1-The Bernoulli and Binomial random variables 

Suppose that a trial, or an experiment, whose outcome can be classified as either a success or a failure is performed.

If we let X = 1 when the outcome is a success and X = 0

when it is a failure, then the probability mass function of X is given by

𝑝(0)=𝑃(𝑋 =0)=1−𝑝

𝑝(1)=𝑃(𝑋 =1)=𝑝

where p, 0≤𝑝≤1, is the probability that the trial is a success.

A random variable X is said to be a Bernoulli random variable 

(after the Swiss mathematician James Bernoulli) if its probability mass function is given by

𝑝(0)=𝑃(𝑋 =0)=1−𝑝

𝑝(1)=𝑃(𝑋 =1)=𝑝

for some p ∈(0, 1).

Suppose now that n independent trials, each of which results in a success with probability p

and in a failure with probability 1 − p, are to be performed.

If X represents the number of successes that occur in the n trials, 

then X is said to be a binomial random variable with parameters (n, p).

Thus, a Bernoulli random variable is just a binomial random variable with parameters (1, p). 

The probability mass function of a binomial random variable having parameters (n, p) is given by

Five fair coins are flipped. If the outcomes are assumed independent,

find the probability mass function of the number of heads obtained.Solution.

If we let X equal the number of heads (successes) that appear, 

then X is a binomial random variable with parameters (𝑛 =5,𝑝 = Hence,  

It is known that screws produced by a certain company will be defective with probability 0.01,

independently of each other. The company sells the screws in packages of 10

and offers a money-back guarantee that at most 1 of the 10 screws is defective.

What proportion of packages sold must the company replace? Solution.

If X is the number of defective screws in a package,

then X is a binomial random variable with parameters (10, 0.01).

Hence, the probability that a package will have to be replaced is

1−𝑃(𝑋 =0)−𝑃(𝑋 =1)

Thus, only 0.4 percent of the packages will have to be replaced.

*Properties of Binomial Random Variables

We will now examine the properties of a binomial random variable with parameters n and p.

To begin, let us compute its expected value and variance. Now, 

Using the identity 

gives 

where Y is a binomial random variable with parameters n − 1, p.

Setting k = 1 in the preceding equation yields 

𝐸(𝑋)=𝑛𝑝

That is, the expected number of successes that occur in n independent trials when

each is a success with probability p is equal to np.

Setting k = 2 in the preceding equation, and using the preceding formula for the expected value

of a binomial random variable yields

𝐸=𝑛𝑝 𝐸(𝑌+1)                       

=𝑛𝑝 [(𝑛−1)𝑝+1]

Since E[X] = np, we obtain

𝑉𝑎𝑟(𝑋)= 𝐸− 

=𝑛𝑝 [(𝑛−1)𝑝+1]−

=𝑛𝑝 (1−𝑝)

Summing up, we have shown the following:

If X is a binomial random variable with parameters n and p,

then 𝐸(𝑋)=𝑛𝑝

𝑉𝑎𝑟(𝑋)= 𝑛𝑝 (1−𝑝)