Under Discrete Probability Distribution we have

  1. Bernoulli Distribution
  2. Binomial Distribution (sampling with replacement)
  3. Hypergeometric Distribution(Sampling without replacement)
  4. Poisson Distribution

Bernoulli Distribution:

  • Discovered by a SWISS Mathematician Jame Bernoulli(1654-1705)
  • Also known as Bernoulli Distribution
  • A successive repetition of an experiment is called Bernoulli Trials.
  • The number of trials is finite and fixed
  • Dealt with dichotomous classification of events(outcomes) either 0 or 1
  • Non-occurrence or Occurrence of an event
  • Occurrence is denoted by p
  • Non- occurrence is denoted by q
  • Outcome of each trial may be either success or failure. Trials are independent. Example: Flipping a coin.
  • When trials are done n times under identical conditions Say Xi for i= 1 to n assumes values 0 or 1 then Xi is Bernoulli  Variate  with   probability  ‘p’
  • Mean of Bernoulli Distribution
  • Here random variable assumes values 1 and 0  with probabilities ‘p’ and ‘q’
  • q = 1-p; Here random variable is called Bernoulli Variable.
  • It has only one parameter ‘p’ that is one constant
  • For different values of p we get different Bernoulli Distribution. (0<p<1)

Example:

  1. a card is drawn from a pack of 52 cards.
  2. The probability of getting a king is 4/52.
  3. Before a second draw, the card drawn is replaced. (drawing with replacement)
  4. But if the card is not replaced, we cannot have binomial distribution.
  5. Tossing of a coin: a head or a tail can be had on a toss of coin ;
  6. Drawing from a pack of cards: a card drawn may be black or red
  7. Inspecting a batch: An item in a batch may be defective or non-defective

When a coin is tossed

  1. you get two possible outcomes. Either Head or Tail
  2. The probability of getting a Head is (p) ½
  3. The probability of getting a Tail is (q= 1-p)  ½
  4. Sum of all probabilities = 1

When two  coins are tossed

The possible sequences are mutually exclusive( only one can happen for each toss). The sum of joint probabilities is equal to one. For The probability of HT and TH we use addition rule P(HT OR TH) =P(HT)+P(TH)= 0.25+0.25=0.50

  1. Probability of two heads up = p*p = p square
  2. Probability of one head and one tail = pq +qp = 2pq
  3. Probability of two tails up = q*q = q square

The sum is

Characteristics of a discrete probability distribution:

  1. For any value of X 0≤P(X)≤1
  2. The values of X are exhaustive. The probability distribution includes all possible values
  3. The values of X are mutually exclusive: Only one value can occur for one experiment
  4. The sum of their probabilities is one ;ƩP() = 1

When three coins are tossed:

Calculation of Mean, Variance and Standard Deviation of Random Variable X:

Let the probability of the person being interviewed  Male – 2/3 and Female – 1/3 . Find mean and SD of the distribution.

  1. p+q = 2/3+1/3 =1
  2. Mean (E(X) = p = 2/3
  3. Var(X) = pq = 2/3*1/3= 2/9
  4. SD(X) = √2/9 =0.4714

Problem:

Find mean, variance and standard deviation of the following discrete probability distribution

Binomial Distribution

  1. When Bernoulli experiment is repeated ‘n’ number of times then it is called Binomial Distribution.
  2. It is a discrete probability distribution.
  3. Probability Distribution which has the  following  probability mass function (p.m.f) is called Binomial Distribution.
  4. Parameters are ‘n’ and ‘p’ and Variable X is discrete and called Binomial variate.  This is denoted by B(n,p)
  5. Assumptions: a)Outcome of the  experiments should be dichotomous Conducted under identical conditions
  6. Experiments must be independent

Formula:

Examples:

  1. Number of defective articles  in sample of 6 drawn  from lot
  2. Number of heads obtained in 3 tosses of a coin
  3. Number of male children in a family of 5  children

Application of Binomial Distribution:

Types of Problems in Binomial Distribution:

  1. Type I: Finding the probability of  events
  2. Type II: Finding the expected values
  3. Type III: Finding the distribution if parameters are give

Type I: Finding the probability of  events

An unbiased coin is tossed  six times. What is the probability that the tosses  will result in

  1. Exactly two heads
  2. At least five heads
  3. At least one head
  4. At most two heads
  5. Not greater than one head
  6. Not less than five heads

  1. Exactly two heads:

Let ‘A’ be the event of getting head.

  • p=1/2
  • q=1/2
  • n=6

Probability of getting exactly two heads is  P(X = 2) =0.234375

Using python : use scipy.stats package  and stats library

use scipy.stats.binom.pmf(k, n, p)

where:

  • k = number of successes (integer or array)
  • n = number of trials (integer)
  • p = probability of success in each trial (float between 0 and 1)

Probability of getting at least 5 heads: means 5 and above

Probability of at least 5 and above (x≥5) is 0.109375

At least one head: means P(x=1..6)

Probability of getting at least one head is 0.98437. Instead of finding P(X=1),P(X=2),P(X=3),P(X=4),P(X=5), and P(X=6)  and sum the out come  we P(X<1) from total probability 1

At most two heads: mean we require P(X= from o to 2)

At most two heads include p(x=0), p(x=1), p(x=2) ie p(x≤ 2). . Probability of getting at most 2 heads: 0.34375

Not greater than one head: means P(x<= 1):

Probability of getting greater than one head ( includes P(x=0) + P(x=1) =0.1093

Not less than five heads (includes p(x=5)+P(x=6)):

Type II: Find the expected values E(X) =\mu

  1. Say a random sample of 5 sachets of coconut oil was examined.
  2. Two were found to be leaking
  3. Dealer received 625 packets each containing 5 sachets.
  4. Find the expected number of packets to contain exactly one sachet leaking?

Solution:

  1. n=5
  2. N= 625
  3. PROBABILITY OF LEAKING p = 2/5
  4. q = 1 –p = 1- 2/5 = 3/5
  5. Expected number of sachets Contains  exactly one leaking

Expected number of packets to contain exactly one leaking sachet is

N * Probability  = N*P(X=1)= 625*162/625 = 162

Formulae  for finding  mean, variance and std for Binomial distribution:

Another Problem:

Researchers found that 70% of the owners accounting  app understand how to program the app. You select Randomly 3 owners. What is the probability that exactly 2 of the owners know how to program the app (ie with p=0.7 , the  probability of selecting exactly 2 owners out of 3 trials. Use S for Success and F for failure

Using Joint Probabilities

Using formula

n = Possible outcome of sequence of 2 success and one failure