Bernoulli Distribution
In this tutorial, we will discuss example on Bernoulli’s distribution.
The discrete random variable $X$ is said to have Bernoulli distribution if its probability mass function is given by
$$ \begin{equation*} P(X=x) = p^x q^{1-x}, \; x=0,1; 0<p,q<1; q=1-p. \end{equation*} $$
Example
The battery from a lot of batteries has 85% chances of non-defective. We select a batter at random from a lot.
a. What is the probability distribution of non-defective battery?
b. What is the probability that selected battery is defective?
c. Find the mean and variance of no. of non-defective battery.
Solution
The two outcomes non-defective (success) and defective (failure) are mutually exclusive.
Let $p$ denote the probability that a battery is non-defective (Success). Given that $p=P(\text{ Battery is non-defective }) =0.85$. Then $q=P(\text{ Battery is defective }) =1-0.85 = 0.15$.
Let $X$ denote the number of non-defective battery (success). Then the random variable $X$ take the value $X=0,1$.
a. The random variable $X\sim Ber(0.85)$. The probability distribution of number of non-defective battery $X$ is
$$ \begin{aligned} P(X=x) &= 0.85^{x} (1-0.85)^{1-x}, \; x=0,1\\ &= 0.85^x0.15^{1-x},\; x=0,1. \end{aligned} $$
b. The probability that selected battery is defective
$$ \begin{aligned} P(X=0)&= 0.85^00.15^{1-0}\\ &= 0.15. \end{aligned} $$
c. The mean of number of non-defective battery is
$$ \begin{aligned} E(X) &= p\\ &= 0.85. \end{aligned} $$
The variance of number of non-defective battery is
$$ \begin{aligned} V(X) &= p(1-p)\\ &= 0.85(1-0.85)\\ &= 0.1275. \end{aligned} $$
