y = i) =
∑^ i
k=
Pr(X = k) Pr(Y = i − k)
∑^ i
k=
e−mmk k!
· e
− m¯ (^) m¯(i−k) (i − k)!
e−(m+ ¯m) i!
∑^ i
k=
i! k!(i − k)! ·^ m
k (^) m¯(i−k)
= e
−(m+ ¯m) i!
∑^ i
k=
i k
mk^ m¯(i−k)
= e
−(m+ ¯m) i!
(m + ¯m)i
Therefore, X + Y has a Poisson distribution with mean m + ¯m.
44 Consider the p.m.f. of a hypergeometric distribution
h(N, k, n, x) =
(k x
)(N −k n−x
(N
n
k! (k−x)!x! ·^
(N −k)! (n−x)!(N −k−n+x)! N! (N −n)!n! =
k!(N − k)!(N − n)! ︸(k −^ x)!(N^ −︷︷^ k^ −^ n^ +^ x)!N^ !︸ (a)
n! x!(n − x)!
Use Sterling formula n! ≈
2 πn · e−nnn^ to approximate the factorial terms in part (a) and we got
(a) ≈
k(N − k)(N − n) √ (k − x)(N − k − n + x)N
· k
k(N − k)N −k(N − n)N −n (k − x)k−x(N − k − n + x)N^ −k−n+xN N
∼ 1 · k
x(N − k)n−x N n =
( (^) k N
)x( (^) N − k N
)n−x = px(1 − p)n−x
Hence, as k,N → ∞,
h(N, k, n, x) ∼
n x
px(1 − p)n−x^ ,
which is binomial(n , p).
