We are interested in the probability P( 3 X 0). As seen from Fi gure 7.2, it
is clear that this probability is equal to the ratio of the shaded area and the unit
total area. Hence,
It is also clear that, owing to uniformity in the distribution, the solution can
be found simply by taking the ratio of the length from 3 to 0 to the total length
of the distribution interval. Stated in general terms, if a random variable X is
uniformly distributed over an interval A, then the probability of X taking
values in a subinterval B is given by
7.1.1 Bivariate Uniform Distribution
Let random variable X be uniformly distributed over an interval (a 1 ,b 1 ), and let
random variable Y be uniformly distributed over an interval (a 2 ,b 2 ). Further-
more, let us assume that they are independent. Then, the joint probability
density function of X and Y is simply
fX(x)
x
–3 0 5
1
— 8
Figure 7. 2 Probability density function, fX (x), of X, in Example 7.1
Some Important Continuous Distributions 193
P 3 X 0 3
1
8
3
8
:
P
XinB
length ofB
length ofA
: 7 : 4
fXY
x;yfX
xfY
y
1
b 1 a 1
b 2 a 2
; fora 1 xb 1 ;anda 2 yb 2 ;
0 ; elsewhere:
8
<
:
7 : 5