Homework 7 Solutions
1.
A Poisson process with 10 accidents per week is equivalently a Poisson process with
' 10 / 7 accidents per day. Thus, given that 7 accidents occurred in a particular week, the
probability that exactly one accident occurred each day of that week is just
P(1 accident per day for 7 days) (e 10 / 7 (10 / 7)) 7 (0.3424) 7 0.0006
10
0.0067 .
P(7 accidents in a week )
0.0901
0.0901
e (10) 7 / 7!
0
2
4
f(x)
6
8
10
For a Poisson process with parameter , the waiting times between arrivals (in this case,
accidents) are independent and exponentially distributed with mean 1/ . In this situation, that
means the waiting times are exponentially distributed with mean 0.1 weeks between accidents,
which corresponds to a mean of 16.8 hours between accidents. However, the distribution of an
exponential random variable is heavily right-skewed – i.e., much of the mass of the pdf is
concentrated below the mean. (See plot below for this exponential distribution over the interval
[0, 1]) Hence, it is far more likely that waiting times less than 16.8 hours occur, and with shorter
waiting times, it is difficult to have only 1 accident a day for 7 straight days.
0.0
0.2
0.4
0.6
x
0.8
1.0
2. Ross 5.37 p. 251
a) If X is uniformly distributed over (-1, 1), then we have
0.5, 1 x 1
f ( x)
otherwise
0,
Hence, the probability that X 0.5 is
P( X 0.5) P( X 0.5) P( X 0.5) 0.25 0.25 0.5 .
b) To find the density function of X :
F X (a) P( X a) P(a X a) a , where 0 a 1 .
Taking the derivative of this with respect to a gives us
dF X (a)
f X (a)
1 , where 0 a 1 .
da
In other words, X is uniformly distributed on (0, 1).
3. Ross 5.41 p. 251
We have that is uniformly distributed on ( / 2, / 2) . If R A sin( ) , then to find the
distribution of R, we just find the cdf for R as a function of x and then take the derivative with
respect to x to obtain the pdf.
x
FR ( x) P( R x) P( A sin( ) x) P sin( )
A
d 1
x
x 1
P arcsin
arcsin
A
A 2
/ 2
dF ( x)
1
f R ( x) R
, where x A
dx
A2 x 2
arcsin(x / A )
4. Ross 6.6 p. 313
5
There are a total of 10 different sequences in which the 5 transistors (2 defective) can be
2
drawn from the bin. Each of these sequences is equally likely to occur, so the joint pdf of N1
and N 2 is just
1
P( N1 i, N 2 j ) , i 1, , 4; j 1, , 5 i
10
5. Ross 6.11 p. 314
The customers can be viewed as draws from a multinomial distribution where p1 0.45 is the
probability of an “ordinary” purchaser, p2 0.15 is the probability of a “plasma” purchaser, and
p3 0.40 is the probability of a browser. Hence, the probability that the store owner sells
exactly 2 ordinary sets and 1 plasma set on a day when 5 customers enter his store is just
5!
(0.45) 2 (0.15)1 (0.4) 2 0.1458 .
2!1!2!
6. Ross 6.10 p. 314
The joint pdf of X and Y is f ( x, y) e ( x y ) , where x and y are both non-negative real numbers.
a) To find the probability that X is less than Y:
e
P( X Y )
( x y )
x y 0
y
dxdy e
( x y )
0 0
y e 2 y
1
dxdy (1 e )e dy e
2 0 2
0
y
y
b) For this part, note that the joint pdf of X and Y can be re-written as
f ( x, y) e x e y f X ( x) f Y ( y) ; that is, X and Y are independent. Hence, P( X a) can
be obtained by integrating e x and ignoring any mention of y (if X and Y were not
independent, we would have to integrate f ( x, y ) over the full range of y and then
integrate x over the interval (0, a)):
a
P( X a) e x dx e x
a
0
1 e a .
0
7. Ross 6.14 p. 314
Let X 1 and X 2 denote respectively the locations of the ambulance and the accident at the
moment the accident occurs. Then the distance is given by X 1 X 2 , but this is equivalently the
range of the two Uniform(0, L) random variables (i.e., X 1 X 2 X ( 2) X (1) ). Using Equation
(6.7), we get as the cdf of
X1 X
F X 1 X 2 (a) P( X 1 X 2 a) P( X ( 2 ) X (1) a ) 2 [ F ( x1 a) F ( x1 )] f ( x1 )dx1
La
2
0
L
L
x2
a 1
2
2
L x1 1
dx1 2
dx1 2 a( L a) 2 Lx1 1
L L
L L
2 La
L
L
La
2
2 L2
( L a) 2
2
(
aL
a
)
L
(
L
a
)
2
L2
L2 2
2
a2
2 aL
2
L
a
a
2
L
L
To obtain the pdf, take the derivative of the cdf with respect to a:
dF X1 X 2 (a) 2 2a 2 a
f X 1 X 2 (a)
2 1 , 0 a L .
da
L L
L L
8. Ross 6.15 p. 314
a) The joint density over the region R must integrate to 1, so we have
1 f ( x, y )dxdy cdxdy cA( R) ,
( x , y )R
( x , y )R
where A(R) is the area of the region R; hence A( R) 1 / c .
b) Since the region R is a square with sides of length 2, A( R ) 4 , and f ( x, y ) 1 / 4 for
1 x 1, 1 y 1 . But this can be re-written as f ( x, y ) f ( x) f ( y ) , where
f ( x) 1 / 2, 1 x 1 , and f ( y ) 1 / 2, 1 y 1 . In other words, X and Y are
independent uniform random variables on the interval (-1, 1).
c) The probability that (X, Y) lies in the circle of radius 1 centered at the origin is just the
area of the circle multiplied by 1/4:
1
1
P( X 2 Y 2 1)
dxdy A(C ) .
4
4
4
C : x 2 y 2 1
0
