Standard Probability Distributions - Advanced Probability Theory for Biomedical Engineers - page 8

ADVANCED PROBABILITY THEORY FOR BIOMEDICAL ENGINEERS
Unlike the preceding distributions, a closed form expression for the Bernoulli CDF is not
easily obtained. Tables A.1-A.3 in the Appendix list values of the Bernoulli CDF for p

=

0 . 05 , 0 . 1 , 0 . 15 ,..., 0 . 5 and n = 5 , 10 , 15, and 20. Let k ∈{ 0 , 1 ,..., n − 1 } and define

n p (1

k

p ) n − .

G ( n , k , p ) = −

= 0

Making the change of variable m = n − yields

p n − m (1

n

n

p ) m G ( n , k , p ) = − .

n − m

m = n − k

Now, since

n
m

n n !

m !( n

=

m )! =

,

n − m −

n
m

p n − m (1 n

m

p n − m (1

− − 1

n n k

p ) m p ) m

G ( n , k , p ) = − − − .

m = 0 m = 0

Using the Binomial Theorem,

G ( n , k , p ) = 1 − G ( n , n − k − 1 , 1 − p ) . (5.22)

This result is easily applied to obtain values of the Bernoulli CDF for values of p > 0 . 5from

Tables A.1-A.3.

Example 5.3.1. The probability that Fargo Polytechnic Institute wins a game is 0.7. In a 15 game
season, what is the probability that they win: (a) at least 10 games, (b) from 9 to 12 games, (c) exactly
11 games? (d) With x denoting the number of games won, find

2
x

η x and σ .

Solution. With x a Bernoulli random variable, we consult Table A.2, using (5.22) with n = 15,

k = 9, and p = 0 . 7, we find

a) P ( x ≥ 10) = 1 − F x (9) = 1 . 0 − 0 . 2784 = 0 . 7216,

b)

P (9

≤ ≤ 12) =

F x (12)

−

F x (8)

= 0 . 8732 − 0 . 1311 = 0 . 7421,

x

c) p x (11) = F x (11) − F x (10) = 0 . 7031 − 0 . 4845 = 0 . 2186.

2
x

d) η x = np = 10 . 5, σ = np (1 − p ) = 3 . 15.