 # 5.1a Probability of an Event that follows a Binomial Distribution

5.1a Probability of an Event that follows a Binomial Distribution

1. For a Binomial Distribution, the probability of obtaining r numbers of successes out of n experiments is given by

$\overline{)P\left(X=r\right)=c{}_{r}.{p}^{r}.{q}^{n-r}}$

where
P = probability
X = discrete random variable
r =  number of success (0, 1, 2, 3, …, n)
n = number of trials
p = probability of success in an experiment (0 < p <1)
q = probability of failure in an experiment (q = 1 – p)

Example 1
Kelvin has taken 3 shots in a shooting practice. The probability that Kelvin strikes the target is 0.6. X represents the number of times kelvin strikes the target.

(a)   List the elements of the binomial discrete random variable X.

(b) Calculate the probability for the occurrence of each of the elements of X.

(c)  Hence, plot a graph to represent the binomial probability distribution of X.

Solution:
(a) X = Number of times Kelvin strikes the target
X = {0, 1, 2, 3}

(b) X ~ B (n, p)
X ~ B (3, 0.6)

$\begin{array}{l}P\left(X=r\right)=c{}_{r}.{p}^{r}.{q}^{n-r}\\ \left(i\right)\text{}P\left(X=0\right)\\ \text{}=C{}_{0}{\left(0.6\right)}^{0}{\left(0.4\right)}^{3}←\overline{)\begin{array}{l}\text{Probability of failure}\\ \text{=}1-0.6=0.4\end{array}}\\ \text{}=0.064\\ \\ \left(ii\right)\text{}P\left(X=1\right)\\ \text{}=C{}_{1}{\left(0.6\right)}^{1}{\left(0.4\right)}^{2}\\ \text{}=0.288\\ \\ \left(iii\right)\text{}P\left(X=2\right)\\ \text{}=C{}_{2}{\left(0.6\right)}^{2}{\left(0.4\right)}^{1}\\ \text{}=0.432\\ \\ \left(iv\right)\text{}P\left(X=3\right)\\ \text{}=C{}_{3}{\left(0.6\right)}^{3}{\left(0.4\right)}^{0}\\ \text{}=0.216\end{array}$