The probability distribution of a discrete random variable is a list of odds associated with each of its possible values. It is also called the probability function or the probability mass function. Basically we can define probability function as a rule that assigns probabilities to the values of random variables. For a discrete random variable X, probability function (probability mass function) must satisfy the following properties,

1. The probability of each value of the discrete random variable X must be between or equal to 0 and 1. 0 ? P (X = x) ? 1 for all
2. The sum of the probabilities of each outcome of the random variable must equal to 1.

No. For all X it satisfies 0 ? P (X = x) ? 1 but the sum of P(X=x)=0,2+0,1+0,4+0,4=1,1>1. So it is not a probability distribution function.

Cumulative distribution function (CDF) of a discrete random variable X, denoted by F (x) and defined as,F(x)=P(X?x)=xi?x?P(X=xi)This expression for cumulative distribution function implies that it’s a cumulative probability value for random variables which are less than or equal to specific random value of xi . Through the definition of CDF, it also supplies probability values through utilizing probability mass function.For a discrete random variable X, CDFF (X) the following properties are shown.

1. 0 ?F (x) ? 1 and
2. If x1 ? x2 then F (x1) ? F (x2)

P(2?X<4)=P(X=2)+P(X=3)=2/10+3/10=0.5

Binomial distribution is a widely employed discrete probability distribution in statistics where a set of independent observations constitutes exactly two disjoint outcomes of a trial. Therefore in binomial distribution, an outcome of a random experiment can be classified under two different categories. For example, when a die is tossed once we observe six different numbers, that is x = 1, 2, 3, 4, 5 and 6. If we classified these outcomes as even or odd numbers then we will get two different outcomes.

A random experiment (trial) with only two possible outcomes is frequently used and called a Bernoulli trial with the following properties:

1. Each trial has only two possible outcomes, such as head and tail, 0 and 1 or success and failure.
2. The trials are independent experiments.
3. The probability of success, denoted by and the probability of failure, denoted by remains constant for all trials (p + q = 1).

The mean (µ) and variance (?2) of the binomial random variable X ~Binomal (n, p) can be calculated from these formulas

µ = E (x) = np and ?² =V (X) = np (1 - p)

At least 3 heads means 3 heads or 4 heads. Thus we have to calculate:

P(H=3)+P(H=4)

P(H=4)=C(4,4)0.5⁴0.5⁰=1/16

P(H=3)=C(4,3)0.5³0.5¹=4/16

P(H=3)+P(H=4)=1/16+4/16=5/16

P(X ? 2)=P(X=0)+P(X=1)+P(X=2)

P(X=0)=C(4,0)*0.6⁰*0.4⁴=0.0256

P(X=1)=C(4,1)*0.6¹*0.4³=0.1536

P(X=2)=C(4,2)*0.6²*0.4²=0.3456

so:

P(X ? 2)=P(X=0)+P(X=1)+P(X=2)=0.0256+0.1536+0.3456=0.5248

P(X=2)=C(3,2)*0.8²*0.2¹=3*0.64*0.2=0.384=38.4%

The Poisson distribution is widely used for discrete probability distribution which is used to model the number of outcomes occurring during a specified time interval or in a definite region. Here, the time interval indicates any length, for instance an hour, a day, or a month. Also, the definite region term means that a piece of line segment, an acre, size of a football field, a volume.

The mean (µ) and variance (?²) of a Poisson random variable X ~ Poisson (?) with parameter ? can be calculated using the following formulas,

µ = E (x) = ?

?² = V (X) = ?

P(X ? 2)=P(X=0)+P(X=1)+P(X=2)

for ?=5

P(X=0)=(e^-5*5⁰)/0!=0.0067

P(X=1)=(e^-5*5¹)/1!=0.0337

P(X=2)=(e^-5*5²)/2!=0.0842

So:

P(X ? 2)=P(X=0)+P(X=1)+P(X=2)=0.124

In binomial distribution the sampling process is carried out with replacement. On the other hand hypergeometric distribution doesn’t involve with independence assumption for each trial and accordingly the sampling process is established on without replacement. Because of these features, hypergeometric distribution is broadly used in various real life applications especially for acceptance sampling in quality control