A discrete probability distribution consists of values a discrete random variable can assume and the corresponding probabilities. Discrete probability distributions can be shown using a graph or a table or can be described by a formula. The probability distribution for the experiment of rolling a die is as follows:

Outcome X |
1 | 2 |
3 |
4 |
5 | 6 |

Probability p(X) |
$\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $\frac{1}{6}$ |

An association of the values of a discrete random variable X and the corresponding probabilities p(x) is said to be a probability distribution if:

- Each probability in the distribution is between 0 and 1. That is, 0 $\leq$ p(x) $\leq$ 1 for all p(x).
- The sum of all probabilities is equal to 1. That is $\sum$p(x) = 1

You may notice that the distribution on the outcomes of a roll of a die satisfies these two conditions.

p(x) = $\frac{1}{6}$ is a positive fraction less than 1.

Sum of all probabilities = $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$ + $\frac{1}{6}$ = $\frac{6}{6}$

= 1.

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Let us determine whether the distribution given below is a probability distribution.

It is evident from the observation that the first requirement for a probability distribution is satisfied as each of the probabilities present in the table is a value between 0 and 1.

To check the second condition, let us find the sum of all probabilities.

$\sum$ p(x) = $\frac{1}{4}$ + $\frac{1}{8}$ + $\frac{1}{16}$ + $\frac{9}{16}$ = $\frac{4+2+1+9}{16}$ = 1.

Hence the given table represents a discrete probability distribution.

The probability distribution of a discrete variable X is given as follows:

Let us find the value of 'm' and determine the probabilities and represent the distribution.

For a discrete probability distribution, the sum of the probabilities = 1

2m + m + $\frac{m}{2}$ + $\frac{m}{4}$ + $\frac{m}{4}$ = 1.

$\frac{8m+4m+2m+m+m}{4}$ = 1

$\frac{16m}{4}$ = 1

4m = 1

m = $\frac{1}{4}$

The probability distribution is given as follows:

Binomial, Poisson, Multinomial and Hypergeometric distributions are some of the discrete probability distributions studied widely.

**Binomial Distribution:**

Binomial distribution consists of the outcomes of a Binomial experiment and the corresponding probabilities. A binomial experiment consists of independent, repeated and a fixed number of trials. Each trial can have only two outcomes or the outcomes can be reduced to two outcomes as success and failure, the probability of success remaining same in all the trials.

The formula for finding the probability of exactly X successes in n trials is given as follows:

P(X) = C(n, x) p^{x }q^{n - x} , where C(n,x) = $\frac{n!}{(n-x)!x!}$

Here, p = Probability of success in a single trial and q = 1- p.

Mean of a Binomial distribution = np

Variance of a Binomial distribution = npq.

**Poisson Distribution:**

Poisson distribution is used when n is large, p is small and the independent trials occur over a period of time. The parameter $\lambda$ is used to describe a Poisson distribution.

The probability of X which occurs in an interval of time, volume or area is given as follows:

P(X) = $\frac{e^{-\lambda }.\lambda ^{x}}{x!}$

where, $\lambda$ is the mean number of occurrences per unit (time, volume, area etc.) and in binomial terms $\lambda$ = np.

In Poisson distribution, both the mean and variance are equal to λ.

X |
1 | 2 | 3 |
4 |

p(X) |
$\frac{1}{4}$ | $\frac{1}{8}$ | $\frac{1}{16}$ | $\frac{9}{16}$ |

It is evident from the observation that the first requirement for a probability distribution is satisfied as each of the probabilities present in the table is a value between 0 and 1.

To check the second condition, let us find the sum of all probabilities.

$\sum$ p(x) = $\frac{1}{4}$ + $\frac{1}{8}$ + $\frac{1}{16}$ + $\frac{9}{16}$ = $\frac{4+2+1+9}{16}$ = 1.

Hence the given table represents a discrete probability distribution.

The probability distribution of a discrete variable X is given as follows:

Let us find the value of 'm' and determine the probabilities and represent the distribution.

X |
0 | 1 | 2 |
3 |
4 |

P(X) |
2m | m | $\frac{m}{2}$ | $\frac{m}{4}$ | $\frac{m}{4}$ |

For a discrete probability distribution, the sum of the probabilities = 1

2m + m + $\frac{m}{2}$ + $\frac{m}{4}$ + $\frac{m}{4}$ = 1.

$\frac{8m+4m+2m+m+m}{4}$ = 1

$\frac{16m}{4}$ = 1

4m = 1

m = $\frac{1}{4}$

The probability distribution is given as follows:

X |
0 | 1 | 2 |
3 |
4 |

P(X) |
$\frac{1}{2}$ |
$\frac{1}{4}$ |
$\frac{1}{8}$ | $\frac{1}{16}$ |
$\frac{1}{16}$ |

Binomial, Poisson, Multinomial and Hypergeometric distributions are some of the discrete probability distributions studied widely.

Binomial distribution consists of the outcomes of a Binomial experiment and the corresponding probabilities. A binomial experiment consists of independent, repeated and a fixed number of trials. Each trial can have only two outcomes or the outcomes can be reduced to two outcomes as success and failure, the probability of success remaining same in all the trials.

The formula for finding the probability of exactly X successes in n trials is given as follows:

P(X) = C(n, x) p

Here, p = Probability of success in a single trial and q = 1- p.

Mean of a Binomial distribution = np

Variance of a Binomial distribution = npq.

Poisson distribution is used when n is large, p is small and the independent trials occur over a period of time. The parameter $\lambda$ is used to describe a Poisson distribution.

The probability of X which occurs in an interval of time, volume or area is given as follows:

P(X) = $\frac{e^{-\lambda }.\lambda ^{x}}{x!}$

where, $\lambda$ is the mean number of occurrences per unit (time, volume, area etc.) and in binomial terms $\lambda$ = np.

In Poisson distribution, both the mean and variance are equal to λ.