NCERT Solutions for Class 12 Maths<\/a> Chapter 13 Probability Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-13-ex-13-1\/<\/p>\nNCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.1<\/h2>\n <\/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
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Question 4. \nEvaluate P(A\u222aB) if 2P(A) = P(B) = \\(\\frac { 5 }{ 13 }\\) and P(A|B) = \\(\\frac { 2 }{ 5 }\\). \nSolution: \n <\/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 <\/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 <\/p>\n
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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 <\/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 <\/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 <\/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
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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 <\/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 <\/p>\n
ii. Let E : both children are girls \nF : atleast on child is a girl \n <\/p>\n
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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 <\/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
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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. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-13-ex-13-1\/ 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":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n