Content text XII - maths - chapter 11 - PROBABILITY( 62-92).pdf
62 PROBABILITY NARAYANAGROUP JEE-MAIN SR.MATHS-VOL-II Random Experiment or Trial (R.E) : A repeatable experiment the results of which are known in advance but exact result will be known only after the experiment is completed is known as random experiment or an experiment, result of which can not be predicted with certainity is called random experiment. a) Tossing a coin b) Rolling a die Drawing a specified number of cards from a well shuffled pack of 52 playing cards etc., Sample Space (S) : The list of all possible outcomes or results of a random experiment is called as sample space for that experiment and it is denoted by S. Ex: i) If a coin is tossed once, the sample space S is given by S = {H,T} ii) If 2 coins are tossed once, the sample space S={HH,HT,TH,TT} iii) If a die is rolled once the sample space S={1,2,3,4,5,6} Elementary event or Indecomposable event: The possible outcome or the result of a random experiment that can not be split further is called an elementary or simple event. Event: An outcome or the result of a random experiment is called an event. Ex: While tossing a coin we may get head or tail upwards. Now getting head upwards is an event or getting tail upwards is also an event. Ex: The event of getting head upwards is an elementary event. Composite event : The possible outcome or the result of a random experiment that can be further split into more than one elementary event is called a composite event. The event can be either an elementary event or a composite event. Ex: Getting a face with odd number of points upwards in rolling of a die is a composite event, since it can be divided further into three elementary events namely: Mutually exclusive event (M.E.E.): The events of a R.E. are said to be M.E. events, if the occurence of any one of them prevents (or) eliminates the occurence of all the other events of that R.E. Two events A and B are said to be M.E. or disjoint when both cannot happen simultaneously in a single trial or experiment. i.e., A B . Ex: In tossing a coin head turning upwards eliminates tail turning upwards in that particular R.E. So getting head & tail are M.E. events. The events 1 2 3 , , ,....., A A A An of a R.E. are said to be M.E. or disjoint events if, , A A i j for i j & 1 , i j n . Note : Whenever it is possible for two or more number of events to happen simultaneously, then those events are said to be compatible. Equally likely events (E.L.E.): The events of a R.E. are said to be E.L. When one does not occur more number of times than any one of the other events i.e., if all the events occur same number of times. The events of a R.E. are said to be equally likely when there is no reason to expect a particular event in preference to any other event. Ex: If an unbiased coin is tossed a large number of times, each face i.e., head and tail can be expected to appear same number of times. Exhaustive events (E.E.): The events of a random experiment are said to be exhaustive, if the occurence of any one of them is a certainity. A set of events is said to be exhaustive, if these include all possible outcomes of the R.E. Ex: In tossing a coin it is certain that either the head or the tail will turn upwards. So head & tail put together are called as exhaustive events. Complementary Events : Suppose A is any event of a random experiment associated with sample space S. The complementary event of an event A denoted by C Aor A is the event given by C A S A Ex: If two coins are tossed, S={HH,HT,TH,TT} PROBABILITY SYNOPSIS
63 PROBABILITY NARAYANAGROUP JEE-MAIN SR.MATHS-VOL-II Let A: occurence of atleast one head Here C A is the event of non - occurrence of atleast one head i.e, C A is the event of getting both tails. Here A={HH,HT,TH}, C A ={TT} Note : Let S be a sample space. The event is called impossible event and the event S is called certain event in S. Favourable Cases: The possible outcomes or results of a R.E. which leads to the occurence of the event A are called as favourable cases for the event A to happen. Ex: In rolling a symmetrical die 1, 3 & 5 are happening for the event A, the event of getting odd number of points upwards. So they are called as favourable cases for the event A. Classical Definition of Probability (or) Mathematical Definition of Probability (or) A priori Probability: The probability of the event A, denoted by P(A) is defined as m n A P A n n S . Where n = The total number of possible outcomes in a R.E. which are M.E., E.L. & collectively exhaustive, m = The number of possible outcomes favourable to A in that R.E., n(S) = The total number of sample points in a sample space S, n(A) = The number of sample points favourable to A in that sample space S. p(A) = m n can also be expressed as "The odds in favour of the event A are m to n - m or m n m " or "The odds against the event A are n-m to m or n m m ". But in both the cases favourable cases total no. of cases P A m m m n m n Now consider m P A n If m = 0, P(A) = 0, then A is said to be an impossible event. If m=n, 1 n P A n , then A is said to be a sure or certain event. The range of P(A) is [0, 1] i.e., 0 1 P A From these we can conclude that the probability of any event is a non-negative rational number which lies between 0 & 1, both inclusive. W.E.1:- A coin is tossed 2 times the probability of getting one head and one tail is Sol: Sample space, S = {HH, HT, TH, TT}; n(S) = 4 Let A be the event of getting 1 head and 1 tail A = {HT, TH} n(A) = 2 ( ) 2 1 ( ) ( ) 4 2 n A p A n S W.E.2:-Two dice are thrown simultaneously. The Probability of getting even numbers on both the dice is Sol: n(S) = 2 6 36 . Let A be the event of getting even numbers on both the diceA = { (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)} ; n(A) = 9 ( ) 9 1 ( ) 36 4 n A p A n S A non-leap year = 52 weeks + 1 day A leap - year = 52 weeks + 2 consecutive days of a week. WE.3:- The probability that a non-leap year will have 53 Fridays is Sol: n(A)=1;n(S)=7 ; p(A)=n(A)/n(S)=1/7 W.E.4:- The probability that a leap year will have only 52 Sundays is Sol: n(A)=5;n(S)=7 ; p(A)=n(A)/n(S)=5/7 Description of Pack of Cards: Pack of cards 52 (13) Spade Club (13) Heart (13) Diamond Black Cards Colours Red Cards (13) Here clubs, spades, diamonds, hearts are four suits.
64 PROBABILITY NARAYANAGROUP JEE-MAIN SR.MATHS-VOL-II Each suit contains 13 cards namely 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A. In every suit J, Q, K are face cards (or court cards) In every suit J, Q, K, A are called honour cards. Jack card (J) is also called knave. The cards bearing the digits 2,3,4,5,6,7,8,9,10 are known as number cards. Ace is treated as 1. Each suit contains 9 numbered cards. In a pack of cards the number of cards with same face value is 4. Odds in favour of A : It is defined as P(A) : P(A) Odds against A : It is defined as P(A): P(A) If P(A) : P(A) = x : y then P(A) = x x y and y P(A) x y Limitations: If n is very very large, we can not determine the probability of the event A with the help of this definition. The probability can not be found when the outcomes are not equally likely. When the outcomes are not M.E., our logic may go wrong. One of the serious drawbacks of this definition is that in defining probability we use the term equally likely i.e., equally probable i.e., probability itself. All the identical trials are to be conducted independently. Again to define the concept of independence, we require the concept of probability. Axiomatic approach to Probability : Let S be a finite sample space associated with a random experiment. Then a real valued function P from the power set of S i.e., P(S) into the real line R is called a probability function on S, if P satisfies the following three axioms. Axiom of Positivity: For every subset A of S, P A 0 i.e., P A A S 0, Axiom of Certainty : P(S) = 1 Axiom of Union or Axiom of Additivity: If 1 2 A A, are two disjoint subsets of S then P A A P A P A 1 2 1 2 . Now the image P(A) of A is called the probability of the event A. Note: In general, if 1 2 3 , , ,....., A A A An are ‘n’ disjoint subsets of S then 1 2 3 ...... P A A A An P A P A P A P A 1 2 3 .... n i.e., 1 1 n n i i i i P A P A U If S consists of ‘n’ equally likely elementary events and Ai is one such elementary event of S, then 1 ; 1, 2,3,...... P A i n i n . Notation :Let A and B be two events in a sample space S. or B does not happen happen Addition Theorem : If A and B are any two compatible events in a sample space S, then the probability of occurence of either the event A or B or both the events A & B i.e., the event of A B is given by P A B P A P B P A B .......... 1 If A and B are M.E. events, P A B 0 , then P A B P A P B .......... 2 From equations (1) & (2) we can conclude that P A B P A P B Further A B A B , P A or P B P A B P A P B
65 PROBABILITY NARAYANAGROUP JEE-MAIN SR.MATHS-VOL-II W.E.6:- In a simulataneous throw of two dice, the porbability of A or B, if A=a sum of 11 points; B=an odd number of points on each die Sol : P (sum 11 or an odd no of points on each die) 2 3 3 11 36 6 6 36 Conditional Event: If A,B are two events in a sample space then the event of happening of B after the happening of event A is called conditional event . It is denoted by B/A Conditional Probability: If A & B are two events in a sample space S such that P A 0 , then the probability of occurence of B, after the occurence of the event A is called the conditional probability of B given A, denoted by either P B A / or B P A or P B A ; and is defined as / P A B n A B P B A P A n A Similarly if P B 0, then / P A B n A B P A B P B n B In general P B A P A B / / W.E.7:- One ticket is selected at random from 50 tickets numbered 00,01,02,.....,49. Then the probability that the sum of the digits on the ticket is 8, given that the product of the digits is zero is (AIEEE 2009) Sol: E1 Event that the product of the digits is zero. E2 Event that the sum of the digits is 8. Favourable cases to E1 00, 01, 02,03,....., 09,10, 20,30, 40 E2 08,17, 26,35, 44 , E E 1 2 08 n E 1 14 , n E 2 5 , n E E 1 2 1 Required probability = 2 1 2 1 1 1 14 E n E E P E n E If A and B are two events such that A B, then P(A) P(B) If A, B are two events of S then P(A B) P(A) P(A B) P(A) P(B) P(exactly one of A, B occurs) = P A B P A B = P A P B P A B 2 P A B P A B P A B P A B If A , B are two events then i) P A B P A B 1 ii) P A B P A B 1 A B B A B & A B A A B For three events A, B and C P A B C P A P B P C P A B P A C P B C P A B C If A, B and C are M.E. events, the probabilities of compound events will be zero, then P A B C P A P B P C If A, B and C are three events then P(exactly one of A, B, C occurs) = P A P B P C P A B 2 2 2 3 P A C P B C P A B C P(exactly two of A, B, C occur) = P A B P A C P B C P A B C 3 P(at least two of A, B, C occurs) = P A B P A C P B C P A B C 2 If A,B and C are mutually exclusive and exhaustive events of the sample space S, then A B C S and P(A)+P(B)+P(C)=P(S)=1 W.E.5:- A die thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then P A B is (AIEEE 2008) Sol: P A B P A P B P A B = 3 4 1 6 1 6 6 6 6