{"id":39873,"date":"2022-12-07T15:57:04","date_gmt":"2022-12-07T10:27:04","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=39873"},"modified":"2022-12-08T11:50:30","modified_gmt":"2022-12-08T06:20:30","slug":"maths-2a-probability-important-questions-long-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-long-answer-type\/","title":{"rendered":"TS Inter Second Year Maths 2A Probability Important Questions Long Answer Type"},"content":{"rendered":"
Students must practice these Maths 2A Important Questions<\/a> TS Inter Second Year Maths 2A Probability Important Questions Long Answer Type to help strengthen their preparations for exams.<\/p>\n Question 1. II) Let A be the event of getting a multiple of 3 then <\/p>\n Question 2. Question 3. <\/p>\n <\/p>\n Question 4. <\/p>\n Proof: <\/p>\n Question 5. <\/p>\n <\/p>\n <\/p>\n Question 6. <\/p>\n <\/p>\n Question 7. <\/p>\n <\/p>\n Question 8. <\/p>\n Question 9. <\/p>\n A die is thrown. B1<\/sub> is chosen if either 1 or 2 turns up. B2<\/sub> is chosen If 3 or 4 turns up and B3<\/sub>, is chosen if 5 or 6 turns up. Having chosen a box In this way, a ball is choosen at random from this box. If the ball drawn Is found to be red, find the probability that it is drawn fron box B2<\/sub>. <\/p>\n Question 10. <\/p>\n <\/p>\n Question 11. <\/p>\n Question 12. <\/p>\n <\/p>\n Question 13. <\/p>\n One box is randomly selected and a ball is drawn from it. If the ball is red, then find the probability that It Is from Box – II. [TS. Mar. 2019] Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Probability Important Questions Long Answer Type to help strengthen their preparations for exams. TS Inter Second Year Maths 2A Probability Important Questions Long Answer Type Question 1. If one ticket is randomly selected from ticket numbers 1 to 30, then find … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[26],"tags":[],"yoast_head":"\nTS Inter Second Year Maths 2A Probability Important Questions Long Answer Type<\/h2>\n
\nIf one ticket is randomly selected from ticket numbers 1 to 30, then find the probability that the number on the ticket
\nis
\nI. a multiple of 5 or 7
\nII. a multiple of 3 or 5 [Mar. ’08]
\nSolution:
\nLet, S be the sample space.
\nNo. of ways of drawing one ticket from 30 ticket
\n\u2234 n(S) = \\({ }^{30} \\mathrm{C}_1\\) = 30
\nI) Let K be the event of getting a multiple of 5 then A = {5, 10, 15, 20, 25, 30}
\n\u2234 n(A) = 6
\n\u2234 P(A) = \\(\\frac{n(A)}{n(S)}=\\frac{6}{30}=\\frac{1}{5}\\)
\nLet B be the event of getting a multiple of 7 then
\nB = {7, 14, 21, 28}
\n\u2234 n(B) = 4
\n\u2234 P(B) = \\(\\frac{n(B)}{n(S)}=\\frac{4}{30}=\\frac{2}{15}\\)
\nA \u2229 R = getting a multiple of 5 and 9 = \u03a6
\n\u2234 n(A \u2229 B) = 0
\nP(A \u2229 B) = \\(\\frac{n(A \\cap B)}{n(S)}=\\frac{0}{30}\\) = 0
\nThe probability that the number on the ticket is a multiple of 5 or 7.
\nAccording to addition theorem on probability,
\nP(A \u222a B) = P(A) + P(\u00df) – P(A \u2229 B)
\n= \\(\\frac{1}{5}+\\frac{2}{15}\\) – 0
\n= \\(\\frac{3+2}{15}=\\frac{5}{15}=\\frac{1}{3}\\).<\/p>\n
\nA = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}
\n\u2234 n(A) = 10
\n\u2234 P(A) = \\(\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{10}{30}=\\frac{1}{3}\\)
\nLet B be the event of getting a multiple of 5 then
\nB = {5, 10, 15, 20, 25, 30}
\n\u2234 n(B) = 6
\n\u2234 P(B) = \\(\\frac{n(B)}{n(S)}=\\frac{6}{30}=\\frac{1}{5}\\)
\nA \u2229 B is the event getting a multiple of 5 and 3 then A \u2229 B = {15, 30}
\n\u2234 n(A \u2229 B) = 2
\n\u2234 P(A \u2229 B) = \\(\\frac{\\mathrm{n}(\\mathrm{A} \\cap \\mathrm{B})}{\\mathrm{n}(\\mathrm{S})}=\\frac{2}{30}=\\frac{1}{15}\\)
\nThe probability that the number on the ticket is a multiple of 3 or 5. According to addition theorem on probability.
\nP(A \u222a B) = P(A) + P(B) – P(A \u2229 B)
\n= \\(\\frac{1}{3}+\\frac{1}{5}-\\frac{1}{15}\\)
\n= \\(\\frac{5+3-1}{15}=\\frac{7}{15}\\).<\/p>\n
\nThe probabilities of three events A, B, C are such that P(A) = 0.3, P(B) = 0.4, P(C) = 0.8, P(A \u2229 B) = 0.08, P(A \u2229 C) = 0.28, P(A \u2229 B \u2229 C) = 0.09 and P(A \u222a B \u222a C) = 0.75. Show that P(B \u2229 C) lies in the interval [0.23, 0.48]. [Board Paper]
\nSolution:
\nGiven that,
\nP(A) = 0.3, P(B) = 0.4, P(C) = 0.8
\nP(A \u2229 B) = 0.08, P(A \u2229 C) = 0.28
\nP(A \u2229 B \u2229 C) = 0.09, and P(A \u222a B \u222a C) \u2265 0.75
\nNow,
\nP(A \u222a B \u222a C) = P(A) + P(B) + P(C) – P(A \u2229 B) – P(B \u2229 C) – P(A \u2229 C) + P(A \u2229 B \u2229 C)
\n= 0.3 + 0.4 + 0.8 – 0.08 – P(B \u2229 C) – 0.28 + 0.09
\n= – 0.27 + 1.5 – P(B \u2229 C)
\n= 1.23 – P(B \u2229 C)
\nSince, P(A \u222a B \u222a C) \u2265 0.75
\n\u21d2 0.75 \u2264 P(A \u222a B \u222a C) \u2264 1
\n\u21d2 0.75 \u2264 1.23 – P(B \u2229 C) \u2264 1
\n\u21d2 0.75 – 1.23 \u2264 P(B \u2229 C) \u2264 1 – 1.23
\n\u21d2 -0.48 \u2264 – P(B \u2229 C) \u2264 – 0.23
\n\u21d2 0.48 \u2265 P(B \u2229 C) \u2265 0.23
\n\u21d2 0.23 \u2264 P(B \u2229 C) \u2264 0.48
\n\u2234 P(B \u2229 C) c [0.23, 0.48]
\n\u2234 P(B \u2229 C) lies in the interval [0.23, 0.48].<\/p>\n
\nThe probabilities of 3 mutually exclisive events are respectively given as \\(\\frac{1+3 \\mathbf{P}}{3}, \\frac{1-\\mathbf{P}}{\\mathbf{4}}, \\frac{\\mathbf{1 – 2 P}}{\\mathbf{2}}\\). Prove that \\(\\frac{1}{3}\\) \u2264 p \u2264 \\(\\frac{1}{2}\\)
\nSolution:
\nSuppose, A, B, C are exclusive event such that
\nP(A) = \\(\\frac{1+3 p}{3}\\)
\nP(B) = \\(\\frac{1-p}{3}\\)
\nP(C) = \\(\\frac{1-2 p}{2}\\)
\nWe know that
\n0 \u2264 P(A) \u2264 1
\n0 \u2264 \\(\\frac{1+3 \\mathrm{p}}{3}\\) \u2264 1
\n0 \u2264 1 + 3p \u2264 3
\n0 \u2264 1 \u2264 3p \u2264 3 – 1
\n– 1 \u2264 3p \u2264 2
\n\\(\\frac{-1}{3} \\leq p \\leq \\frac{2}{3}\\) …………….(1)
\n0 \u2264 P(B) \u2264 1
\n0 \u2264 \\(\\frac{1-p}{4}\\) \u2264 1
\n0 \u2264 1 – p \u2264 4
\n0 – 1 \u2264 1 – p – 1 \u2264 4 – 1
\n– 1 \u2264 – p \u2264 3
\n1 \u2265 p \u2265 – 3
\n– 3 \u2264 p \u2264 1 ……………(2)
\n0 \u2264 \\(\\frac{1-2 p}{2}\\) \u2264 1
\n0 \u2264 1 – 2p \u2264 2
\no – 1 \u2264 1 – 2p – 1 \u2264 2 – 1
\n– 1 \u2264 – 2p \u2264 1
\n\\(+\\frac{1}{2}\\) p \u2265 \\(\\frac{1}{2}\\)
\n\\(\\frac{1}{2}\\) \u2264 p \u2264 \\(\\frac{1}{2}\\)
\nSince, A, B, C are mutually exclusive events then 0 \u2264 P(A \u222a B \u222a C) \u2264 1
\n0 \u2264 P(A) + P(B) + P(C) \u2264 1
\n0 \u2264 \\(\\frac{1+3 p}{3}+\\frac{1-p}{4}+\\frac{1-2 p}{2}\\) \u2264 1
\n0 \u2264 \\(\\frac{4+12 p+3-3 p+6-12 p}{12}\\) \u2264 1
\n0 \u2264 \\(\\frac{13-3 p}{12}\\) \u2264 1
\n0 \u2264 13 – 3p \u2264 12
\n0 – 13 \u2264 – 3p \u2264 12 – 13
\n– 13 \u2264 – 3p \u2264 – 1<\/p>\n
\nState and prove addition theorem on probability [May \u201814, \u201812, \u201809, \u201808, ’07, \u201806, \u201805, Mar. ’14. \u201811. \u201807; AP – Mar. \u201818. \u201817; TS – Mar.\u201918. May \u201816]
\nSolution:
\nAddition theorem on probability:
\nIf A, B are two events in a sample space, S then P(A \u222a B) = P(A) + P(B) – P(A \u2229 B)<\/p>\n
\nIn a sample space
\nA = {a1<\/sub>, a2<\/sub>, …………………. a1<\/sub>, a1+1<\/sub>, ……………. am<\/sub>)
\nB = {a1+1<\/sub> ……… am, am+1<\/sub> …………………… an<\/sub>}
\nA \u2229 B = {a1+1<\/sub>, ………………., am<\/sub>}
\nA \u222a B = {a1<\/sub>, a2<\/sub>, ……….. a1<\/sub>, a1+1<\/sub>, am<\/sub>, am+1<\/sub>, …………..am<\/sub>}
\nAccording to definition of probability
\nP(A) = \\(\\sum_{i=1}^m\\) P(ai<\/sub>)
\nP(B) = \\(\\sum_{\\mathbf{i}=l+1}^{\\mathrm{n}}\\) P(ai<\/sub>)
\nP(C) = \\(\\sum_{\\mathbf{i}=l+1}^{\\mathrm{n}}\\) P(ai<\/sub>)<\/p>\n
\nIf A, B, C are 3 independent events of an experiment such that \\(P\\left(A \\cap B^C \\cap C^C\\right)=\\frac{1}{4}\\), \\(P\\left(A^C \\cap B \\cap C^C\\right)=\\frac{1}{8}\\), \\(P\\left(A^C \\cap B \\cap C^C\\right)=\\frac{1}{4}\\) then P(A), P(B) and P(C). [Mar. ’10, AP – May 2016; TS – Mar. 2015]
\nSolution:
\nGiven that,
\nA, B, C are three independent events then \\(A^{\\mathrm{C}}, \\mathrm{B}^{\\mathrm{C}}, \\mathrm{C}^{\\mathrm{C}}\\) are also independent events.<\/p>\n
\nState and prove multiplication theorem on probability. [TS – May ’15, May ’10, Mar. ’04]
\nSolution:
\nMultiplication theorem of probability: (or) Theorem of compound probability:
\nLet A, B be two events in a sample s pace, S such that P(A) \u2260 0, P(B) \u2260 0 then
\nI) P(A \u2229 B) = P(A). \\(\\mathrm{P}\\left(\\frac{\\mathrm{B}}{\\mathrm{A}}\\right)\\)
\nII) P(A \u2229 B) = P(B). \\(\\mathrm{P}\\left(\\frac{\\mathrm{A}}{\\mathrm{B}}\\right)\\)
\nProof:
\nLet, n(A), n(B), n(A \u2229 B), n(S) be the number of sample points in A, B, A \u2229 B, S respectively.
\nThen,<\/p>\n
\nState and prove Baye\u2019s theorem on probability,
\nStatement: If A1<\/sub>, A2<\/sub>, ………….., An<\/sub>, are mutually exclusive and exhaustive events in a sample space, S such that P(A1<\/sub>) > 0 for i = 1, 2, …, n and E is any event with P(E) > 0 then \\(P\\left(\\frac{A_K}{E}\\right)=\\frac{P\\left(A_K\\right) \\cdot P\\left(\\frac{E}{A_K}\\right)}{\\sum_{i=1}^n P\\left(A_i\\right) P\\left(\\frac{E}{A_i}\\right)}\\) for K = 1, 2, ………….., n. [Mar. 12, \u201809, May \u201805 AP – Mar. May, ’15, ’16, TS – Mar. \u201817, \u201816, AP – Mar. 2019]
\nSolution:
\nProof :
\nSince, A1<\/sub>, A2<\/sub>, …………….. An<\/sub> are mutually exclusive and exhaustive events in a sample space, S it follows that \\(\\bigcup_{i=1}^n\\) Ai<\/sub> = S and
\nA1<\/sub>, A2<\/sub>, ………….. An<\/sub> are mutually disjoint.
\nNow,
\nE \u2229 A1<\/sub>, E \u2229 A2<\/sub>, ……………….. E \u2229 An<\/sub> are mutually disjoint.
\n\u2234 P(E) = P(E \u2229 S)<\/p>\n
\nSuppose that an urn B1<\/sub> contains two white and 3 black balls and another urn B2<\/sub> contains 3 white and 4 black balls. One urn is selected at random and a ball is drawn from it. If the ball drawn is found black, find the pobability that the urn choosen was B1<\/sub>.
\nSolution:
\nLet E1<\/sub>, E2<\/sub> denote the events of selecting mens B1<\/sub> and B2<\/sub> respectively then
\nP(E1<\/sub>) = \\(\\frac{1}{2}\\),
\nP(E2<\/sub>) = \\(\\frac{1}{2}\\)
\nLet B denote the event that the ball choosen from the selected men is black then<\/p>\n
\nThree boxes B1<\/sub>, B2<\/sub> and B3<\/sub> contain balls with different colours as shown below.<\/p>\n
\nSolution:
\nLet E1<\/sub>, E2<\/sub>, E3<\/sub> denote the events of selecting boxes B1<\/sub>, B2<\/sub>, B3<\/sub> respectively.
\n\u2234 P(E1<\/sub>) = \\(\\frac{2}{6}=\\frac{1}{3}\\)
\nP(E2<\/sub>) = \\(\\frac{2}{6}=\\frac{1}{3}\\)
\nP(E3<\/sub>) = \\(\\frac{2}{6}=\\frac{1}{3}\\)
\nLet f denote the event that the ball choosen from the selected box is red.<\/p>\n
\nAn urn contains \u2018w\u2019 white halls and \u2018b\u2019 black balls. Two players Q and R alternatively draw a ball with replacement from the urn. The player that draws a white ball first wins the game. If Q begins the game, find the probability of his winning the game.
\nSolution:
\nLet w denote the event of drawing a white ball in any draw then
\nP(w) = \\(\\frac{w_{C_1}}{(w+b)_{C_1}}=\\frac{w}{w+b}\\)
\nLet B denote the event of drawing a black ball in any draw then
\nP(B) = \\(\\frac{\\mathrm{b}_{\\mathrm{C}_1}}{(\\mathrm{w}+\\mathrm{b})_{\\mathrm{C}_1}}=\\frac{\\mathrm{b}}{\\mathrm{w}+\\mathrm{b}}\\)
\nThe probability of Q wins the game.
\n= P(w \u222a BBw \u222a BBBBw \u222a ……………. )
\n= P(w) + P(BBw) + P(BBBB)w) + ……………..
\n= P(w) + P(B)P(B)P(w) + P(B) P(B) P(B) P(B) P(w) + …………….
\n= P(w) [1 [P(B)2<\/sup>+ [P(B)]4<\/sup> + ……………….]<\/p>\n
\nThree urns have the following composition of balls:
\nurn I : 1 white, 2 black
\nurn II: 2 white, 1 black
\nurn III: 2 white, 2 black
\nOne of the urns is selected at random and a ball is drawn. it turns out of two be white. Find the probability that It came
\nfrom urn III. [AP – Mar. 2017] [May ’13]
\nSolution:
\nLet, A1<\/sub>, A2<\/sub>, A3<\/sub> be the events of selecting urn – I, urn – II, urn – III respectively then
\nP(A1<\/sub>) = \\(\\frac{1}{3}\\),
\nP(A2<\/sub>) = \\(\\frac{1}{3}\\),
\nP(A3<\/sub>) = \\(\\frac{1}{3}\\)
\nNow A1<\/sub>, A2<\/sub>, A3<\/sub> are mutually exclusive and exhaustive events.
\nLet E be the event of drawing a white ball from the selected urn.<\/p>\n
\nA person is known to speak truth 2 out of 3 times. He throws a die and reports that it is 1. Find the probability that it is actually 1.
\nSolution:
\nLet, A be the event that 1 occurs when a die is thrown.
\n\u2234 P(A) = \\(\\frac{1}{6}\\)
\nLet, E be the event that the man reports that it is 1. Since, the man speaks the truth 2 out of 3 times.<\/p>\n
\nThree boxes numbered I, II, III contain the balls as follows:<\/p>\n
\nSolution:
\nLet A1<\/sub>, A2<\/sub>, A3<\/sub> be the events of drawing a ball from the box numbered I, II, III respectively and E be the event of drawing a red ball from the selected box.
\nP(A1<\/sub>) = \\(\\frac{1}{3}\\),
\nP(A2<\/sub>) = \\(\\frac{1}{3}\\),
\nP(A3<\/sub>) = \\(\\frac{1}{3}\\) and
\nA1<\/sub>, A2<\/sub>, A3<\/sub> are mutually exclusive and exhaustive events.
\nP(E\/A1<\/sub>) = \\(\\frac{3}{6}=\\frac{1}{2}\\)
\nP(E\/A2<\/sub>) = \\(\\frac{1}{4}\\)
\nP(E\/A3<\/sub>) = \\(\\frac{3}{12}=\\frac{1}{4}\\)
\nP(A2<\/sub>\/E) = P(A2<\/sub>) . P(E\/A2<\/sub>) P(A1<\/sub>) . P(E\/A1<\/sub>) + P(A2<\/sub>) . P(E\/A2<\/sub>) + P(A3<\/sub>) P(E\/A3<\/sub>)
\n= \\(\\frac{\\frac{1}{3} \\times \\frac{1}{4}}{\\frac{1}{3} \\times \\frac{1}{2}+\\frac{1}{3} \\times \\frac{1}{4}+\\frac{1}{3} \\times \\frac{1}{4}}\\) = \\(\\frac{\\frac{1}{4}}{\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{4}}=\\frac{1}{4}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"