Bernoulli trail for ECE department 2nd year pptx

ECE 2nd year

1. Bernoulli TraiL Binomial Law Significance of Binomial Distribution from Binomial Law. Subsequent Example . By Anup Barman(34900323057) Anupam Biswas(34900323058) Nipa Kundu(34900323059) Pritam Das(34900323060)

2. INTRODUCTION The Binomial Law, also known as the Binomial Distribution, is a fundamental concept in probability theory. It tells us how likely it is to get a certain number of “successes” in a series of independent trials, where each trial has only two possible outcomes: “success” and “failure.” Fixed number of trials (n): You have a predetermined number of independent trials (e.g., flipping a coin 10 times).
Two possible outcomes: Each trial has only two distinct outcomes, traditionally labeled “success” (S) and “failure” (F). Constant probability of success (p): The probability of success (p) remains the same for each independent trial (e.g., a fair coin has p = 0.5 for heads).
Independent trials: The outcome of one trial doesn’t affect the outcome of any other trial.

3. Bernoulli Trail Bernoulli’s trials are a type of experiment where you repeat the same thing over and over again independently. Each time, you’re looking for a specific outcome, which we call event A. The chance of this outcome happening is the same each time you do the experiment, no matter how many times you repeat it. The experiment itself only has two possible outcomes, either event A happens, or it doesn’t. It’s like flipping a coin where you can either get heads or tails. The idea of Bernoulli’s trials comes from Jacob Bernoulli, a famous mathematician from Switzerland who came up with the concept way back in the 17th century. People use Bernoulli’s trials a lot in math and statistics to understand things that happen in the real world. For example, if you wanted to figure out the odds of getting heads or tails when flipping a coin, you could use Bernoulli’s trials.

4. Examples of Bernoulli’s Trials The most common example of the Bernoulli trials is flipping a coin. Each flip of the coin has only two possible outcomes: Heads and Tails. If we consider the Head to be a success, then automatically the tail becomes a failure and vice versa is also true. Other than this, rolling a die to get a specific number is also an example of Bernoulli’s Trials. Here if consider getting a desired number to be a success then any other outcome other than the desired number becomes a failure. In this case, each roll of the dice is a Bernoulli’s Trial. Checking emails: Suppose you have an inbox full of emails, and you want to check whether a particular email has arrived or not. Each time you check your inbox, you’re looking for the same event, which is the arrival of that particular email. Checking your inbox is also an example of a Bernoulli trial. Flipping a light switch: Suppose you have a light switch that may either turn on or off, and you want to see if it works. Each time you flip the switch, you’re conducting a Bernoulli’s trial, and you’re looking for a particular event, which is whether the light turns on or not.

5. Theorem Related to Bernoulli’s Trials Statement: If the probability of occurrence of an event (probability of success) in a single trial of Bernoulli’s experiment is p, then the probability that the event occurs exactly r times out of n independent trials is equal to nCr qn – r pr, where q = 1 – p, the probability of failure of the event. Probability of r success in n Trials = nCr qn – r pr where, p is the Probability of Success, q = 1 – p is the Probability of Failure, n is the Number of Independent trials, and r is the number of times an event occurred.

6. Binomial Law Binomial Theorem is a theorem that is used to find the expansion of algebraic identity (ax + by)n. We can easily find the expansion of (x + y)2, (x + y)3, and others but finding the expansion of (x + y)21 is a tedious task and this task can easily be achieved using the Binomial Theorem or Binomial Expansion. As the Binomial theorem is used to find the expansion of two terms it is called the Binomial Theorem. The exponent value in the Binomial expansion can also be a negative value or fraction value. Let’s learn about the Binomial theorem, its expansion, formula, and others in detail in this article.

7. What is Binomial Theorem? Binomial Theorem is used to solve binomial expressions in a simple way. This theorem was first used somewhere around 400 BC by Euclids a famous Greek mathematician. It gives an expression to calculate the expansion of algebraic expression ( a+b )n. The terms in the expansion of the following expression are exponent terms and the constant term associated with each term is called the coefficient of terms. Binomial Theorem Statement The Binomial theorem for the expansion of ( a+b )n is stated as, (a + b)n = nC0 anb0 + nC1 an-1 b1 + nC2 an-2 b2 + …. + nCr an-r br + …. + nCn a0bn where n>0 and the nCk is the binomial coefficient.

8. Significance of Binomial Distribution from Binomial Law. 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.

9. Binomial Probability Distribution, Negative Binomial Distribution 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.

10. 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 P(x:n,p) =  n C x  p x  (1-p) n-x Or P(x:n,p) =  n C x  p x  (q) n-x

11. Subsequent Example 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) =  5 C4 p 4  q 5-4  = 5!/4! 1! × (½) 4 × (½) 1  = 5/32 P(x = 5) =  5 C5 p 5  q 5-5  = (½) 5  = 1/32 Therefore, P(x ≥ 4) = 5/32 + 1/32 = 6/32 = 3/16

12. Subsequent Example 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) =  5 C 1  (½) 5. = 5/32 P(x=2) =  5 C2 p 2  q 5-2  = 5! / 2! 3! × (½) 2 × (½) 3  = 5/16 Therefore, P(X ≤ 2) = 1/32 + 5/32 + 5/16 = 1/2

13. reference Erwin Kreyszig, Advanced Engineering Mathematics, 9 th Edition, John Wiley & Sons, 2006.
(ii) P. G. Hoel, S. C. Port and C. J. Stone, Introduction to Probability Theory, Universal Book Stall, 2003 (Reprint).
(iii) S. Ross, A First Course in Probability, 6 th Ed., Pearson Education India, 2002.
(iv) W. Feller, An Introduction to Probability Theory and its Applications, Vol. 1, 3 rd Ed., Wiley, 1968.
(v) N.P. Bali and Manish Goyal, A text book of Engineering Mathematics, Laxmi Publications, Reprint, 2010.