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Chapter 4 Problems
- (a) Ω = the collection of all possible ways to select three from the 16 grapefruits so described. Note that |Ω| =
3
(b) We need to compute f( 0 ), f( 1 ), f( 2 ), and f( 3 ), since f(x) = 0 for all other values of x. Now
f( 0 ) =
3
3
Similarly, f( 1 ) =
1
×
2
3
f( 2 ) =
2
×
1
3
, and
f( 3 ) =
3
3
As an aside, we have also
E(X) =
1 ×
2 ×
3 ×
5(a). (i) Want c ·
x= 1 2 x/x! = 1 , that is to say that c is the reciprocal of ∑∞ x= 1 2 x/x!. By the Taylor–McLaurin formula ∑∞ x= 0 2 x/x! = e (^2). Therefore,
x= 1 2 x/x! = e (^2) − 1 , and hence c = 1 /(e (^2) − 1 ). (ii) Want c = 1 /
x= 1 p x. Since ∑∞ x= 0 p x (^) = 1 /( 1 − p), ∑∞ x= 1 p x (^) = 1 /( 1 − p) − 1 = p/( 1 − p). Therefore, c = ( 1 − p)/p. (iii) As before, c = 1 /
x= 1 p xx− (^1). Now
∑^ ∞ x= 1
px x
∑^ ∞
x= 1
∫ (^) p
0
ux−^1 du =
∫ (^) p
0
∑^ ∞
x= 1
ux−^1 du =
∫ (^) p
0
du 1 − u = ln
1 − p
Therefore, c is the reciprocal of ln( 1 /( 1 − p)). (iv) c = 1 /
x= 1 x − (^2) , which happens to be 6 /π (^2) , since the sum can be shown to be π^2 / 6. (v) c is the reciprocal of
x= 1 [x(x^ +^1 )] − (^1). Because
1 x(x + 1 )
x
x + 1
∑N
x= 1
x(x + 1 )
∑^ N
x= 1
x
∑^ N
x= 1
x + 1
=
N
N + 1
N + 1
Let N → ∞ to find that
x= 1 [x(x^ +^1 )]−^1 =^1 , and hence^ c^ =^1.
- Yes, yes, and yes.
- Condition on the position of the first tails tossed. Let Tj denote the event that the first tail occurs on the jth toss. Then,
Pr(An) = Pr(An | T 1 ) Pr(T 1 ) + Pr(An | T 2 ) Pr(T 2 ) + Pr(An | T 3 ) Pr(T 3 ),
since Pr(An ∩ T 4 ) = Pr(An ∩ T 5 ) = · · · = 0. You should check that Pr(Tk) = ( 1 / 2 )k. Therefore,
Pr(An) =
Pr(An | T 1 ) +
Pr(An | T 2 ) +
Pr(An | T 3 ).
Because the tosses are independent from one another, the conditional probab. of An, given that the first toss is tails, is the same as Pr(An− 1 ). Similarly, Pr(An | T 2 ) = Pr(An− 2 ) and Pr(An | T 3 ) = Pr(An− 3 ). There- fore, Pr(An) = 1 2 Pr(An− 1 ) + 1 4 Pr(An− 2 ) + 1 8 Pr(An− 3 ). (∗)
Because Pr(A 1 ) = Pr(A 2 ) = 0 and Pr(A 3 ) = 1 / 8 , the preceding inductive formula solves Pr(An) for all n > 4 , viz., (a) Pr(A 4 ) = 12 Pr(A 3 ) = 1 / 16 ; (b) Pr(A 5 ) = 12 Pr(A 4 ) + 14 Pr(A 3 ) = 1 / 16 ; (c) Pr(A 6 ) = 12 Pr(A 5 ) + 14 Pr(A 4 ) + 18 Pr(A 3 ) = 1 / 16 , (d) Pr(A 7 ) = 12 Pr(A 6 ) + 14 Pr(A 5 ) + 18 Pr(A 4 ) = 7 / 128 , etc. [It is true that (∗) can be solved in terms of the root of a cubic equa- tion, using the generating-function methods of chapter 3, but (∗) is a very useful formula.]
