Bernoulli trials - Columbia University

Bernoulli trials

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

X = 1 when the outcome is a success 0 when outcome is a failure

If p is the probability of a success then the pmf is, p(0) =P(X=0) =1-p p(1) =P(X=1) =p

A random variable is called a Bernoulli random variable if it has the above pmf for p between 0 and 1.

Expected value of Bernoulli r. v.:

E(X) = 0*(1-p) + 1*p = p

Variance of Bernoulli r. v.:

E(X 2) = 02*(1-p) + 12*p = p Var(X) = E(X 2) - (E(X)) 2 = p - p2 = p(1-p)

Ex. Flip a fair coin. Let X = number of heads. Then X is a Bernoulli random variable with p=1/2. E(X) = 1/2 Var(X) = 1/4

Binomial random variables

Consider that n independent Bernoulli trials are performed. Each of these trials has probability p of success and probability (1-p) of failure.

Let X = number of successes in the n trials.

p(0) = P(0 successes in n trials) = (1-p)n p(1) = P(1 success in n trials) = (n 1)p(1-p)n-1 p(2) = P(2 successes in n trials) = (n 2)p2(1-p)n-2

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...... p(k) = P(k successes in n trials) = (n k)pk(1-p)n-k

A random variable is called a Binomial(n,p) random variable if it has the pmf, p(k) = P(k successes in n trials) = (n k)pk(1-p)n-k for k=0,1,2,....n.

Valid pmf: sum_{k=0}^n p(k) = 1 sum_{k=0}^n (n k)pk(1-p)n-k = (p+(1-p))n = 1

Ex. A Bernoulli(p) random variable is binomial(1,p)

Ex. Roll a dice 3 times. Find the pmf of the number of times we roll a 5.

X = number of times we roll a 5 (number of successes) X is binomial(3,1/6)

p(0) = (3 0)(1/6) 0(5/6)3 = 125/216 p(1) = (3 1)(1/6) 1(5/6)2 = 75/216 p(2) = (3 2)(1/6) 2(5/6)1 = 15/216 p(3) = (3 3)(1/6) 3(5/6)0 = 1/216

Ex. Screws produced by a certain company will be defective with probability .01 independently of each other. If the screws are sold in packages of 10, what is the probability that two or more screws are defective?

X = number of defective screws. X is binomial(10,0.01)

P(X 2) = 1 - P(X ................
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