3.5    The Binomial Distribution

1. have a fixed number n of Bernoulli trials (success (s) or failure (f) for each)
2. the trials are independent, probability of success on each trial is same (as before, denote probability of success as  p,  probability of failure as  q)
3. X = total number of successes in the n trials.  (possible values are 0 to n)

ex:

Test with 3 questions; have probability of .8 getting any of questions correct. Let  X  = the number of questions you get correct. Then X is a binomial random variable, with  n = 3  (3 trials) and  p = .8  (probability of success on each trial), since it counts the total number of successes in a fixed number of trials. A sample space giving the 8 possible outcomes of your exam is shown below; beneath each outcome is the corresponding value of X; beneath this is the probability of that particular outcome (as before, these are found using the multiplication rule for independent events).



From the above we can get the probability density function, given in the following table:
 

f(x)  =    ,    x  =  0, 1, 2, ..., n

m_X(t)  =  (pe^t + q)ⁿ

From this, we can compute the first and second moments, and use these to compute the mean and variance:

E(X)  =   = 
=  n (pe^t + q)^n-1 (pe^t) |_t=0      (by the chain rule)
=  n (p + q)^n-1 (p)   =   np     (since  p + q  =  1)
E(X²)  =   = 
=   n (n-1) p²  +  np  (when the smoke clears, using the product and chain rules to differentiate; do this as an exercise!)

var(X) = E(X²) - E(X)²   =   np(1-p)     (exercise)

m  =  np
var(X)  =  npq  =  np(1-p)
(so the standard deviation   s  =   )

ex:

Suppose you take a test like the one above, in which the probability of your getting any one of the questions correct is .8, but suppose now the test has 20 questions.  What’s the expected number of questions you'll get correct?
Let  X = the number of questions out of the 20 you get correct; then X is a binomial random variable, with n = 20 and p = .8. Then  E(X)  =  np  =  (20)(.8)  =  16,  i.e., you'd expect to get 16 of the 20 correct.

What’s the probability you'll get exactly 1 wrong (i.e., 19 correct)?

P(X = 19)  =  f(19)  =    =   .058
What's the probability you'll get 14 or fewer correct?
We want   P(X <= 14)  =  F(14)
there's no convenient formula for F(x) for a binomial random variable. We could compute F(14) by summing f(1) + f(2) + ... + f(14), but this is tedious. Fortunately, values of F(x) for binomial random variables have been tabulated for various values of n and p; your text has tables in the back (p. 720-724) giving values for n = 1 through 20 and p = .1 through .9. Using the table on p. 724 for n = 20 and p = .8, we find
P(X <= 14)  =  F(14)  =  .1958;
thus there's only about a 20% chance you'd get 14 or fewer correct.

Note:  Though the tables only go up to n = 20, we'll find there's a quick way for us to approximate the value of probabilities for larger values of n using the normal distribution (to be discussed shortly).
 
 

ex:

A binary packet consisting of 64 bits is sent over a communications line.  Because of noise, some of the bits will be corrupted when they are sent (i.e., they'll be sent as 1's or 0's, but will be received as the opposite). Suppose the the probability that any 1 bit is corrupted during transmission is p = 02.

What’s the expected # of bad bits in a packet?
Let  X = # of corrupted bits received out of the 64 sent. Then X is a binomial random variable, with n = 64 and p = .02 (here, "success" is interpretted as a bit being corrupted). Thus

E(X)  =  np  =  64*(.02)  =  1.28
On average, we would expect 1.28 bits to be corrupted per packet.

What’s the probability the message comes through correctly (i.e., # bad bits = 0)?

P(X=0)  =    =  .27
So only about 1/4 of the 64-bit packets sent would be expected to come through without any errors.

In this situation, since most of the pakets will be corrupted, but usually only by about 1 bit, it would make sense to use an error-correcting code to transmit the data which can correct for errors in 1 or 2 bits.