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BERNOULLI TRIALS and the BINOMIAL DISTRIBUTION

Module by: Ewa Paszek

Summary: This course is a short series of lectures on Introductory Statistics. Topics covered are listed in the Table of Contents. The notes were prepared by Ewa Paszek and Marek Kimmel. The development of this course has been supported by NSF 0203396 grant.

BERNOULLI TRIALS AND THE BINOMIAL DISTRIBUTION

A Bernoulli experiment is a random experiment, the outcome of which can be classified in but one of two mutually exclusive and exhaustive ways, mainly, success or failure (e.g., female or male, life or death, nondefective or defective).

A sequence of Bernoulli trials occurs when a Bernoulli experiment is performed several independent times so that the probability of success, say, p, remains the same from trial to trial. That is, in such a sequence we let p denote the probability of success on each trial. In addition, frequently q=1−p q=1−p denote the probability of failure; that is, we shall use q and 1−p 1−p interchangeably.

Bernoulli distribution

Let X be a random variable associated with Bernoulli trial by defining it as follows:

X(success)=1 and X(failure)=0.

That is, the two outcomes, success and failure, are denoted by one and zero, respectively. The p.d.f. of X can be written as

f( x )= p x ( 1−p ) 1−x , f( x )= p x ( 1−p ) 1−x , (1)

and we say that X has a Bernoulli distribution. The expected value of is

μ=E( X )= ∑ X=0 1 x p x ( 1−p ) 1−x =( 0 )( 1−p )+( 1 )( p )=p, μ=E( X )= ∑ X=0 1 x p x ( 1−p ) 1−x =( 0 )( 1−p )+( 1 )( p )=p, (2)

and the variance of X is

σ 2 =Var( X )= ∑ x=0 1 ( x−p ) 2 p x ( 1−p ) 1−x = p 2 ( 1−p )+ ( 1−p ) 2 p=p( 1−p )=pq. σ 2 =Var( X )= ∑ x=0 1 ( x−p ) 2 p x ( 1−p ) 1−x = p 2 ( 1−p )+ ( 1−p ) 2 p=p( 1−p )=pq. (3)

It follows that the standard deviation of X is σ= p( 1−p ) = pq . σ= p( 1−p ) = pq .

In a sequence of n Bernoulli trials, we shall let X i X i denote the Bernoulli random variable associated with the ith trial. An observed sequence of n Bernoulli trials will then be an n-tuple of zeros and ones.

Binomial Distribution

In a sequence of Bernoulli trials we are often interested in the total number of successes and not in the order of their occurrence. If we let the random variable X equal the number of observed successes in n Bernoulli trials, the possible values of X are 0,1,2,…,n. If x success occur, where x=0,1,2,...,n x=0,1,2,...,n , then n-x failures occur. The number of ways of selecting x positions for the x successes in the x trials is ( n x )= n! x!( n−x )! . ( n x )= n! x!( n−x )! . Since the trials are independent and since the probabilities of success and failure on each trial are, respectively, p and q=1−p q=1−p , the probability of each of these ways is p x ( 1−p ) n−x . p x ( 1−p ) n−x . . Thus the p.d.f. of X, say f( x ) f( x ) , is the sum of the probabilities of these ( n x ) ( n x ) mutually exclusive events; that is,

f( x )=( n x ) p x ( 1−p ) n−x ,x=0,1,2,...,n. f( x )=( n x ) p x ( 1−p ) n−x ,x=0,1,2,...,n.

These probabilities are called binomial probabilities, and the random variable X is said to have a binomial distribution.

Summarizing, a binomial experiment satisfies the following properties:
  1. A Bernoulli (success-failure) experiment is performed n times.
  2. The trials are independent.
  3. The probability of success on each trial is a constant p; the probability of failure is q=1−p q=1−p .
  4. The random variable X counts the number of successes in the n trials.

A binomial distribution will be denoted by the symbol b( n,p ) b( n,p ) and we say that the distribution of X is b( n,p ) b( n,p ) . The constants n and p are called the parameters of the binomial distribution, they correspond to the number n of independent trials and the probability p of success on each trial. Thus, if we say that the distribution of X is b( 12,14 ), b( 12,14 ), we mean that X is the number of successes in n =12 Bernoulli trials with probability p= 1 4 p= 1 4 of success on each trial.

Example 1

In the instant lottery with 20% winning tickets, if X is equal to the number of winning tickets among n =8 that are purchased, the probability of purchasing 2 winning tickets is

f( 2 )=P( X=2 )=( 8 2 ) ( 0.2 ) 2 ( 0.8 ) 6 =0.2936. f( 2 )=P( X=2 )=( 8 2 ) ( 0.2 ) 2 ( 0.8 ) 6 =0.2936.

The distribution of the random variable X is b( 8,0.2 ) b( 8,0.2 ) .

Example 2

Leghorn chickens are raised for lying eggs. If p =0.5 is the probability of female chick hatching, assuming independence, the probability that there are exactly 6 females out of 10 newly hatches chicks selected at random is

( 10 6 ) ( 1 2 ) 6 ( 1 2 ) 4 =P( X≤6 )−P( X≤5 )=0.8281−0.6230=0.2051. ( 10 6 ) ( 1 2 ) 6 ( 1 2 ) 4 =P( X≤6 )−P( X≤5 )=0.8281−0.6230=0.2051.

Since P( X≤6 )=0.8281 P( X≤6 )=0.8281 and P( X≤5 )=0.6230, P( X≤5 )=0.6230, which are tabularized values, the probability of at least 6 females chicks is

∑ x=6 10 ( 10 x ) ( 1 2 ) x ( 1 2 ) 10−x =1−P( X≤5 )=1−0.6230=0.3770 . ∑ x=6 10 ( 10 x ) ( 1 2 ) x ( 1 2 ) 10−x =1−P( X≤5 )=1−0.6230=0.3770 .

Example 3

Suppose that we are in those rare times when 65% of the American public approve of the way the President of The United states is handling his job. Take a random sample of n =8 Americans and let Y equal the number who give approval. Then the distribution of Y is b( 8,0.65 ). b( 8,0.65 ). To find P( Y≥6 ) P( Y≥6 ) note that

P( Y≥6 )=P( 8−Y≤8−6 )=P( X≤2 ), P( Y≥6 )=P( 8−Y≤8−6 )=P( X≤2 ),

where X=8−Y X=8−Y counts the number who disapprove. Since q=1−p=0.35 q=1−p=0.35 equals the probability if disapproval by each person selected, the distribution of X is b( 8,0.35 ) b( 8,0.35 ) . From the tables, since P( X≤2 )=0.4278 P( X≤2 )=0.4278 it follows that P( Y≥6 )0.4278. P( Y≥6 )0.4278. Similarly,

P( Y≤5 )=P( 8−Y≥8−5 )=P( X≥3 )=1−P( X≤2 )=1−0.4278=0.5722 P( Y≤5 )=P( 8−Y≥8−5 )=P( X≥3 )=1−P( X≤2 )=1−0.4278=0.5722

and

P( Y=5 )=P( 8−Y=8−5 )=P( X=3 )=P( X≤3 )−P( X≤2 )=0.7064−0.4278=0.2786. P( Y=5 )=P( 8−Y=8−5 )=P( X=3 )=P( X≤3 )−P( X≤2 )=0.7064−0.4278=0.2786.

RECALL THAT:
if n is a positive integer, then ( a+b ) n = ∑ x=0 n ( x n ) b x a n−x . ( a+b ) n = ∑ x=0 n ( x n ) b x a n−x . Thus the sum of the binomial probabilities, if we use the above binomial expansion with b=p b=p and a=1−p a=1−p , is ∑ x=0 n ( n x ) p x ( 1−p ) n−x = [ ( 1−p )+p ] n =1, ∑ x=0 n ( n x ) p x ( 1−p ) n−x = [ ( 1−p )+p ] n =1,

A result that had to follow from the fact that f( x ) f( x ) is a p.d.f. We use the binomial expansion to find the mean and the variance of the binomial random variable X that is b( n,p ) b( n,p ) . The mean is given by

μ=E( X )= ∑ x=0 n x n! x!( n−x )! p x ( 1−p ) n−x . μ=E( X )= ∑ x=0 n x n! x!( n−x )! p x ( 1−p ) n−x . (4)

Since the first term of this sum is equal to zero, this can be written as

μ= ∑ x=0 n n! ( x−1 )!( n−x )! p x ( 1−p ) n−x . μ= ∑ x=0 n n! ( x−1 )!( n−x )! p x ( 1−p ) n−x . (5)

because x/x!=1/( x−1 )! x/x!=1/( x−1 )! when x>0. x>0.

To find the variance, we first determine the second factorial moment E[ X( X−1 ) ] E[ X( X−1 ) ] :

E[ X( X−1 ) ]= ∑ x=0 n x( x−1 ) n! x!( n−x )! p x ( 1−p ) n−x . E[ X( X−1 ) ]= ∑ x=0 n x( x−1 ) n! x!( n−x )! p x ( 1−p ) n−x . (6)

The first two terms in this summation equal zero; thus we find that

E[ X( X−1 ) ]= ∑ x=2 n n! ( x−2 )!( n−x )! p x ( 1−p ) n−x . E[ X( X−1 ) ]= ∑ x=2 n n! ( x−2 )!( n−x )! p x ( 1−p ) n−x .

After observing that x( x−1 )/x!=1/( x−2 )! x( x−1 )/x!=1/( x−2 )! when x>1 x>1 . Letting k=x−2 k=x−2 , we obtain

E[ X( X−1 ) ]= ∑ x=0 n−2 n! k!( n−k−2 )! p k+2 ( 1−p ) n−k−2 .=n(n−1) p 2 ∑ x=0 n−2 ( n−2 )! k!( n−2−k )! p k ( 1−p ) n−2−k . E[ X( X−1 ) ]= ∑ x=0 n−2 n! k!( n−k−2 )! p k+2 ( 1−p ) n−k−2 .=n(n−1) p 2 ∑ x=0 n−2 ( n−2 )! k!( n−2−k )! p k ( 1−p ) n−2−k .

Since the last summand is that of the binomial p.d.f. b( n−2,p ) b( n−2,p ) , we obtain E[ X( X−1 ) ]=n( n−1 ) p 2 . E[ X( X−1 ) ]=n( n−1 ) p 2 .

Thus, σ 2 =Var( X )=E( X 2 )− [ E( X ) ] 2 =E[ X( X−1 ) ]+E( X )− [ E( X ) ] 2 =n( n−1 ) p 2 +np− ( np ) 2 =−n p 2 +np=np( 1−p ). σ 2 =Var( X )=E( X 2 )− [ E( X ) ] 2 =E[ X( X−1 ) ]+E( X )− [ E( X ) ] 2 =n( n−1 ) p 2 +np− ( np ) 2 =−n p 2 +np=np( 1−p ).

Summarizing,

if X is b( n,p ) b( n,p ) , we obtain

μ=np, σ 2 =np( 1−p )=npq,σ= np( 1−p ) . μ=np, σ 2 =np( 1−p )=npq,σ= np( 1−p ) .

Note That:
When p is the probability of success on each trial, the expected number of successes in n trials is np, a result that agrees with most of our intuitions.

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