{"id":21709,"date":"2020-01-28T14:18:18","date_gmt":"2020-01-28T08:48:18","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=21709"},"modified":"2021-07-02T15:26:27","modified_gmt":"2021-07-02T09:56:27","slug":"2nd-puc-maths-question-bank-chapter-13-ex-13-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-13-ex-13-2\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 13 Probability Ex 13.2"},"content":{"rendered":"

Students can Download Maths Chapter 13 Probability Ex 13.2 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Maths Question Bank Chapter 13 Probability Ex 13.2<\/h2>\n

2nd PUC Maths Probability NCERT Text Book Questions and Answers Ex 13.2<\/h3>\n

Question 1.
\nIf \\(\\mathbf{P}(\\mathbf{A})=\\frac{3}{5} \\text { and } \\mathbf{P}(\\mathbf{B})=\\frac{1}{5}, \\text { find } \\mathbf{P}(\\mathbf{A} \\cap \\mathbf{B})\\) ,find P(A\u2229B) if A and B are independent events.
\nAnswer:
\nIf two events A and B are independents P (A\u2229B) = P(A). P(B)
\n\\((A \\cap B)=\\frac{3}{5} \\times \\frac{1}{5}=\\frac{3}{25}\\)<\/p>\n

Question 2.
\nTwo cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
\nAnswer:
\nThere are 26 black cards out 52 cards. When cards are drawn without replacement,
\n\\(P(\\text { both black cards })=\\frac{26 \\mathrm{C}_{2}}{52 \\mathrm{C}_{2}}=\\frac{26 \\times 25}{52 \\times 51}=\\frac{25}{102}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nA box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
\nAnswer:
\nSince selection is without replacement
\n\"2nd<\/p>\n

Question 4.
\nA fair coin and an unbiased die are tossed.Let A be the event \u2018head appears on the coin\u2019 and B be the event \u20183 on the die\u2019.Check whether A and B are independent events or not.
\nAnswer:
\n\"2nd<\/p>\n

Question 5.
\nA die marked 1, 2, 3 in red and 4, 5, 6 in green is tossed. Let A be the event, \u2018the number is event, and B be the event, \u2018the number is red\u2019. Are A and B independent?
\nAnswer:
\nA : number is even, B : The number is red
\n\"2nd<\/p>\n

Question 6.
\nLet E and F be events with P(E) \\({ P }({ E })=\\frac { 3 }{ 5 } ,{ P }({ F })=\\frac { 3 }{ 10 } { \\quad and\\quad }{ P }({ E }\\cap { F })=\\frac { 1 }{ 5 } \\). Are E and F independent?
\nAnswer:
\n\"2nd
\nhence they are not independent.<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nGiven that the events A and B are such that \\(P(A)=\\frac { 1 }{ 2 } ,P(A\\cup B)=\\frac { 3 }{ 5 } \\quad { and }\\quad P(B)=p\\) Find p if they are
\n(i) mutually exclusive
\n(ii) independent.
\nAnswer:
\n(i) If the events are mutually exclusive
\n\"2nd<\/p>\n

Question 8.
\nLet A and B be independent events with
\nP(A) = 0.3 and P (B) = 0.4. Find
\n(i) P(A\u2229B)
\n(ii)P(A\u222aB)
\n(iii) P (A|B)
\n(iv) P (B|A)
\nAnswer:
\n\"2nd<\/p>\n

Question 9.
\nIf A and B are two events such that
\n\\(P(A)=\\frac { 1 }{ 4 } ,{ P }({ B })=\\frac { 1 }{ 2 } { and }{ P }({ A }\\cap { B })=\\frac { 1 }{ 8 } \\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nEvents A and B are such that \\({ P }({ E })=\\frac { 3 }{ 5 } ,{ P }({ F })(B)=\\frac { 7 }{ 12 } \\) P (not A or not B) = \\(\\frac{1}{4}\\). State whether A and B are independent ?
\nAnswer:
\n\"2nd<\/p>\n

Question 11.
\nGiven two independent events A and B such that P (A) = 0.3, P (B) = 0.6. Find
\n(i) P (A and B)
\nAnswer:
\ni.e. P (A\u2229B) = P (A) x P (B) = .3 x .6 = .18<\/p>\n

(ii) P (A and not B)
\nAnswer:
\nP(A and not B) = \\(P(A \\cap \\bar{B})\\)
\n= P(A) x \\(P(\\bar{B})\\)\u00a0\u21d2 P (A) x [1-P(B)]
\n= .3 x (1 – 0.6) = .3 x .4 = .12<\/p>\n

(iii) P (A or B)
\nAnswer:
\nP (A or B) = P (A\u222aB)
\n= P (A) + P (B) – P (A\u2229B)
\n= .3+ .6-.18 = .9 -.18 = .72<\/p>\n

(iv) P (neither A nor B)
\nAnswer:
\nP (neither A nor B) = \\(P(\\bar{A} \\cap \\bar{B})\\)
\n= \\(P(\\overline{A \\cup B})\\)
\n= 1 -P(A\u222aB)= 1 -.72 = .28<\/p>\n

Question 12.
\nA die is tossed thrice. Find the probability of getting an odd number at least once.
\nAnswer:
\nE : an odd number at least once
\n\\(\\overline{\\mathrm{E}}\\) : not an odd number
\n\\(\\overline{\\mathrm{E}}\\) : an even number on all times
\nwe get even all time 3 x 3 x 3 = 27
\n\\(P(E)=1-P(\\bar{E})=1-\\frac{27}{216}=\\frac{7}{8}\\)<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nTwo balls are drawn at random With replacement from a box containing 10 black and 8 red balls. Find the probability that
\n(i) both balls are red.
\n(ii) first ball is black and second is red
\n(iii) one of them is black and other is red
\nAnswer:
\n\"2nd<\/p>\n

Question 14.
\nProbability of solving specific problem independently by A and B are \\(\\frac{1}{2} \\text { and } \\frac{1}{3}\\) respectively. If both try to solve the problem independently, find the probability that
\n(i) the problem is solved
\n(ii) exactly one of them solves the problem.
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

Question 15.
\nOne card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent ?
\n(i) E : \u2018the card drawn is a spade\u2019
\nF : \u2018the card drawn is a ace\u2019
\n(ii) E : \u2018the card drawn is black\u2019
\nF : \u2018the card drawn is a king\u2019
\n(iii) E : \u2018the card drawn is a king or queen\u2019
\nF : \u2018the card drawn is a queen or jack\u2019.
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 16.
\nIn a hostel, 60% of the students read Hindi news paper, 40% read English news paper and 20% read both Hindi^and English news papers. A student is selected at random.
\n(a) Find the probability that she reads neither Hindi nor English news papers.
\n(b) If she reads Hindi news paper, find the probability that she reads English news paper.
\n(c) If she reads English news paper, find the probability that she reads Hindi news paper.
\nAnswer:
\n(a) 20% read neither Hindi nor english
\n\"2nd<\/p>\n

Choose the correct answer in Exercises 17 and 18.<\/p>\n

Question 17.
\nThe probability of obtaining an even prime number on each die, when a pair of dice is rolled is
\n(A) 0
\n(B) \\(\\frac{1}{3}\\)
\n(C) \\(\\frac{1}{12}\\)
\n(D) \\(\\frac{1}{36}\\)
\nAnswer:
\nEven Prime no is (2,2)
\nP (even prime no) = \\(\\frac{1}{36}\\)
\nOption \u2018D\u2019<\/p>\n

Question 18.
\nTwo events A and B will be independent, if
\n(A) A and B are mutually exclusive
\n(B) P (A’B’) = [1 – P(A)] [1 – P(B)]
\n(C) P (A) = P (B)
\n(D) P (A) + P (B) = 1
\nAnswer:
\nP(A’B’) = P(A’\u2229B’)
\n= \\(P\\bar { A\\cup B } \\)
\n= 1 – P (A\u222aB)
\n= 1 – P(A) – P(B) – P(A) – P(B)
\n= 1 – P (A) – P (B) (1 – P (A)]
\n= [1 – P (A)] [1 -P(B)
\nOption \u2018B\u2019<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 13 Probability Ex 13.2 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Maths Question Bank Chapter 13 Probability Ex 13.2 2nd PUC Maths Probability NCERT Text Book …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/21709"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=21709"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/21709\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=21709"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=21709"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=21709"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}