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# Chapter 3 probability

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### Chapter 3 probability

1. 1. Probability The term probability refers to indicate the likelihood that some event will happen. For example, „there is high probability that it will rain tonight‟. We conclude probability based on some observations or measurements. m Mathematically, probability is P n Where „m‟ is the no of favourable trials and „n‟ is the total no of trials.
2. 2. PROBABILITY.. Sex (M:F)distribution in human being is 1:1. So probability of a child to be born as M or F is 50% ie 0.5.
3. 3. PROBABILIT Introduction A random variable is a variable whose actual value is determined by chance operations. When you flip a coin the outcome may be a head or tail. Possibility of appearing head or tail is 50%-50%. Probability of an albino offspring from two heterozygous parents or the probability of an individual being <6‟ tall are also determined as outcomes of random variables. Random variables may be discrete or continuous.
4. 4. DEFINITION OF PROBABILITY If a trial results in „n‟ exhaustive, mutually exclusive and equally likely cases and „m‟ of the favourable to an event A, then the probability of p of the happening of A is given by Favourable number of cases m p Total number of cases n This gives the numerical measure of probability. Obviously “p” be a positive number not greater than unity, so that 0≤ p ≤1.
5. 5. PROBABILITY.. Assume an expt consist of one through of a 6 sided die. Possible results are 1, 2, 3, 4, 5 and 6. Each of those possible results is a simple event. The probability of each of those events is 1/6 ie P(E1) = P(E2) = P(E3) = P(E4) = P(E5) = P(E6) =1/6. This is shown in Table 1.
6. 6. TABLE 1. Observation Event (Ei) P(Ei) 1 E1 P(E1) = 1/6 2 E2 P(E2) = 1/6 3 E3 P(E3) = 1/6 4 E4 P(E4) = 1/6 5 E5 P(E5) = 1/6 6 E6 P(E6) =1/6
7. 7. PROBABILITY DENSITY Random FUNCTION (PDF)1OR 2 variable:x 3 4 5 6 1/6 1/6 1/6 1/6 PROBABILITY DISTRIBUTION Density: f(x) 1/6 1/6 1.A fair 6-sided dice is rolled with the discrete random variable X representing the number obtained per roll. Give the density function for this variable. Values of X not listed in the table are presumed to be impossible to occur and their corresponding values of f are 0. For example f(0.1)=0, f(-3)=0. Notice that f 1
8. 8. Density: f(x) 2 3 4 1 2 3 4 5 6 5 4 3 2 1 36 36 36 36 36 36 36 36 36 36 36 2.A fair 6-sided dice is rolled twice with the discrete random variable X representing the sum of the numbers obtained on both rolls. Give the pdf of this variable. 5 6 7 8 9 10 11 12 PDF Random variable x There are 6x6 different outcomes possible, so each has a probability of 1/36. The possible values range from 2 to 12. While 2 (1+1) and 12 (6+6) can occur only one way, all other values occur at least two ways eg. A 3 is either a 1 then a 2 or a 2 then a 1; a 4 is either a 1 then a 3, a 3 then a 1, or two 2‟s.
9. 9. PROBABILITY DENSITY FUNCTION… A sum of 10 in two trial may appear in three ways such as 5 + 5 = 10 ii) 4 + 6 = 10 iii) 6 + 4 = 10 i) A sum of 7 in two trial may appear in six ways such as i) 5+2=7 ii) 2+5=7 iii) 4 + 3 =7 iv) 3+4=7 v) 6+1=7 vi) 1+6=7
10. 10. LAWS OF PROBABILITY 1. Additive law of probability 2. Multiplicative law of probability
11. 11. ADDITIVE LAW OF PROBABILITY One can use either the density function table or graph to find the probability of various outcomes. For example, P(X = 10) = 3/36= 1/12 and P(X = 10 or 11) = 3/36 + 2/36 = 5/36 10 and 11 are mutually exclusive events, so application of the general addition law leads to summation of the individual probabilities. This law is known as ADDITIVE LAW of probability
12. 12. WHAT IS MUTUALLY EXCLUSIVE ????? Cases are said to be mutually exclusive if the occurrence of one of them excludes the occurrence of all the others. For example, in tossing an unbiased coin, the cases of appearing “head” and “tail” are mutually exclusive.
13. 13. MULTIPLICATIVE LAW OF PROBABILITY Consider k sets of elements of size n1, n2,….nk. If one element is randomly chosen from each set, then the total no of different results is n1n2n3…nk . Example: Consider 3 pens with animals marked as Pen 1: 1,2,3 Pen 2: A, B, C Pen 3: x,y
14. 14. MULTIPLICATIVE LAW The possible triplets with one animal taken from each pen are 1Ax, 1Ay, 1Bx, 1By, 1Cx, 1Cy OF 2Ax, 2Ay, 2Bx, 2By, 2Cx, 2Cy PROBABILITY… 3Ax, 3Ay, 3Bx, 3By, 3Cx, 3Cy The number of possible triplets is : n!= (3)(3)(2)=18
15. 15. PERMUTATIONS From a set of n elements, the number of ways those n elements can be rearranged , ie put in different orders, is the permutation of n elements Pn = n! The symbol n! (factorial of n) denotes the product of all natural numbers from 1 to n n! = (1)(2)(3)….(n) By definition 0! = 1.
16. 16. PROBLEM 1: In how many ways can three animals, x, y and z be arranged in triplets ? The number of permutations of n =3 elements P(3) = 3! = (1)(2)(3) = 6. The six possible triplets: xyz, xzy, yxz, yzx, zxy, zyx More generally, we can define permutations of n elements taken k at a time in particular order as Pn,k = n! (n k )!
17. 17. EXAMPLE In how many ways can three animals, x, y, z be arranged in pairs such that the order in the pairs is important (xz is different than zx)? Pn,k = (3 3!2 )!= 6 The six possible pairs are: xy xz yx yz xz zy
18. 18. MULTIPLICATIVE Problem: A bag contains 4 white and 5 red balls. Two balls are drawn successively at random from the bag. What is the probability that both the balls are white when the drawings are made i)with replacement? ii)without replacement? LAW OF PROBABILITY.. Soln: Let A be the event that the first ball is white and B be the event that the second ball is also white. i) Since the drawings are made with replacement , the two events become independent Hence P(AB) = P(A) P(B) = ( 4/9)x(4/9) = 16/81 ii) Since the drawings are made without replacement the events become dependent. Hence P(AB) = P(A) P(B/A) = 4/9x3/8 = 12/72
19. 19. COMPOUND EVENTS A compound event is an event composed of two or more events. Consider two events A and B. The compound event such that both events A and B occur is intersection of the events, denoted by A ∩ B . The compound such that either event A or event B occurs is called the union of events, denoted by A U B. The probability of an intersection is P(A ∩ B) and the probability of union is P(A U B). Also P(A U B) = P(A) + P(B) – P(A∩ B)