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Sathyabama Institute of Science and Technology
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This document from Sathyabama Institute of Science and Technology covers fundamental probability concepts, including theorems, conditional probability, and independent evetns. It features example problems to enhance understanding.
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Sathyabama Institute of Science and Technology Important Theorems Theorem 1: Probability of impossible event is zero. Proof: Let S be sample space (certain events) and be the impossible event. Certain events and impossible events are mutually exclusive. P(S ) = P(S) +...
Sathyabama Institute of Science and Technology Important Theorems Theorem 1: Probability of impossible event is zero. Proof: Let S be sample space (certain events) and be the impossible event. Certain events and impossible events are mutually exclusive. P(S ) = P(S) + P() (Axiom 3) S=S P(S) = P(S) + P() P() = 0, hence the result. Theorem 2: If A is the complementary event of A, P( A ) 1 P( A) 1. Proof: Let A be the occurrence of the event A be the non-occurrence of the event. Occurrence and non-occurrence of the event are mutually exclusive. P( A A ) P( A) P( A ) A A S P( A A ) P( S ) 1 1 P( A) P( A ) P( A ) 1 P( A) 1. Theorem 3: (Addition theorem) If A and B are any 2 events, P(A B) = P(A) + P(B) P(A B) P(A) + P(B). Proof: We know, A AB AB and B A B AB P( A) P( AB ) P( AB) and P( B) P( A B) P( AB) (Axiom 3) P( A) P( B) P( AB ) P( AB) P( A B) P( AB) P( A B) P( A B) P(A B) = P(A) + P(B) P(A B) P(A) + P(B). Note: The theorem can be extended to any 3 events, A,B and C P(A B C) = P(A) + P(B) +P(C) P(A B) P(B C) P(C A) + P(A B C) Theorem 4: If B A, P(B) P(A). Proof: A and AB are mutually exclusive events such that B AB A P( B AB ) P( A) P( B) P( AB ) P( A) (Axiom 3) P ( B ) P ( A) Conditional Probability The conditional probability of an event B, assuming that the event A has happened, is denoted by P(B/A) and defined as P( A B) P( B / A) , provided P(A) 0 P( A) Page no 2 Sathyabama Institute of Science and Technology Product theorem of probability Rewriting the definition of conditional probability, We get P( A B) P( A) P( A / B) The product theorem can be extended to 3 events, A, B and C as follows: P( A B C ) P( A) P( B / A) P(C / A B) Note: 1. If A B, P(B/A) = 1, since A B = A. P( B) 2. If B A, P(B/A) P(B), since A B = B, and P( B), P( A) As P(A) P(S) = 1. 3. If A and B are mutually exclusive events, P(B/A) = 0, since P(A B) = 0. 4. If P(A) > P(B), P(A/B) > P(B/A). 5. If A1 A2, P(A1/B) P(A2/B). Independent Events A set of events is said to be independent if the occurrence of any one of them does not depend on the occurrence or non-occurrence of the others. If the two events A and B are independent, the product theorem takes the form P(A B) = P(A) P(B), Conversely, if P(A B) = P(A) P(B), the events are said to be independent (pair wise independent). The product theorem can be extended to any number of independent events, If A1 A2 A3 ….. An are n independent events, then P(A1 A2 A3 ….. An) = P(A1) P(A2 ) P(A3 )….. P(An) Theorem 4: If the events A and B are independent, the events A and B are also independent. Proof: The events A B and A B are mutually exclusive such that (A B) ( A B) = B P(A B) + P( A B) = P(B) P( A B) = P(B) P(A B) = P(B) P(A) P(B) (A and B are independent) = P(B) [1 P(A)] = P( A ) P(B). Theorem 5: If the events A and B are independent, the events A and B are also independent. Proof: P( A B ) = P A B = 1 P(A B) = 1 [ P(A) + P(B) P(A B)] (Addition theorem) = [1 P(A)] P(B) [1 P(A)] = P( A )P( B ). Page no 3 Sathyabama Institute of Science and Technology Problem 1: From a bag containing 3 red and 2 balck balls, 2 ball are drawn at random. Find the probability that they are of the same colour. Solution : Let A be the event of drawing 2 red balls B be the event of drawing 2 black balls. P(A B) = P(A) + P(B) 3C 2 2C 2 3 1 2 = = 5C 2 5C 2 10 10 5 Problem 2: When 2 card are drawn from a well-shuffled pack of playing cards, what is the probability that they are of the same suit? Solution : Let A be the event of drawing 2 spade cards B be the event of drawing 2 claver cards C be the event of drawing 2 hearts cards D be the event of drawing 2 diamond cards. 13C 2 4 P(A B C D) = 4 =. 52C 2 17 Problem 3: When A and B are mutually exclusive events such that P(A) = 1/2 and P(B) = 1/3, find P(A B) and P(A B). Solution : P(A B) = P(A) + P(B) = 5/6 ; P(A B) = 0. Problem 4: If P(A) = 0.29, P(B) = 0.43, find P(A B ), if A and B are mutually exclusive. Solution : We know A B = A P(A B ) = P(A) = 0.29 Problem 5: A card is drawn from a well-shuffled pack of playing cards. What is the probability that it is either a spade or an ace? Solution : Let A be the event of drawing a spade B be the event of drawing a ace P(A B) = P(A) + P(B) P(A B) 13 4 1 4 = . 52 52 52 13 Page no 4 Sathyabama Institute of Science and Technology Problem 6: If P(A) = 0.4, P(B) = 0.7 and P(A B) = 0.3, find P( A B ). Solution : P( A B ) = 1 P(A B) = 1 [P(A) + P(B) P(A B)] = 0.2 Problem 7: If P(A) = 0.35, P(B) = 0.75 and P(A B) = 0.95, find P( A B ). Solution : P( A B ) = 1 P(A B) = 1 [P(A) + P(B) P(A B)] = 0.85 Problem 8: A lot consists of 10 good articles, 4 with minor defects and 2 with major defects. Two articles are chosen from the lot at random(with out replacement). Find the probability that (i) both are good, (ii) both have major defects, (iii) at least 1 is good, (iv) at most 1 is good, (v) exactly 1 is good, (vi) neither has major defects and (vii) neither is good. Solution : 10C2 3 (i) P( both are good) = 16C2 8 2C2 1 (ii) P(both have major defects) = 16C2 120 10C1 6C1 10C2 7 (iii) P(at least 1 is good) = 16C2 8 10C0 6C2 10C1 6C1 5 (iv) P(at most 1 is good) = 16C2 8 10C16C1 1 (v) P(exactly 1 is good) = 16C2 2 14C2 91 (vi) P(neither has major defects) = 16C2 120 6C2 1 (vii) P(neither is good) = . 16C2 8 Problem 9: If A, B and C are any 3 events such that P(A) = P(B) = P(C) = 1/4, P(A B) = P(B C) = 0; P(C A) = 1/8. Find the probability that at least 1 of the events A, B and C occurs. Solution : Since P(A B) = P(B C) = 0; P(A B C) = 0 P(A B C) = P(A) + P(B) +P(C) P(A B) P(B C) P(C A) + P(A B C) 3 1 5 = 00 . 4 8 8 Page no 5 Sathyabama Institute of Science and Technology Problem 10: A box contains 4 bad and 6 good tubes. Two are drawn out from the box at a time. One of them is tested and found to be good. What is the probability that the other one is also good? Solution : Let A be a good tube drawn and B be an other good tube drawn. 6C2 1 P(both tubes drawn are good) = P(A B) = 10C2 3 P( A B ) 1/ 3 5 P(B/A) = = (By conditional probability) P( A ) 6 / 10 9 Problem 11: In shooting test, the probability of hitting the target is 1/2, for a, 2/3 for B and ¾ for C. If all of them fire at the target, find the probability that (i) none of them hits the target and (ii) at least one of them hits the target. Solution : Let A, B and C be the event of hitting the target. P(A) = 1/2, P(B) = 2/3, P(C) = 3/4 P( A ) = 1/2, P( B ) = 1/3, P( C ) = 1/4 P(none of them hits) = P( A B C ) = P( A ) P( B ) P( C ) = 1/24 P(at least one hits) = 1 P(none of them hits) = 1 (1/24) = 23/24. Problem 12: A and B alternatively throw a pair of dice. A wins if he throws 6 before B throws 7 and B wins if he throws 7 before A throws 6. If A begins, show that his chance of winning is 30/61. Solution : Let A be the event of throwing 6 B be the event of throwing 7. P(throwing 6 with 2 dice) = 5/36 P(throwing 7 with 2 dice) = 1/6 P(not throwing 6) = 31/36 P(not throwing 7) = 5/6 A plays in I, III, V,……trials. A wins if he throws 6 before Be throws 7. P(A wins) = P(A A B A A B A B A …… ) = P(A) + P( A B A) + P( A B A B A) + …… 5 31 5 5 31 5 2 5 = 36 36 6 36 36 6 36 30 = 61 Problem 13: A and B toss a fair coin alternatively with the understanding that the first who obtain the head wins. If A starts, what is his chance of winning? Page no 6 Sathyabama Institute of Science and Technology Solution : P(getting head) = 1/2 , P(not getting head) = 1/2 A plays in I, III, V,……trials. A wins if he gets head before B. P(A wins) = P(A A B A A B A B A …… ) = P(A) + P( A B A) + P( A B A B A) + …… 1 1 1 1 1 1 2 1 = 2 2 2 2 2 2 2 2 = 3 Problem 14: A problem is given to 3 students whose chances of solving it are 1/2 , 1/3 and 1/4. What is the probability that (i) only one of them solves the problem and (ii) the problem is solved. Solution : P( A solves) = 1/2 P(B) = 1/3 P(C) = 1/4 P( A ) = 1/2, P( B ) = 2/3, P( C ) = 3/4 P(none of them solves) = P( A B C ) = P( A ) P( B ) P( C ) = 1/4 P(at least one solves) = 1 P(none of them solves) = 1 (1/4) = 3/4. Baye’s Theorem Statement: If B1, B2, B3, ….Bn be a set of exhaustive and mutually exclusive events associated with a random experiment and A is another event associated with Bi, then P ( Bi ) P ( A / Bi ) P ( Bi / A) n P( Bi ) P( A / Bi ) i 1 Proof : B1 B2 B3 B4 B5 ….. Bn A The shaded region represents the event A, A can occur along with B 1, B2, B3, ….Bn that are mutually exclusive. AB1, AB2, AB3, …, ABn are also mutually exclusive. Also A = AB1 AB2 AB3 … ABn P(A) = P(AB1 ) + P(AB2 ) + P(AB3 ) + …+P(ABn) n = P( ABi ) i 1 n = P( Bi ) P( A / Bi ) (By conditional probability) i 1 Page no 7 Sathyabama Institute of Science and Technology P( Bi ) P( A / Bi ) P( Bi ) P( A / Bi ) P(Bi/A) = =. P( A ) n P( Bi ) P( A / Bi ) i 1 Problem 15: Ina bolt factory machines A, B, C manufacture respectively 25%, 35% and 40% of the total. Of their output 5%, 4% and 2% are defective bolts. A bolt is drawn at random from the produce and is found to be defective. What are the probabilities that it was manufactured by machines A, B and C. Solution : Let B1 be bolt produced by machine A B2 be bolt produced by machine B B3 be bolt produced by machine C Let A/B1 be the defective bolts drawn from machine A A/B2 be the defective bolts drawn from machine B A/B3 be the defective bolts drawn from machine C. P(B1) = 0.25 P(A/B1) = 0.05 P(B2) = 0.35 P(A/B2) = 0.04 P(B3) = 0.40 P(A/B3) = 0.02 Let B1/A be defective bolts manufactured by machine A B2/A be defective bolts manufactured by machine B B3/A be defective bolts manufactured by machine C 3 P( A) P( Bi ) P( A / Bi ) = (0.25) (0.05) + (0.35) (0.04) + (0.4) i 1 (0.02) = 0.0345 P( B1 ) P( A / B1 ) P(B1/A) = = 0.3623 P( A) P ( B2 ) P ( A / B2 ) P(B2/A) = = 0.405 P( A) P( B3 ) P( A / B3 ) P(B3/A) = = 0.231. P( A) Problem 16 : The first bag contains 3 white balls, 2 red balls and 4 black balls. Second bag contains 2 white, 3 red and 5 black balls and third bag contains 3 white, 4 red and 2 black balls. One bag is chosen at random and from it 3 balls are drawn. Out of three balls two balls are white and one is red. What are the probabilities that they were taken from first bag, second bag and third bag. Solution : Let P(selecting the bag) = P(Ai) = 1/3 , i = 1, 2, 3. 3 3C2 2C1 6 P(A/B1) = P( A) P( Bi ) P( A / Bi ) = 9C3 84 i 1 0.0746 2C2 3C1 3 P(A/B2) = 10C3 120 Page no 8 Sathyabama Institute of Science and Technology 3C2 4C1 12 P(A/B3) = 9C3 84 P( B1 ) P( A / B1 ) P(B1/A) = = 0.319 P( A) P ( B2 ) P ( A / B2 ) P(B2/A) = = 0.4285 P( A) P( B3 ) P( A / B3 ) P(B3/A) = = 0.638 P( A) Page no 9 Sathyabama Institute of Science and Technology Random Variable Random Variable: A random variable is a real valued function whose domain is the sample space of a random experiment taking values on the real line. Discrete Random Variable: A discrete random variable is one which can take only finite or countable number of values with definite probabilities associated with each one of them. Probability mass function: Let X be discrete random variable which assuming values x1 , x2 ,..., xn with each of the values, we associate a number called the probability P X xi p xi , i 1, 2,..., n this is called the probability of xi satisfying the following conditions i. pi 0 i i.e., pi ' s are all non-negative n ii. p i 1 i p1 p2 ... pn 1 i.e., the total probability is one. Continuous random variable: A continuous random variable is one which can assume every value between two specified values with a definite probability associated with each. Probability Density Function: A function f is said to be the probability density function of a continuous random variable X if it satisfies the following properties. i. f x 0; x ii. f x dx 1. Distribution Function or Cumulative Distribution Function i. Discrete Variable: A distribution function of a discrete random variable X is defined as P X x P xi . xi x ii. Continuous Variable: A distribution function of a continuous random variable X is defined x as F x P X x f x dx. Mathematical Expectation The expected value of the random variable X is defined as i. If X is discrete random variable E X xi p xi where p x is the probability i 1 function of x. 1 Page no 10 Sathyabama Institute of Science and Technology ii. If X is continuous random variable E X xf x dx where f x is the probability density function of x. Properties of Expectation: 1. If C is constant then E C C Proof: Let X be a discrete random variable then E x xp x Now E C Cp x n C p x since p i p1 p2 ... pn 1 i 1 C 2. If a, b are constants then E ax b aE x b Proof: Let X be a discrete random variable then E x xp x Now E ax b ax b p x axp x bp x n a xp x b p x since p i p1 p2 ... pn 1 i 1 aE x b 3. If a and b are constants then Var ax b a 2Var x Proof: Var ax b E ax b E ax b 2 E ax b aE x b 2 E a x E x 2 2 a E x E x 2 2 a Var x . 2 4. If a is constant then Var ax a 2Var x Proof: Var ax E ax E ax 2 E ax aE x 2 E a 2 x E x 2 a 2 E x E x 2 a Var x . 2 2 Page no 11 Sathyabama Institute of Science and Technology 5. Prove that Var x E x 2 E x 2 Proof: Var ( x) E x E x 2 E x 2 E x 2 xE x 2 E x 2 x 2 2 E x 2 E 2 E 2 x E x 2 2 2 E x E x 2 2 2 2 E x2 2 Var ( x) E x 2 E x 2 Moment Generating Function (m.g.f) A moment generating function of a random variable X (about origin) is defined e f x dx , if x is continuous tx as M X t E e tX e p x , if x is discrete tx Properties of Moment Generating Function 1. M cx t M x ct Proof: M cx t E ecxt E e x ct M x ct 2. M x c t ect M x t Proof: M x c t E e x c t E (e xt ect ) ect M x t 3. M ax b t ebt M x at Proof: M ax b t E e ax b t E eaxt ebt ebt E e x at ebt M x at 4. If X and Y are independent random variables then M x y t M x t .M y t 3 Page no 12 Sathyabama Institute of Science and Technology Proof: M x y t E e x y t E e xt e yt E e xt E e yt M x y t M x t M y t Problem.1 If the probability distribution of X is given as X : 1 2 3 4 P X : 0.4 0.3 0.2 0.1 Find P 1/ 2 X 7 / 2 X 1 Solution: P 1/ 2 X 7 / 2 X 1 P 1/ 2 X 7 / 2 / X 1 P X 1 P X 2 or 3 P X 2,3 or 4 P X 2 P X 3 P X 2 P X 3 P X 4 0.3 0.2 0.5 5 . 0.3 0.2 0.1 0.6 6 Problem.2 A random variable X has the following probability distribution X : 2 1 0 1 2 3 P X : 0.1 K 0.2 2 K 0.3 3K a) Find K , b) Evaluate P X 2 and P 2 X 2 b) Find the cdf of X and d) Evaluate the mean of X. Solution: a) Since P X 1 0.1 K 0.2 2K 0.3 3K 1 6K 0.6 1 6K 0.4 0.4 1 K 6 15 b) P X 2 P X 2, 1, 0 or 1 P X 2 P X 1 P X 0 P X 1 1 1 1 2 10 15 5 15 3 2 6 4 15 1 30 30 2 P 2 X 2 P X 1, 0 or 1 P X 1 P X 0 P X 1 4 Page no 13 Sathyabama Institute of Science and Technology 1 1 2 15 5 15 1 3 2 6 2 15 15 5 c) The distribution function of X is given by F x defined by X x P( X x) F x P( X x) -2 1 1 F x P( X 2) 10 10 -1 1 1 F x P( X 1) 15 6 0 2 11 F x P( X 0) 10 30 1 2 1 F x P( X 1) 15 2 2 3 4 F x P( X 2) 10 5 3 3 F x P X 3 1 15 d) Mean of X is defined by E X xP x 1 1 1 2 3 1 E X 2 1 0 1 2 3 10 15 5 15 10 5 1 1 2 3 3 16 . 5 15 15 5 5 15 Problem.3 A random variable X has the following probability function: X : 0 1 2 3 4 5 6 7 2 2 2 P X : 0 K 2 K 2 K 3K K 2 K 7 K K Find (i) K , (ii) Evaluate P X 6 , P X 6 and P 0 X 5 (iii). Determine the distribution function of X. (iv). P 1.5 X 4.5 X 2 (v). E 3 x 4 , Var (3x 4) 1 (vi). The smallest value of n for which P X n . 2 Solution: 7 (i) Since P X 1, x 0 K 2K 2K 3K K 2 2K 2 7K 2 K 1 10K 2 9K 1 0 1 K or K 1 10 5 Page no 14 Sathyabama Institute of Science and Technology 1 As P X cannot be negative K 10 (ii) P X 6 P X 0 P X 1 ... P X 5 1 2 2 3 1 81 ... 10 10 10 10 100 100 Now P X 6 1 P X 6 81 19 1 100 100 Now P 0 X 5 P X 1 P X 2 P X 3 P X 4 K 2K 2K 3K 8 4 8K . 10 5 (iii) The distribution of X is given by F x P X x X x P( X x) F x P( X x) 0 0 F x P ( X 0) 0 1 1 1 F x P( X 1) 10 10 2 2 3 F x P( X 2) 10 10 3 2 5 F x P( X 3) 10 10 4 3 8 F x P( X 4) 10 10 5 1 81 F x P X 5 100 100 6 2 83 F x P( X 6) 100 100 7 17 F x P ( X 7) 1 100 P x 3 P x 4 (iv) P 1.5 X 4.5 X 2 1 P x 0 P x 1 P x 2 5 5 10 3 7 1 10 (v) E x xp ( x) 1 2 2 3 1 2 17 1 2 3 4 5 6 7 10 10 10 10 100 100 100 E x 3.66 E x2 x2 p x 6 Page no 15 Sathyabama Institute of Science and Technology 1 2 2 3 1 2 17 12 22 32 42 52 62 72 10 10 10 10 100 100 100 E x 16.8 2 Mean E x 3.66 Variance E x 2 E x 2 16.8 3.66 2 3.404 1 (vi) The smallest value of n for which P X n is 4 2 Problem.4 The probability mass function of random variable X is defined as P X 0 3C 2 , P X 1 4C 10C 2 , P X 2 5C 1 , where C 0 , and P X r 0 if r 0,1, 2. Find (i). The value of C. (ii). P 0 X 2 x 0 . (iii). The distribution function of X. 1 (iv). The largest value of x for which F x . 2 Solution: x2 (i) Since p x 1 x 0 p 0 p 1 p 2 1 3C 2 4C 10C 2 5C 1 1 7C 2 9C 2 0 2 C 1, 7 C 1 is not applicable 2 C 7 The Probability distribution is X : 0 1 2 12 16 21 P X : 49 49 49 P 0 x 2 x 0 (ii) P 0 x 2 x 0 P x 0 P 0 x 2 P x 1 P x 0 P x 1 P X 2 16 16 P 0 x 2 49 x 0 16 21 37 49 49 (iii). The distribution function of X is 7 Page no 16 Sathyabama Institute of Science and Technology X F X x P X x 12 0 F 0 P X 0 0.24 49 12 16 1 F 1 P X 1 P X 0 P X 1 0.57 49 49 12 16 21 2 F 2 P X 2 P X 0 P X 1 P X 2 1 49 49 49 1 (iv) The Largest value of x for which F x P X x is 0. 2 Problem.5 x ; x 1, 2,3, 4,5 If P x 15 0 ; elsewhere Find (i) P X 1or 2 and (ii) P 1/ 2 X 5 / 2 x 1 Solution: i) P X 1 or 2 P X 1 P X 2 1 2 3 1 15 15 15 5 1 5 P X X 1 1 2 2 ii) P X / x 1 5 2 2 P X 1 P X 1or 2 X 1 P X 1 P X 2 1 P X 1 2 /15 2 /15 2 1 . 1 1/15 14 /15 14 7 Problem.6 A continuous random variable X has a probability density function f x 3x 2 , 0 x 1. Find ' a ' such that P X a P X a . Solution: 1 Since P X a P X a , each must be equal to because the probability is 2 always 1. 1 P X a 2 a 1 f x dx 0 2 a a 1 x3 1 0 a . 2 3 3 x dx 3 2 3 0 2 8 Page no 17 Sathyabama Institute of Science and Technology 1 1 3 a 2 Problem.7 Cxe x ; if x 0 A random variable X has the p.d.f f x given by f x Find the value 0 ; if x 0 of C and cumulative density function of X. Solution: Since f x dx 1 Cxe x dx 1 0 C x e x e x 1 0 C 1 xe x ; x 0 f x 0 ;x 0 Cumulative Distribution of x is x x F x f x dt xe x dx xe x e x xe x e x 1 x 0 0 0 = 1 1 x e x , x 0. Problem.8 1 x 1 ; 1 x 1 If a random variable X has the p.d.f f x 2. Find the mean and 0 ; otherwise variance of X. Solution: 1 1 1 Mean=1 xf x dx x x 1 dx x 2 x dx 1 1 1 2 1 2 1 1 1 x3 x 2 1 2 3 2 1 3 1 1 x x3 1 1 4 2 x f x dx x x dx 2 1 3 2 1 2 1 2 4 3 1 1 1 1 1 1 2 4 3 4 3 1 2 1 . 2 3 3 2 Variance 2 1 9 Page no 18 Sathyabama Institute of Science and Technology 1 1 3 1 2 = . 3 9 9 9 Problem.9 A continuous random variable X that can assume any value between X 2 and X 5 has a probability density function given by f ( x) k (1 x). Find P X 4 . Solution: k (1 x) , 2 x 5 Given X is a continuous random variable whose pdf is f x . 0 , Otherwise 5 Since f x dx 1 k (1 x)dx 1 2 5 (1 x) 2 k 1 2 2 (1 5) 2 (1 2) 2 k 1 2 2 9 k 18 1 2 27 2 k 1 k 2 27 2(1 x) ,2 x 5 f x 27 0 , Otherwise 4 2 27 2 P( X 4) (1 x) dx 4 2 (1 x)2 2 (1 4)2 (1 2) 2 2 25 9 2 16 16 . 27 2 2 27 2 2 27 2 2 27 2 27 Problem.10 2e2 x ; x 0 A random variable X has density function given by f x . Find m.g.f 0 ; x 0 Solution: M X t E e tx e f x dx e tx tx 2e 2 x dx 0 0 2 e t 2 x dx 0 e t 2 x 2 2 ,t 2. t 2 0 2t Problem.11 2 x, 0 x b The pdf of a random variable X is given by f x . For what value of b is 0, otherwise f x a valid pdf? Also find the cdf of the random variable X with the above pdf. 10 Page no 19 Sathyabama Institute of Science and Technology Solution: 2 x, 0 x b Given f x 0, otherwise b Since f x dx 1 2 x dx 1 0 b x2 2 1 2 0 b 2 0 1 b =1 2 x, 0 x 1 f x 0, otherwise x x x2 x F x P( X x) f x dx 2 xdx 2 x 2 , 0 x 1 0 0 2 0 x x F x P( X x) f x dx 0 dx 0 , x 0 0 1 x F x P( X x) f x dx f x dx f x dx 0 1 1 0 1 x2 x 0 dx 2 x dx 0 dx = 2 1 , x 1 0 1 2 0 0, x0 F x x2 , 0 x 1 1, x 1 Problem.12 K , x A random variable X has density function f x 1 x 2. Determine K 0 , Otherwise and the distribution functions. Evaluate the probability P x 0 . Solution: Since f x dx 1 K 1 x 2 dx 1 dx K 1 1 x2 K tan 1 x 1 K 1 2 2 K 1 11 Page no 20 Sathyabama Institute of Science and Technology 1 K x x K F x f x dx 1 x dx 2 1 1 tan x 2 1 F x tan 1 x , x 2 1 1 tan x dx P X 0 1 x 1 0 2 0 1 1 1 tan 0 . 2 2 Problem.13 Ke3 x , x 0 If X has the probability density function f x find K , 0 , otherwise P 0.5 X 1 and the mean of X. Solution: Since f x dx 1 Ke 3 x dx 1 0 e3 x K 1 3 0 K 1 3 K 3 1 1 e3 e1.5 P 0.5 X 1 f x dx 3 e 3 x dx 3 e e 1.5 3 0.5 0.5 3 Mean of X E x xf x dx 3 xe 3 x dx 0 0 e3 x e3 x 3 1 1 3x 1 3 9 0 9 3 1 Hence the mean of X E X . 3 Problem.14 If X is a continuous random variable with pdf given by 12 Page no 21 Sathyabama Institute of Science and Technology Kx in 0 x 2 2 K in 2 x 4 f x . Find the value of K and also the cdf F x . 6 K Kx in 4 x 6 0 elsewhere Solution: Since F x dx 1 2 4 6 Kxdx 2Kdx 6k kx dx 1 0 2 4 x 2 2 6 x2 6 K 2 x 2 6 x 1 4 2 0 4 2 4 K 2 8 4 36 18 24 8 1 8K 1 1 K 8 x We know that F x f x dx x If x 0 , then F x f x dx 0 x If x 0, 2 , then F x f x dx 0 x F x f x dx f x dx 0 0 x 0 x 1 0dx Kxdx 0dx xdx 0 80 x x2 x2 F x , 0 x 2 16 0 16 0 2 x If x 2, 4 , then F x f x dx f x dx f x dx 0 2 0 2 x 0dx Kxdx 2Kdx 0 2 2 x2 x 2 x x x 1 dx dx 0 8 2 4 16 0 4 2 1 x 1 4 4 2 x 4 x 1 F x ,2 x4 4 16 4 13 Page no 22 Sathyabama Institute of Science and Technology 0 2 4 x If x 4, 6 , then F x 0dx Kxdx 2Kdx K 6 x dx 0 2 4 2 4 x x 1 1 dx dx 6 x dx 0 8 2 4 4 8 2 x x2 x 6x x2 4 16 0 4 2 8 16 4 1 1 6 x x2 1 3 1 4 2 8 16 4 16 8 12 x x 2 48 16 16 x 12 x 20 2 F x ,4 x 6 16 0 2 4 6 If x 6 , then F x 0dx Kxdx 2Kdx K 6 x dx 0dx 0 2 4 6 F x 1 , x 6 0 ;x0 2 x ;0 x 2 16 1 F x x 1 ;2 x4 4 1 16 20 12 x x ; 4 x 6 2 1 ;x 6 Problem.15 2 x , 0 x 1 A random variable X has the P.d.f f x 0 , Otherwise 1 1 1 3 1 Find (i) P X (ii) P x (iii) P X / X 2 4 2 4 2 Solution: 1/ 2 1 1/ 2 1/ 2 x2 2 1 1 (i) P x 2 f x dx 0 0 2 xdx 2 2 0 8 4 1/ 2 1 1 1/ 2 x2 1/ 2 (ii) P x f x dx 2 xdx 2 4 2 1/ 4 1/ 4 2 1/ 4 1 1 1 1 3 2 . 8 32 4 16 16 3 1 3 P X X P X 1 (iii) P X / X 2 3 4 4 4 2 1 1 P X P X 2 2 14 Page no 23 Sathyabama Institute of Science and Technology 1 3 1 1 x2 9 7 P X f x dx 2 xdx 2 1 4 3/ 4 3/ 4 2 3/ 4 16 16 1 1 1 1 x2 1 3 P X f x dx 2 xdx 2 1 2 1/ 2 1/ 2 2 1/ 2 4 4 7 3 1 7 4 7 P X / X 16 . 4 2 3 16 3 12 4 Problem.16 1 2x e ,x 0 Let the random variable X have the p.d.f f x 2.Find the moment 0 , otherwise. generating function, mean & variance of X. Solution: M X t E etx etx f x dx etx e x / 2 dx 1 0 2 1 t x 1 e 2 1 t x 1 1 1 e 2 dx , if t . 20 2 1 1 2t 2 2 t 0 d 2 E X M X t 2 2 dt t 0 1 2t t 0 d2 8 E X 2 2 M X t 3 8 dt t 0 1 2t t 0 Var X E X 2 E X 8 4 4. 2 Problem.17 The first four moments of a distribution about x 4 are 1,4,10 and 45 respectively. Show that the mean is 5, variance is 3, 3 0 and 4 26. Solution: Given 1 1, 2 4, 3 10, 4 45 r r th moment about to value x 4 Here A 4 Here Mean A 1 4 1 5 2 Variance 2 2 1 4 1 3. 3 3 3 321 2 1 10 3 4 1 2 1 0 3 15 Page no 24
