{"id":32678,"date":"2021-08-30T16:27:06","date_gmt":"2021-08-30T10:57:06","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=32678"},"modified":"2021-08-30T16:27:06","modified_gmt":"2021-08-30T10:57:06","slug":"ncert-solutions-for-class-12-maths-chapter-13-ex-13-1","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-13-ex-13-1\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 13 Probability Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.1<\/h2>\n

\"NCERT<\/p>\n

Question 1
\nGiven that E and Fare events such that P (E) = 0.6, P (F) = 0.3 and P(E\u2229F) = 0.2 find P(E|F) and P (F|E).
\nSolution:
\nGiven: P (E)=0.6, P (F)=0.3, P (E \u2229 F)=0.2
\nP(E|F) = \\(\\frac { P(E\\cap F) }{ P(E) } =\\frac { 0.2 }{ 0.3 } =\\frac { 2 }{ 3 }\\)
\nP(F|E) = \\(\\frac { P(E\\cap F) }{ P(E) } =\\frac { 0.2 }{ 0.6 } =\\frac { 1 }{ 3 }\\)<\/p>\n

Question 2
\nCompute P(A|B) if P(B)=0.5 and P (A\u2229B) = 0.32.
\nSolution:
\nGiven: P(B) = 0.5, P(A\u2229B) = 0.32
\nP(A|B) = \\(\\frac { P(A\\cap B) }{ P(B) } =\\frac { 0.32 }{ 0.50 } =\\frac { 32 }{ 50 }\\) = 0.64<\/p>\n

Question 3.
\nIf P (A) = 0.8, P (B)=0.5 and P(B\/A)=0.4, find
\n(i) P(A\u2229B)
\n(ii) P(A\/B)
\n(iii)P(A\u222aB)
\nSolution:
\n(i) P(B\/A) = \\(\\frac { P(A\\cap B) }{ P(A) } \\Rightarrow 0.4=\\frac { P(A\\cap B) }{ 0.8 }\\)
\n\u2234 P(A\u2229B) = 0.4 x 0.8 = 0.32<\/p>\n

(ii) P(A\/B) = \\(\\frac { P(A\\cap B) }{ P(B) } =\\frac { 0.32 }{ 0.5 } =\\frac { 32 }{ 50 } =\\frac { 16 }{ 25 }\\)<\/p>\n

(iii) P(A\u222aB) = P(A) + P(B) – P(A\u2229B)
\n= 0.8 + 0.5 – 0.32 = 1.30 – 0.32 = 0.98<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nEvaluate P(A\u222aB) if 2P(A) = P(B) = \\(\\frac { 5 }{ 13 }\\) and P(A|B) = \\(\\frac { 2 }{ 5 }\\).
\nSolution:
\n\"NCERT<\/p>\n

Question 5.
\nIf P(A) = \\(\\frac { 6 }{ 11 }\\),P(B) =\\(\\frac { 5 }{ 11 }\\) and P(A\u222aB) = \\(\\frac { 7 }{ 11 }\\),
\nFind
\n(i) P(A\u2229B)
\n(ii) P(A|B)
\n(iii) P(B|A)
\nSolution:
\n\"NCERT<\/p>\n

Question 6.
\nA coin is tossed three times, where
\n(i) E: head on third toss F: heads on first two tosses.
\n(ii) E: at least two heads F : at most two heads
\n(iii) E: at most two tails F: at least one tail
\nSolution:
\nThe sample space S = {HHH, HHT, THH, TTH, THH, THT, TTT}
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nTwo coins are tossed once
\n(i) E: tail appears on one coin F: one coin shows head
\n(ii) E: no tail appears F: no head appears.
\nSolution:
\nThe Sample space S = {HH, TH, HT, TT}
\n\"NCERT<\/p>\n

Question 8.
\nA die is thrown three times.
\nE: 4 appears on the third toss
\nF: 6 and 5 appears respectively on first two tosses.
\nSolution:
\n\"NCERT<\/p>\n

Question 9.
\nMother, father and son line up at random for a family picture:
\nE: son on one end, F: father in middle
\nSolution:
\nMother (m), Father (f) and son (s) line up at random E : son on one end: {(s, m, f), (s, f, m), (f, m, s), (m, f, s)}
\nF: Father in middle: {(m, f,s), (s, f, m), (s, f, m)}
\nE\u2229F = {(m, f, s), (s, f, m)}
\n\"NCERT<\/p>\n

Question 10.
\nA Mack and a red die are rolled.
\n(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
\n(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
\nSolution:
\nLet S be the sample space. Then
\nn(S) = 36
\na. E : event of getting a sum greater than 9
\nE = {(4, 6), (5, 5), (5, 6), (6,4), (6, 5), (6, 6)}
\nF: event of getting 5 on black die.
\nF = {(5, 1), (5, 2), (5, 3), (5,4), (5, 5), (5, 6))
\nE \u2229 F = {(5, 5), (5, 6))
\n\\(\\mathrm{P}(\\mathrm{F})=\\frac{6}{36}, \\mathrm{P}(\\mathrm{E} \\cap \\mathrm{F})=\\frac{2}{36}\\)
\n\\(P(E \\mid F)=\\frac{P(E \\cap F)}{P(F)}=\\frac{\\left(\\frac{2}{36}\\right)}{\\left(\\frac{6}{36}\\right)}=\\frac{2}{6}=\\frac{1}{3}\\)<\/p>\n

b. E : event of getting a sum 8
\nE = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
\nF : event of getting a number less than 4 on red die
\nF= {(1, 1),(2, 1),(3, 1),(4, 1),(5, 1),(6, 1), (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (1,3), (2,3), (3, 3), (4, 3), (5, 3), (6, 3)}
\nE \u2229 F = {(5,3),(6,2)}
\n\\(P(E \\cap F)=\\frac{2}{36}, P(F)=\\frac{18}{36}\\)
\n\u2234 P(E|F) = \\(\\frac{P(E \u2229 F)}{P(F)}\\) = \\(\\frac{(\\frac{2}{36})}{(\\frac{18}{36})}\\) = \\(\\frac{1}{9}\\)<\/p>\n

\"NCERT<\/p>\n

Question 11
\nA fair die is rolled. Consider events E = {1,3,5} F = {2,3} and G = {2,3,4,5}, Find
\n(i) P(E|F) and P(F|E)
\n(ii) P(E|G) and P(G|E)
\n(iii) P((E \u222a F) |G) and P (E \u2229 F)|G)
\nSolution:
\n\"NCERT<\/p>\n

Question 12.
\nAssume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
\n(i) the youngest is a girl,
\n(ii) at least one is girl?
\nSolution:
\nLet B : The elder boy
\nG : The elder girl
\nb : The younger boy
\ng : The younger girl
\nThe sample space S = {(B, h), (B, g), (G, b),(G, g)}<\/p>\n

i. Let E : both children are girls
\nF : the youngest is a girl
\n\"NCERT<\/p>\n

ii. Let E : both children are girls
\nF : atleast on child is a girl
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 13.
\nAn instructor has a question bank consisting of 300 easy True\/ False questions, 200 difficult True\/ False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
\nSolution:
\n\"NCERT<\/p>\n

Question 14.
\nGiven that the two numbers appearing on throwing two dice are different. Find the probability of the event \u2018the sum of numbers on the dice is 4\u2019.
\nSolution:
\nLet S be the sample space.
\n\u2234 n(S) =36
\nLet E : getting different numbers on the dice
\n\u2234 n(E) = 30 \u2234P(E) = \\(\\frac { 30 }{ 36 }\\)
\nLet F : getting the sum of numbers on the dice is 4
\n\u2234 F= {(1, 3), (2, 2), (3, 1))
\nn(F) = 3 \u2234 P(F) = \\(\\frac { 3 }{ 36 }\\)
\nE \u2229 F = {(1,3),(3, 1)}
\nn(E \u2229 F) = 2 \u2234 P(E \u2229 F) = \\(\\frac{2}{36}\\)
\n\\(\\mathrm{P}(\\mathrm{F} \\mid \\mathrm{E})=\\frac{\\mathrm{P}(\\mathrm{E} \\cap \\mathrm{F})}{\\mathrm{P}(\\mathrm{E})}=\\frac{\\left(\\frac{2}{36}\\right)}{\\left(\\frac{30}{36}\\right)}=\\frac{1}{15}\\)<\/p>\n

Question 15.
\nConsider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event \u2018the coin shows a tail\u2019 given that \u2018at least one die shows a 3\u2019.
\nSolution:
\nThe sample space S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, H). (1, T), (2, H). (2, T), (4, H). (4, T), (5, H), (5, T), (6, 1),(6, 2),(6, 3), (6,4),(6, 5),(6, 6))
\nLet E: coin shows tail
\nF : atleast one die shows 3
\nNow E \u2229 F = \u03a6 or P(E \u2229 F) = 0
\nP(E|F) = 0 since P(E \u2229 F) = 0<\/p>\n

\"NCERT<\/p>\n

Question 16.
\nIf P(A) = \\(\\frac { 1 }{ 2 }\\), P (B) = 0 then P (A | B) is
\n(a) 0
\n(b) \\(\\frac { 1 }{ 2 }\\)
\n(c) not defined
\n(d) 1
\nSolution:
\nP(A) = P(B) = 0
\n\u2234 P(A\u2229B) = 0
\n\u2234 P(A|B) = \\(\\frac { P(A\\cap B) }{ P(B) } =\\frac { 0 }{ 0 }\\)
\nThus option C is correct.<\/p>\n

Question 17.
\nIf A and B are events such that P(A | B) = P(B | A) then
\n(a) A\u2282B but A\u2260B
\n(b) A = B
\n(c) A\u2229B = \u03c6
\n(d) P(A) = P(B)
\nSolution:
\nP(A | B) = P(B | A)
\nThus option (d) is correct.
\n\\(\\frac { P(A\\cap B) }{ P(B) } =\\frac { P(B\\cap A) }{ P(A) }\\)
\n\u21d2 P(A) = P(B)<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.1 Question 1 Given that E and Fare events such that P (E) = 0.6, P (F) = 0.3 and P(E\u2229F) = …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-13-ex-13-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.1 Question 1 Given that E and Fare events such that P (E) = 0.6, P (F) = 0.3 and P(E\u2229F) = … NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Read More »","og_url":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-13-ex-13-1\/","og_site_name":"NCERT Solutions","article_publisher":"https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","article_published_time":"2021-08-30T10:57:06+00:00","og_image":[{"url":"https:\/\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png"}],"twitter_card":"summary_large_image","twitter_creator":"@ncertsolguru","twitter_site":"@ncertsolguru","twitter_misc":{"Written by":"Prasanna","Est. reading time":"9 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Organization","@id":"https:\/\/ncertsolutions.guru\/#organization","name":"NCERT Solutions","url":"https:\/\/ncertsolutions.guru\/","sameAs":["https:\/\/www.facebook.com\/NCERTSolutionsGuru\/","https:\/\/www.instagram.com\/ncertsolutionsguru\/","https:\/\/www.linkedin.com\/in\/ncertsolutionsguru\/","https:\/\/in.pinterest.com\/ncertsolutionsguru\/","https:\/\/en.wikipedia.org\/wiki\/National_Council_of_Educational_Research_and_Training","https:\/\/twitter.com\/ncertsolguru"],"logo":{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/#logo","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2020\/05\/NCERT-Solutions.png?fit=460%2C45&ssl=1","width":460,"height":45,"caption":"NCERT Solutions"},"image":{"@id":"https:\/\/ncertsolutions.guru\/#logo"}},{"@type":"WebSite","@id":"https:\/\/ncertsolutions.guru\/#website","url":"https:\/\/ncertsolutions.guru\/","name":"NCERT Solutions","description":"NCERT Solutions for Class 1 to 12","publisher":{"@id":"https:\/\/ncertsolutions.guru\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/ncertsolutions.guru\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-13-ex-13-1\/#primaryimage","inLanguage":"en-US","url":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","contentUrl":"https:\/\/i1.wp.com\/ncertsolutions.guru\/wp-content\/uploads\/2021\/01\/NCERT-Solutions-Guru.png?fit=170%2C17&ssl=1","width":170,"height":17,"caption":"NCERT Solutions"},{"@type":"WebPage","@id":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-13-ex-13-1\/#webpage","url":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-12-maths-chapter-13-ex-13-1\/","name":"NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 - 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