Tiêu đề 13. PROBABILITY.pdf ✅

TOPIC-1 Conditional Probability and Multiplication Theorem on Probability Revision Notes 1. Basic Definition of Probability: Let S and E be the sample space and event in an experiment respectively. Then, Probability = Number of Favourable Events Total number of Elementary Events = n E n S ( ) ( ) 0 £ n(E) £ n(S) 0 £ P(E) £ 1 Hence, if P(E) denotes the probability of occurrence of an event E, then 0 £ P(E) £ 1 and P E( ) = 1 – P(E) such that P E( ) denotes the probability of non-occurrence of the event E. z Note that P E( ) can also be represented as P(E’). 2. Mutually Exclusive Or Disjoint Events: Two events A and B are said to be mutually exclusive if occurrence of one prevents the occurrence of the other i.e., they can’t occur simultaneously. In this case, sets A and B are disjoint i.e., A Ç B = f. 3. Independent Events: Two events are independent if the occurrence of one does not affect the occurrence of the other. 4. Exhaustive Events: Two or more events say A, B and C of an experiment are said to be exhaustive events, if (a) their union is the total sample space i.e., A È B È C = S (b) the event A, B and C are disjoint in pairs i.e., A Ç B = f, B Ç C = f and C Ç A = f. (c)P(A) + P(B) + P(C) = 1. z If A and B are mutually exclusive events, then we always have P(A Ç B) = f [As n(A Ç B) = n(f) = 0] \ P(A È B) = P(A) + P(B). z If A, B and C are mutually exclusive events, then we always have P(A È B È C) = P(A) + P(B) + P(C) MNEMONICS Concept: Independent and Mutually exclusive events. I Is not ME ME Is not I Here, I: Independent Events ME: Mutually Exclusive events 5. Conditional Probability: By the conditional probability, we mean the probability of occurrence of event A when B has already occurred. The ‘conditional probability of occurrence of event A when B has already occurred’ is sometimes also called as probability of occurrence of event A w.r.t. B. P(A|B) = P A B P B ( ) ( ) ∩ , B 1 f i.e., P(B) 1 0 P(B|A) = P A B P A ( ) ( ) ∩ , A 1 f i.e., P(A) 1 0 P A( |B) = P A B P B ( ) ( ) ∩ , P(B) 1 0 P A( |B) = P A B P B ( ) ( ) ∩ , P B( ) 1 0 P A( |B) = P A B P B ( ) ( ) ∩ , P B( ) 1 0 z P A( |B P ) ( + = A B| ) 1, . B ≠ φ [Board 2024,23,22,20] PROBABILITY LEARNING OUTCOMES After going through this Chapter, the student would be able to learn:  Conditional Probability  Multiplication Theorem of Probability  Bayes' Theorem  Probability Distribution  Mean of Random Variable. 13 CHAPTER LIST OF TOPICS Topic-1:Conditional Probability and Multipli- cation Theorem on Probability Topic-2:Bayes’ Theorem Topic-3:Random Variable and its Probability Distributions
Probability Example-1 A family has two children. What is the probability that both the children are boys given that at least one of them is a boy ? Sol. Let g and b, respectively denote a girl and a boy. So, sample space of the experiment is S = {(b, b), (g, b), (b, g), (g, g)} \ n(S) =4 Let events E = Both the children are boys F = At least one of the child is a boy Then E = {(b, b)} \ n(E) =1 and F = {(b, b), (g, b), (b, g)} \ n(F) =3 Also, E ∩ F = {(b, b)} \ n(E ∩ F) =1 Thus, P(F) = n F n S ( ) ( ) = 3 4 and P(E ∩ F) = n E F n S ( ) ∩ ( ) = 1 4 Therefore, P(E | F) = P E F P F ( ) ∩ ( ) = = 1 4 3 4 1 3 Example-2 Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn at random. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number? Sol. Let events A= Number on the card drawn is even B= Number on the card drawn is greater than 3 Sample space of the experiment is S= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} \ n(S) = 10 Then A = {2, 4, 6, 8, 10}, B = {4, 5, 6, 7, 8, 9, 10} and A ∩ B = {4, 6, 8, 10} \ n(A) = 5, n(B) = 7 and n(A ∩ B) = 4 Thus, P(A) = 5 10 , p(B) = 7 10 and p(A ∩ B) = 4 10 Now, P(A | B) = P A B P B ( ) ∩ ( ) = = 4 10 7 10 4 7 NOTE : EVENTS AND SYMBOLIC REPRESENTATIONS: Verbal description of the event Equivalent set notation Event A A Not A A or A' A or B (occurrence of atleast one A or B) A ∪ B or A + B A and B (simultaneous occurrence of both A and B) A ∩ B or AB A but not B (A occurs but B does not) A ∩ B or A – B Neither A nor B A ∩ B Atleast one A, B or C A ∪ B ∪ C All the three A, B and C A ∩ B ∩ C KEY FORMULAE (a) P A( ) ∪ = B P( ) A P + − ( ) B P( ) A B ∩ = i.e., ( P A or B P ) (A P ) () ( + − B P A B and ) (b) P A( ) ∪ ∪B C = + P A( ) P B( ) + − P C( ) P A( ) ∩ − B P( ) B C ∩ − P C( ) ∩ + A P( ) A B ∩ ∩ C (c) P A( ) ∩ = BP BP ( ) only = − ( ) B A = = P B( ) but not A P( ) B P − ∩ ( ) A B (d) P A( ) ∩ = B P( ) only A P = − ( ) A B = = P A( ) but not B P( ) A P − ∩ ( ) A B (e) P A( ) ∩ = B P( ) neither n A B or = −1 P A( ) ∪ B KEY-TERMS Sample Space: A set in which all of the possible outcomes of a statistical experiment are represented as points. Event: Event is a subset of a sample space. e.g.,: Event of getting odd outcome in a throw of a die. KEY-FACTS z Probability is originated from a gambler’s dispute in 1654 concerning the division of a stake between two players whose game was interrupted before it close. z The probability of living 110 years or more is about 1 in 7 million. z If you are in the group of 23 people, there is a 50% chance that 2 of them share a birthday. If you are in a group of 70 people, that probability jumps to over 99%.
MATHEMATICS, Class-XII Very Short Answer Type Questions (1 mark each) 1. Two cards are drawn at random and one by one without replacement from a well shuffled pack of 52 playing cards, find the probability that one card is red and the other is black. U [Delhi Set-2, 2020-21] Concept Applied Multiplication theorem probability Topper's Answer, 2020 Sol. 2. The probabilities of A and B solving a problem independently are 1 3 and 1 4 respectively. If both of them try to solve the problem independently, what is the probability that the problem is solved? R [SQP 2020-21] 3. The probability that it will rain on any particular day is 50%. Find the probability that it rains only on first 4 days of the week. R&U [SQP 2020-21] Sol. P(rain on any particular day) = 50% = 50 100 = 1 2 P(rain on first four days of week) = 1 2 1 1 2 4 3       −       = 1 2 1 2 4 3             = 1 2 7       = 1 128 [Marking Scheme SQP 2020-21] Commonly Made Error Some students forget to take the probability for not raining on the remaining three days and give the answer as 1 16 . Read the question twice to sort out the hidden elements in it. Answering Tip Short Answer Type Questions-I (2 marks each) 1. Two cards are drawn at random from a pack of 52 cards one-by-one without replacement. What is the probability of getting first card red and second card Jack? U [SQP 2022-23] Sol. The required probability = P[(The first is a red jack card and the second is a jack card) or (The first is a red non-jack card and the second is a jack card)]1 = 2 52 3 51 24 52 4 51 . . + = 1 26 1 [Marking Scheme SQP, 2022-23] Commonly Made Error Some students forget to take the probability (The first is a red non-jack card and the second is a jack card) and solve only getting first card red and second card Jack Read the question twice to sort out the hidden elements in it. Most of the time this type of question follow two condition. Answering Tip 2. Events A and B are such that 1 7 1 ( ) , ( ) and ( ) 2 12 4 PA PB = = PA B ∪ = . Find whether the events A and B are independent or not. R [Delhi Set-1, 2022] Sol. Given P(A) = 1 2 , P(B) = 7 12 and P(A È B) = 1 4 For A and B are independent P(A Ç B) = P(A).P(B) ...(i) Now, P(A È B) = P A( ) ∩ B Þ P(A È B) = 1 – P(A Ç B) Þ P(A Ç B) = 1 – P(A È B) Þ P(A Ç B) = 1 1 4 − = 3 4 ...(ii) 1 Now, P(A).P(B) = 1 2 7 12 × = 7 24 ...(iii) Since, from eqs. (ii) & (iii) P(A Ç B) 1 P(A).P(B) Therefore, events A and B are not independent. 1 [Marking Scheme Delhi Set-1, 2022] 3. A box B1 contains 1 white ball and 3 red balls. Another box B2 contains 2 white balls and 3 red balls. If one ball is drawn at random from each of the boxes B1 and B2, then find the probability that the two balls drawn are of the same colour. U [Delhi Set-1, 2022] Sol. Box B1 1 White Balls 3 Red Balls Box B2 2 White Balls 3 Red Balls \ P(Required) = P(Both are white) + P(Both are red) = 1 4 2 5 3 4 3 5 ×+× = 2 20 9 20 + = 11 20 [Marking Scheme Delhi Set-1, 2022] This Question is for practice and its solution is given at the end of the chapter. SUBJECTIVE TYPE QUESTIONS