Discrete Probability Distribution Examples - Word Problems

# Discrete Probability Distribution Examples

If a variable can be assigned any value between two values then it a continuous variable. If only a fixed number of values can be taken such as 2, 3, 4, 5, then it is known as discrete variable. The mean of a discrete probability distribution can be written as:
$\mu =\sum_{x=i}^{n}xP(x)$
The standard deviation of a probability distribution can be given as:
$\sigma = \sqrt{(X-\mu )^{2}P(X)}$
Discrete probability distributions are of various forms. Prominent of them are binomial distribution, hypergeometric distribution, poisson's distribution, multinomial distribution.

## Word Problems

Problem 1:

Find the mean value of variable X from the given probability distribution.

 X 0 1 2 P(X) 0.3 0.2 0.5

Solution:

Mean of variable X,

$\mu =\sum_{x=i}^{n}XP(X)$

$\mu = 0\times 0.3 + 1\times 0.2+2\times 0.5=1.2$

Hence, mean value of this probability distribution is 1.2.
Problem 2:

In a draw, only numbers 3, 5 and 6 can come and given is the probability distribution for these numbers. Find the mean of the numbers obtained in the draw.

 X 3 5 6 P(X) 0.6 0.3 0.1

Solution:

Let the numbers obtained be X. The mean of X can be derived using the formula:

$\mu =\sum_{x=i}^{n}XP(X)$

$\mu = 3\times 0.6 + 5\times 0.3+6\times 0.1= 3.9$

Mean of numbers got in the draw is 3.9.
Problem 3:

For a variable X, the mean is 1.1 and probability distribution is given below.

 X 0 1 2 P(X) 0.2 0.5 0.3

Find the standard deviation of X.

Solution:

The standard deviation of X can be obtained using the formula,

$\sigma = \sqrt{(X-\mu )^{2}P(X)}$

$\sigma = \sqrt{(0-1.1 )^{2}\times 0.2+(1-1.1 )^{2}\times 0.5+(2-1.1 )^{2}\times 0.3}$

$\sigma = \sqrt{0.242 + 0.005 + 0.243} = \sqrt 0.49$

$\sigma = 0.7$

Standard deviation is 0.7.
Problem 4:

Make the probability distribution table for getting a head if two coins are tossed together.

Solution:

Sample space, S = {HH, HT, TH, TT}

Numbers of heads can be 0, 1 or 2.

The table for probability distribution will be,

 X 0 1 2 P(X) $\frac{1}{4}$ $\frac{2}{4}$ $\frac{1}{4}$