Probability Distribution PDF

**Probability Distribution** **1. Binomial Distribution** The binomial distribution arises from a repeated random experiment which has two possible outcomes: a. That an event will occur; b. That an event will not occur. In the experiment of tossing a fair coin repeatedly, there are two possib...

**Probability Distribution** **1. Binomial Distribution** The binomial distribution arises from a repeated random experiment which has two possible outcomes: a. That an event will occur; b. That an event will not occur. In the experiment of tossing a fair coin repeatedly, there are two possible outcomes namely: a head or not a head. Each repetition of tossing of the fair coin is called a trial. The two possible outcomes of the random experiment are usually called success and failure. If the event that a head shown up is a success, the event that a head does not show up is a failure, or if the event that a head shows up is a failure, then the event that a head does not show up is a success. Part of our theoretical assumptions is that the probability of success or failure in one trial is not influenced in anyway by the probability of success or failure of a previous trial. We shall denote the probability of success by p and the probability of failure by q. Since the two events; success or failure are complementary, p + q = 1 or q = 1 -- p. Also, the probability of success is the same for each the trials. The independent trials which we have been discussing are usually called Bernoulli trials. if we denote the probability that in *n* Bernoulli trials, we have *r* successes and *n -- r* failures by P(*x = r}* then p {*x = r} is* defined as: P~r~ {*x = r) =^n^C~r~*p^r^q^n-r^.....where *^n^C~r~*= [\$\\frac{n!}{\\left( n - r \\right)!r!}\$]{.math.inline} The parameters of the binomial distribution are n and p. The standard discrete probability distributions that are consequent to these processes are the Binomial and the Poisson distribution. We will now look into the conditions that characterise these processes, and examine the standard distributions associated with the processes. This will enable us to identify situations for which these distributions apply. Properties of the Binomial Distribution If we denote the mean and standard deviation of a binomial distribution by µ and [*σ*]{.math.inline} respectively, then: a. Mean µ = np. b. Standard deviation [*σ*]{.math.inline} = [\$\\sqrt{\\text{npq\\ }}\$]{.math.inline} c. Variance [*σ*^2^]{.math.inline}= npq Example Find the probability that when a fair coin is tossed three times, a head shows up twice. Solution. Let p be the probability of a head shows up. Let q be the probability that a tail shows up. Let P~r~ = {x*=2*} be the probability that two heads show up in three show up in three trials. Then: p = ½, q = 1 -- ½ = ½ *Pr*{*x*=2} = ^3^C~2~ (½)^2^ (½) = [\$\\frac{3!\\ }{2!1!}\$]{.math.inline} x (½)^2^ x (½) = 3 x ¼ x ½ = 3/8 **Binomial Probability Function** The binomial distribution probability is given as: Where: n = Sample size x = 0, 1, 2,......., n p = Probability of a success q = (1-p) = Probability of failure Note: X should not exceed 20. Looking at the binomial distribution formula, it is clear that it has two parameters. These are n and p. Therefore, the mean of a binomial distribution is np. If the mean of a binomial distribution is np then the standard deviation of a binomial distribution is given as:  The principles upon which binomial distributions are applicable are, in the cases of repeated trials. **Examples of trials in binomial distribution:** 1\. Probability that 5 out of 15 students will pass the introductory statistics course; 2\. Probability that half the patients will die in the emergency ward; 3\. Sampling from a finite population with replacement; 4\. Sampling from infinite population without replacement. *In-text Question 1: Two out of the three distributions are discrete. What are they?* *Answer* *Both the binomial distribution and the Poisson distribution are discrete distributions* **2. Poisson Distribution** Poisson distribution is a discrete random variable, which is often useful when dealing with the number of occurrences of events over a specified period of time or after a certain interval of time. When the number of trials is very large and the probability of success is comparatively very small, the binomial distribution may not be a very suitable model for random experiments with repeated trials. A discrete probability distribution which is more suitable for this purpose is what is called the Poisson distribution and is defined as: P~r~ (*x*) = [\$\\frac{\\lambda\^{x}e\^{- \\lambda}}{x!}\$]{.math.inline}*x* = 0,1,2,... Where λ = *np*; is the mean number of successes and *e* = 2.718. Example: A labour expert on the phenomenon of strikes conducted research. The researcher found out that the mean number of strikes in a manufacturing industry was 3.4 per month. What is the probability that during a given month, there will be? a\. No strike at all b\. More than 2 strikes c\. Exactly 4 strikes Solution:  **Properties of the Poisson Distribution** If we denote the mean and standard deviation of a Poisson distribution by µ and [σ ]{.math.inline}respectively, then: Mean µ = λ Standard deviation [σ ]{.math.inline}= [\$\\sqrt{\\text{λ\\ }}\$]{.math.inline} Variance [*σ*²]{.math.inline} = λ Example 0.2% of the corks produced by a machine were found to be detective. Find the probability that 5 out of 1000 corks would be defective. Solution Let *p* be the probability of success, i.e. that a cork would be defective, then: P = [\$\\frac{2}{1000\\ }\$]{.math.inline} = 0.002, n = 1000 λ = np = 1000 x 0.002 = 2 We observe that the probability of success is very small and the number of trials is comparatively large, hence, the Poisson distribution will be a suitable probability model. Hence: *Pr*(*x* = 5) *=* [\$\\frac{2\^{5}e\^{- 2}}{5!}\$]{.math.inline} = [\$\\frac{32\\ }{120e\^{\\lambda}}\$]{.math.inline} = [\$\\frac{32\\ }{886.7}\$]{.math.inline} = 0.036 *In-text Question 1:* What are the two possible outcomes of a random experiment? *Answer* *The two possible outcomes of the random experiment are usually called success and failure.* **3 Normal Distribution** Empirically, most statistical distributions of continuous nature can be described by a function called the normal distribution function. The function takes the form *P*(*x*) = [\$\\frac{1}{\\text{σ\\ }\\sqrt{2\\pi}}{- \\ ½}\^{(\\frac{x - \\mu}{\\sigma})²}\$]{.math.inline} Where [*σ*]{.math.inline} is the standard deviation, µ is the mean and *e =* 2.718? Some random variables that have normal distributions are: i. The masses of students in a school; ii. The yields of agricultural produce in a farm iii. The heights of plants, etc. The graphical representation of a normal distribution function is a bell-shaped curve shown below: *In-text Question 2 Which distribution is continuous?* *Answer* *The normal distribution is a continuous distribution.* **Binomial Distribution** In probability theory and statistics, the binomial distribution is the discrete probability distribution that gives only two possible results in an experiment, either Success or Failure. For example, if we toss a coin, there could be only two possible outcomes: heads or tails, and if any test is taken, then there could be only two results: pass or fail. This distribution is also called a binomial probability distribution. There are two parameters n and p used here in a binomial distribution. The variable 'n' states the number of times the experiment runs and the variable 'p' tells the probability of any one outcome. Suppose a die is thrown randomly 10 times, then the probability of getting 2 for anyone throw is ⅙. When you throw the dice 10 times, you have a binomial distribution of n = 10 and p = ⅙. Learn the formula to calculate the two-outcome distribution among multiple experiments along with solved examples here in this article. **Binomial Probability Distribution** In binomial probability distribution, the number of 'Success' in a sequence of n experiments, where each time a question is asked for yes-no, then the boolean-valued outcome is represented either with success/yes/true/one (probability p) or failure/no/false/zero (probability q = 1 − p). A single success/failure test is also called a Bernoulli trial or Bernoulli experiment, and a series of outcomes is called a Bernoulli process. For n = 1, i.e. a single experiment, the binomial distribution is a Bernoulli distribution. The binomial distribution is the base for the famous binomial test of statistical importance. **Negative Binomial Distribution** In probability theory and statistics, the number of successes in a series of independent and identically distributed Bernoulli trials before a particularised number of failures happens. It is termed as the negative binomial distribution. Here the number of failures is denoted by 'r'. For instance, if we throw a dice and determine the occurrence of 1 as a failure and all non-1's as successes. Now, if we throw a dice frequently until 1 appears the third time, i.e., r = three failures, then the probability distribution of the number of non-1s that arrived would be the negative binomial distribution. Binomial Distribution Examples As we already know, binomial distribution gives the possibility of a different set of outcomes. In real life, the concept is used for: Finding the quantity of raw and used materials while making a product. Taking a survey of positive and negative reviews from the public for any specific product or place. By using the YES/ NO survey, we can check whether the number of persons views the particular channel. To find the number of male and female employees in an organisation. The number of votes collected by a candidate in an election is counted based on 0 or 1 probability. **Binomial Distribution Formula** The binomial distribution formula is for any random variable X, given by; P(x:n,p) = nCx px (1-p)n-x Or P(x:n,p) = nCx px (q)n-x Where, n = the number of experiments x = 0, 1, 2, 3, 4,... p = Probability of Success in a single experiment q = Probability of Failure in a single experiment = 1 -- p The binomial distribution formula can also be written in the form of n-Bernoulli trials, where nCx = n!/x!(n-x)!. Hence, P(x:n,p) = n!/\[x!(n-x)!\].px.(q)n-x **Binomial Distribution Mean and Variance** For a binomial distribution, the mean, variance and standard deviation for the given number of success are represented using the formulas Mean, μ = np Variance, σ2 = npq Standard Deviation σ= √(npq) Where p is the probability of success q is the probability of failure, where q = 1-p **Binomial Distribution Vs Normal Distribution** The main difference between the binomial distribution and the normal distribution is that binomial distribution is discrete, whereas the normal distribution is continuous. It means that the binomial distribution has a finite amount of events, whereas the normal distribution has an infinite number of events. In case, if the sample size for the binomial distribution is very large, then the distribution curve for the binomial distribution is similar to the normal distribution curve. **Properties of Binomial Distribution** The properties of the binomial distribution are: There are two possible outcomes: true or false, success or failure, yes or no. There is 'n' number of independent trials or a fixed number of n times repeated trials. The probability of success or failure remains the same for each trial. Only the number of success is calculated out of n independent trials. Every trial is an independent trial, which means the outcome of one trial does not affect the outcome of another trial. **Binomial Distribution Examples and Solutions** Example 1: If a coin is tossed 5 times, find the probability of: \(a) Exactly 2 heads \(b) At least 4 heads. Solution: \(a) The repeated tossing of the coin is an example of a Bernoulli trial. According to the problem: Number of trials: n=5 Probability of head: p= 1/2 and hence the probability of tail, q =1/2 For exactly two heads: x=2 P(x=2) = 5C2 p2 q5-2 = 5! / 2! 3! × (½)2× (½)3 P(x=2) = 5/16 \(b) For at least four heads, x ≥ 4, P(x ≥ 4) = P(x = 4) + P(x=5) Hence, P(x = 4) = 5C4 p4 q5-4 = 5!/4! 1! × (½)4× (½)1 = 5/32 P(x = 5) = 5C5 p5 q5-5 = (½)5 = 1/32 Therefore, P(x ≥ 4) = 5/32 + 1/32 = 6/32 = 3/16 Example 2: For the same question given above, find the probability of: a\) Getting at most 2 heads Solution: P (at most 2 heads) = P(X ≤ 2) = P (X = 0) + P (X = 1) + + P (X = 2) P(X = 0) = (½)5 = 1/32 P(X=1) = 5C1 (½)5.= 5/32 P(x=2) = 5C2 p2 q5-2 = 5! / 2! 3! × (½)2× (½)3 = 5/16 Therefore, P(X ≤ 2) = 1/32 + 5/32 + 5/16 = 1/2 Example 3: A fair coin is tossed 10 times, what are the probability of getting exactly 6 heads and at least six heads. Solution: Let x denote the number of heads in an experiment. Here, the number of times the coin tossed is 10. Hence, n=10. The probability of getting head, p ½ The probability of getting a tail, q = 1-p = 1-(½) = ½. The binomial distribution is given by the formula: P(X= x) = nCxpxqn-x, where = 0, 1, 2, 3,... Therefore, P(X = x) = 10Cx(½)x(½)10-x \(i) The probability of getting exactly 6 heads is: P(X=6) = 10C6(½)6(½)10-6 P(X= 6) = 10C6(½)10 P(X = 6) = 105/512. Hence, the probability of getting exactly 6 heads is 105/512. \(ii) The probability of getting at least 6 heads is P(X ≥ 6) P(X ≥ 6) = P(X=6) + P(X=7) + P(X= 8) + P(X = 9) + P(X=10) P(X ≥ 6) = 10C6(½)10 + 10C7(½)10 + 10C8(½)10 + 10C9(½)10 + 10C10(½)10 P(X ≥ 6) = 193/512. Practice Problems Solve the following problems based on binomial distribution: The mean and variance of the binomial variate X are 8 and 4 respectively. Find P(X\