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Unit 11 Binomial Distribution IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP 1 Unit 11 Binomial Distribution.

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Presentation on theme: "Unit 11 Binomial Distribution IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP 1 Unit 11 Binomial Distribution."— Presentation transcript:

1 Unit 11 Binomial Distribution IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP 1 Unit 11 Binomial Distribution

2 IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP2 Introducing binomial distribution Consider the following random variables: a) X : no.of “6” obtained in 10 rolls of a fair die. b) X : no. of tails obtained in 100 tosses of a fair coin. c) X : no. of defective light bulb in a batch of 1000. d) X : no. of boys in a family of 5 children.

3 Unit 11 Binomial Distribution IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP3 In each case, a basic experiment is repeated a number of times. For example, the basic experiment in case (a) is rolling the die once.

4 Unit 11 Binomial Distribution IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP4 The following are common characteristics of the random variables in cases (a) to (d): 1) The number of trials n of the basic experiment is fixed in advance. 2) Each trial has two possible outcomes which may be called “success” and “failure”. 3) The trials are independent. 4) The probability of success is fixed.

5 Unit 11 Binomial Distribution IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP5 11.1 Binomial random variable A random variable X defined to be the number of successes among n trials called a binomial random variable if the properties (1) to (4) are satisfied. Mathematically, we write X ~ Bin(n,  ), where n = no. of trials, and  = prob. of success.

6 Unit 11 Binomial Distribution IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP6 The p.d. of a binomial r.v. If X ~ Bin(n,  ), then p(x) = P(X = x) = n C x  x (1-  ) n-x where x = 0, 1, 2, …, n.

7 Unit 11 Binomial Distribution IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP7 Example 11. 1-1 A fair coin is tossed 8 times. Find the probability of obtaining 5 heads. Let X be the number of heads obtained in 8 tosses. Then X ~ Bin(8, 1/2). P(5 heads) = = 7/32

8 Unit 11 Binomial Distribution IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP8 Example 11. 1-2 (1/2) There are 10 multiple-choice questions in a test and each question has 5 options. Suppose a student answers all 10 questions by randomly picking an option in each question. Find the probability that (a) he will answer six questions correctly, (b) he will get at least 3 correct answers.

9 Unit 11 Binomial Distribution IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP9 Example 11. 1-2 (2/2) Let X be the number of correct answers he will get. Then X ~ Bin(10, 0.2). (a) P(X = 6) = 10 C 6 (0.2) 6 (1-0.2) 10-6 = 0.00551 (b)P(at least 3 correct answers) = 1 – P(X = 0) – P(X = 1) – P(X = 2) = 1 – 10 C 0 (0.8) 10 – 10 C 1 (0.2)(0.8) 9 – 10 C 2 (0.2) 2 (0.8) 8 = 0.322

10 Unit 11 Binomial Distribution IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP10 Example 11. 1-3 (1/3) Binary digits 0 and 1 are transmitted along a data channel in which the presence of noise results in the fact that each digit may be wrongly received with a probabilty of 0.00002. Each message is transmitted in blocks of 2000 digits. (a)What is the probabilty that at least one digit in a block is wrongly received? (b)If a certain message has a length of 20 blocks, find the probability that 2 or more blocks are wrongly received.

11 Unit 11 Binomial Distribution IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP11 Example 11. 1-3 (2/3) (a) Let X be the number of digits wrongly received in a block of 2000 digits. Then X ~ Bin(2000, 0.00002). P(X  1)= 1 – P(X = 0) = 1 – 0.00002 2000 = 0.0392

12 Unit 11 Binomial Distribution IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP12 Example 11. 1-3 (3/3) (b) Let Y be the number of block that are wrongly received among the 20 blocks. Then Y~ Bin(20, 0.0392). P(Y  2) = 1 – P(Y = 0) – P(Y = 1) = 1 – (1 – 0.0392) 20 – 20 C 1 (0.0392)(1 – 0.0392) 19 = 0.184


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