A Probability Distribution is a rule that assigns probabilities to each element of a set of events that may occur. Types of Probability Distribution (i) Discrete (ii) Continuous If a random variable is a discrete variable, its probability distribution is called a discrete probability distribution. Ex , Suppose if we toss a coin are two times. This simple experiment can have four possible outcomes: HH, HT, TH, and TT. Now, let the random variable X represent the number of Heads that result from this experiment. The random variable X can only take on the values 0, 1, or 2, so it is a discrete random variable. Possible values of x Probability of each value 0 1 1/4 2/4 2 1/4 Continuous Probability Distribution Continuous Probability Distribution is that in which the random variable can be any number within some given range of values. 5-5.Height of students (in feet) 5-5.8 5.00 PROBABILITY 4/10 3/10 3/10 .8-6.5 5. m. . A Discrete Probability Distribution is known as Probability mass function( p.d.f) and a continuous Probability Distribution is known as Probability density function( p.f). Binomial 2. Poisson 3.1. Normal . It was discovered by mathematician James bernouli. . It is the discrete Probability Distribution. It is the theoretical distribution of discrete random variables. Bernoulli. say n.The Binomial Distribution is associated with the name of J. ‘hit’ or ‘miss’. i. ‘good’ or ‘defective’. ii) Each trial is independent of other trials. q = The probability of failure. The probability of success . survive or die. probability of an outcome for any particular trial is not influenced by the outcomes of the other trials. iii) Each trial has two possible outcomes. iv) These above conditions will satisfy if we toss a coin.e. p . The features of Binomial Distribution is as follows: i) It is a random process which is performed under the same conditions for a fixed and finite number of trials. such as ‘success’ or ‘failure’. . remains constant from trial to trial . where q = 1-p. which are mutually exclusive. . TTT.A fair coin is tossed 3 times and we are interested in finding the probability of exactly two heads.HTH THH Favorable Outcomes HHT. Total outcomes 8: HHH.HTH THH P = 3/8. TTH. HHT. Therefore we will consider head as success and tail as failure with corresponding probabilities p and q respectively. THT. HTT . As the number of tosses increases(say 20 0r 50 times). Here an easy method is required and hence we use binomial formula. it becomes more and more difficult to calculate the probability. . . .f( x) =P(x)= n C x p x q n-x n = the number of trials p= the probability of a success on a trial q = the probability of a failure on a trial X = the number of successes in n trials x = 0. 1. . 2. . n . If 3 balls are drawn at random one by one with replacement from a bag containing 8 white and 15 red balls. balls are drawn without replacement . A coin is tossed six times . . 2. find the probability that i) exactly 2 workers suffer from the disease ii) not more than 2 workers suffer from the disease. what is the probability of obtaining four or more heads? The incidence of a certain disease is such that on the average 20% of workers suffer from it. (Assuming the distribution fits binomial) If 10 workers are selected at random.1. It is believed that 20% of the employees in an office are usually late. . If 10 employees report for duty on a given day. 0.32 .2. (c) At least 3 employees are late.88. (b) At most 3 employees are late.Q3. 0. what is the probability that: (a) Exactly 3 employees are late. 0. The limit of credit sales of different buyers is decided by the manufacturer based upon the perception of the goodwill of the individual buyer.Q4. The manufacturer observed that 30% of the buyers take more than a month in making the payment. In a particular city if goods are sold to 5 buyers on credit.1323.837 . 0. The market practice is such that goods are sold on one month’s credit in the domestic market. what is the probability that (i) Exactly 3 buyers will delay the payment beyond one month (ii) At most 2 buyers will delay ? 0.A company manufactures motor parts. 0.354.47 . After the privatisation of the power sector in Delhi.Q5. On testing of meters it was found that 10% of the meters were defective and run faster. what is the probability that (i) one meter is defective. (ii) at least one meter is defective? 0. consumers often complain that new meters installed by the private power companies are defective and run faster. a test check was conducted on 6 meters. Ina housing society. qn-1.2. n.n.p 2.p 1q n-2. P(x) 0 1 2 .Let n denotes no.p2 2q nc n np q0 npn .p2 n-1.3…. x takes the values 0. n 0q n.p0 0 1.1. . of trials in a binomial distribution .n(n-1) /2qn-2. p denotes probability of success and q denotes probability of failure. X P(x) nc nc nc x. . n. P(x) ∑P(x) ∑x..p2 +…..p + …….qn-2.+ pn-1 = np( p+q) n-1 =np(1) n-1 = np ..p + n(n-1) . q n + 1..pn =np qn-1 +(n-1) .qn-1.Arithmetic Mean= ∑x.+ n. P(x)=0.qn-2.. Mean= np Variance=npq . is 6.Q1.The mean of a binomial distribution is 40 and standard deviation is 6. Q2. Q3. Calculate n.In eight throws of a die 1 or 6 is considered as success.Bring out the fallacy . Find the mean number of success and the SD. . The mean is 10 and its s. in the following statement. if any. p and q.d. . If 7 bombs are dropped at the bridge .Q4. Two bombs are enough to destroy a bridge. The probability of a bomb hitting a target is 2/5. If there are 5 choices for each question. A student appearing in a multiple choice test answers 10 questions. 0. purely by guessing.007. What is the probability that 6 or more answers will be correct. find the probability that the bridge is destroyed. Q5. where the total no of observation (n)is large and their chance of Occurrence(p) is low. Poisson.A second important discrete probability Distribution is the Poisson Distribution. named after the French mathematician S. . This distribution is used to describe the behaviour of events. . It is used by the quality control departments of manufacturing industries to count the number of defects found in a lot. The number of accidents that occur on a given highway during a given time period. The number of printing mistakes in a page of a book The number of earthquakes in Delhi in a decade. The number of deaths of the policyholders recorded by the LIC . 1.mx x! e=base of natural logarithm whose value is 2.2. P(x) = e-m.Poisson distribution is given by.3. m= mean = np and m >0 . x =0. of occurrences of an event.7183 x= no. Mean= m Variance=m . If 100 items are packed in each box.On the average. one in 400 items is defective. what is the probability that any given box will contain : i) no defective ii) less than two defectives iii) one or more defectives iv) more than 3 defectives . i) P (x=0) = e-m. m= np = 0. p = 1/400.Here. probability of defective item which is very low. n= 100 .25 .no of items packed in the box which is quite large. average number of defectives in a box of 100 items.mx = x! ii) P (x<2) = P (x=0) + P (x=1) . Customers arrive randomly at a retail counter at an average rate of 10 per hour. ii) Exactly one customer arrives in any particular minute. calculate i) No customer arrives in any particular minute. Hint: m= 10 . Assuming a Poisson Distribution. calculate the probability that in a factory employing 300 workers. there will be at least two fatal accidents in a year.25 =0.7787 . Hint: e -0.Assuming that the probability of a fatal accident in a factory during a year is 1/1200. who produces medicine bottles. The bottles are packed in boxes containing 500 bottles.1% of the bottles are defective. A drug manufacturer buys 100 boxes from the producer of bottles.6065) Ans61 &9 . Using Poisson distribution find out how many boxes will contain (i) No defective (ii) At least 2 defectives (e .A manufacturer .5 = 0.0. finds that 0. 13.2.If the probability of a defective bolt is 0.4 . find (i) The mean (ii) The standard deviation in a total of 900 bolts. (i) (ii) (iii) n p np=m α 0 . . (.4)2 .6)2 = .130 4 .4)1 .6)0 = .4)0 .6)3 = .346 6 .4)3 .346 4 .6 . (. (. (. (. (.4)4 .4. (.154 1 .026 p=0.X 0 1 2 3 4 P(X) 1 .6)1 = . (.6)4 = . q= 0. (. (. 0.05 P(X) .35 0.346 . p=.25 0.346 .05 0 0 1 2 3 4 X 0 1 2 3 4 n=4 .3 0.154 .2 0.026 .130 .15 0.1 0. σ= standard deviation of the normal distribution. Hence. normal distribution is approximation to binomial distribution. μ = Mean of the normal distribution. σ ) . called normal distribution. A random variable follows normal distribution with mean μ and standard deviation σ. X ~ N (μ . the curve becomes smooth. X takes continuous values.When no of trials become infinite. ∞ <x<∞ σ = standard deviation of the normal distribution Π = constant.1416.7183 μ = Mean of the normal distribution .5066 e=2.. √2 Π =2. 22/7=3. μ . σ -3 -2 -1 0 1 2 3 .σ) A random variable X can be transformed to a standardized normal variable Z by subtracting the mean and divided by standard deviation. Z=x.X∽N(μ. Probability Function given by .7% within 3 standard deviations of the mean. Bell-shaped Symmetric about mean Continuous Never touches the x-axis Total area under curve is 1. This is the Empirical Rule mentioned earlier. and 99. Data values represented by x which has mean µ and standard deviation σ.00 Approximately 68% lies within 1 standard deviation of the mean. 95% within 2 standard deviations. . calculate how many workers have a salary between Rs. Suppose the salary of workers in a company follows normal distribution.400 and Rs.Find out the area between z=0 and z=1.54 2.650. If the average salary is Rs.500 with a standard deviation of Rs.1. 11618 .100. 15thousand. Assuming the distribution to be normal .140 thousand and Rs.The average daily sales of 500 branch offices is 150 thousand and the standard deviation Rs. indicate how many branches have sales between: (i) Rs.120 thousand and Rs.145 thousand. (ii) Rs.165 thousand 174 &295 . A workshop produces 2000 units per day. Assuming the distribution to be normal . find out how many units are expected to weigh less than 142 kg? 1770 . The average weight of units is 130 kg with a standard deviation of 10 kg. it was found that lifetime of the fans was normally Distributed with an average life of 2.040 hours and standard deviation of 60 hours. On the basis of the information estimate the number of fans that is expected to run for (a) more than 2.150 Hours (b) less than 1.960 hours.000 electric fans manufactured by a company.As a result of tests on 20. 672 & 1836 . If on a certain day. city’s traffic is disrupted for about 3 hours( 180 minutes) on an average with a standard deviation of 45 minutes.Delhi’s Traffic police claims that whenever any rally is organized in the city. traffic in the city is seriously disrupted. a rally is organized in the city what is the probability that: (a) Traffic was disrupted up to 2 hours. . (b) Traffic was disrupted up to 5 hours.9962 . It is believed that the disruption of traffic is normally distributed. On the day of rally.0918 0.