STAT 305B Fall 2007
Homework 4–Solutions
•Problem 1:
(a) Let {N1, N2}be the nuts selected, a=1
2in. bolt and b=9
16 in. bolt, then the sample
space contains all the possible outcomes which are A1={a, a},A2={a, b}and
A3={b, b}. I assume here that the order of N1and N2does not matter. See part (b)
for the probability associated with each outcome.
(b) Let X=
1 if {N1, N2}=A1
2 if {N1, N2}=A2
3 if {N1, N2}=A3
, then by the condition given, the random variable X
is defined by x1 2 3
f(x) 0.09 0.42 0.49 .
(c) The sample space contains all the possible pairs of the nut and bolt . P(A)=P({1
2in.
pair or 9
16 pair})=0.3×0.4 + 0.7×0.6 = 0.54.
(d) If nuts of the same size are not distinguishable, then the total number of distinct
samples are only 11, i.e. from no 1
2in. nut to 10 1
2in. nuts. If the nuts of the same size
are distinguishable, then the total number of distinct samples are
30
070
10+30
170
9+···+30
1070
0=100
10 .
(e) Considering that the order of the selected iterms does not matter, P(B) = (30
10)
(100
10 ).In this
problem, if considering the order of the selcted items does matter, the following answer
is also valid, P(B) = 30×29×···×21
100×99×···×91 .
•Problem 2:
(a) k= 0.7, as k+ 0.1 + 0.2 = 1 by Axioms. The arguments are the sample space
S={1,2,3}, so P(S) = P({X= 1 or X= 2 or X= 3}) = 1. Since {X= 1},
{X= 2}and {X= 3}are mutually exclusive, P({X= 1 or X= 2 or X= 3}) =
P({X= 1}) + P({X= 2}) + P({X= 3}).
(b) P(C) = P(X= 1 or X= 2) = 0.1 + 0.2 = 0.3.
(c) P(D) = 1.
(d) P(E) = P(T > 0) = P(X > 2.5) = P(X= 3) = 0.7.
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