{"id":39697,"date":"2022-12-07T12:11:25","date_gmt":"2022-12-07T06:41:25","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=39697"},"modified":"2022-12-07T15:55:31","modified_gmt":"2022-12-07T10:25:31","slug":"maths-2a-probability-important-questions-short-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/","title":{"rendered":"TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type"},"content":{"rendered":"

Students must practice these Maths 2A Important Questions<\/a> TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type to help strengthen their preparations for exams.<\/p>\n

TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type<\/h2>\n

Question 1.
\nIf four fair coins are tossed simultaneously then find the probability that two heads and 2 tails appear. [Mar. ’02, ,01]
\nSolution:
\nLet, S be the sample space.
\n4 coins are tossed simultaneously then total no. of ways = 24<\/sup> = 16.
\n\u2234 n(S) = 16
\nLet, A be the event of getting two heads and two tails then n(A) = \\({ }^4 C_2=\\frac{4.3}{1.2}\\) = 6
\n[HHTT, HTHT, HTTH, THTH, TTHH, THHT]
\n\u2234 Probability of getting 2 heads and 2 tails is P(A) = \\(\\frac{n(A)}{n(S)}=\\frac{6}{16}=\\frac{3}{8}\\).<\/p>\n

Question 2.
\nFind the probability that a non-leap year contains
\nI) 53 Sundays
\nII) 52 Sundays only [Mar. ’12, ’09]
\nSolution:
\nA non-leap year contains 365 days in which there are 52 weeks and one day extra.
\n52 weeks contains 52 \u00d7 7 = 364 days and
\nleft out one day may be either Sunday, Mon, Tue, Wed, Thu, Fri or Sat.
\n\u2234 n(S) = 7
\nI) For a non-leap year to contain 53 Sundays, we have only one possibility to the have 365th day, a Sunday.
\n\u2234 n(A) = i
\n\u2234 Probability of getting 53 Sundays in a non-leap year is P(A) = \\(\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{1}{7}\\)<\/p>\n

II) For a non-leap year to contain 52 Sundays we have 6 possibilities for 365th day be a day other than Sunday.
\n\u2234 n(A) = 6
\n\u2234 Probability of getting 52 Sundays in a non-leap year is P(A) = \\(\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{6}{7}\\).<\/p>\n

\"TS<\/p>\n

Question 3.
\nFor any two events A and B show that P(\\(\\mathbf{A}^c \\cap \\mathbf{B}^{\\mathrm{c}}\\)) = 1 + P(A \u2229 B) – P(A) – P(B). [March ’05]
\nSolution:
\nLet A, B are any two sets.
\nthen \\(\\mathbf{A}^c \\cap \\mathbf{B}^{\\mathrm{c}}\\) = \\((A \\cup B)^c\\)
\nLHS = \\(\\mathrm{P}\\left(\\mathrm{A}^{\\mathrm{c}} \\cap \\mathrm{B}^{\\mathrm{c}}\\right)\\)
\n= \\(\\left[(A \\cup B)^c\\right]\\)
\n= 1 – P(A \u222a B)
\n= 1 – [P(A) + P(B) – P(A \u2229 B)]
\n= 1 + P(A \u2229 B) – P(A) – P(B) = RHS
\n\u2234 \\(\\mathrm{P}\\left(\\mathrm{A}^{\\mathrm{c}} \\cap \\mathrm{B}^{\\mathrm{c}}\\right)\\) = 1 + P(A \u2229 B) – P(A) – P(B).<\/p>\n

Question 4.
\nTwo persons A and B are rolling a die on the condition that the person who gets 3 will win the game. If A starts the game, then find the probabilities of A and respectively to win the game. [TS – May 2015; Board Paper]
\nSolution:
\nTwo persons A and B rolling a die.
\nP = Probability of getting 3 = \\(\\frac{1}{6}\\)
\nq = 1 – p
\n= 1 – \\(\\frac{1}{6}\\) = \\(\\frac{5}{6}\\)
\nProbability of success, P = \\(\\frac{1}{6}\\)
\nProbability of failure, q = \\(\\frac{5}{6}\\)
\n‘A’ may win the game in either in I trial or in III trial or V trial etc.
\nProbability of A win = p + qqp + qqqqp + …………..
\n= p + q2<\/sup>p + q4<\/sup>p + …………..
\n= p(1 + q2<\/sup> + q4<\/sup> + …………….. )
\n= p\\(\\left(\\frac{1}{1-q^2}\\right)\\) (\u2235 S\u221e<\/sub> = \\(\\frac{a}{1-r}\\))
\n= \\(\\frac{p}{1-q^2}\\)
\n= \\(\\frac{\\frac{1}{6}}{1-\\frac{25}{36}}=\\frac{\\frac{1}{6}}{\\frac{36-25}{36}}=\\frac{\\frac{1}{6}}{\\frac{11}{36}}=\\frac{6}{11}\\)
\nProbability of B win = 1 – P(A)
\nP(B) = 1 – \\(\\frac{6}{11}\\) = \\(\\frac{5}{11}\\)
\nP(A) = \\(\\frac{6}{11}\\),
\n\u2234 P(B) = \\(\\frac{5}{11}\\)<\/p>\n

Question 5.
\nA, B, C are three newspapers from a city. 20% of the population read A, 16% read B, 14% read C, 8% both A and B, 5% both A and C, 4% both B and C and 2% all the three. Find the percentage of the population who read atleast one newspaper.
\nSolution:
\nA, B, C are three newspapers from a city.
\nGiven, P(A) = \\(\\frac{20}{100}\\)
\nP(B) = \\(\\frac{16}{100}\\)
\nP(C) = \\(\\frac{14}{100}\\)
\nP(A \u2229 B) = \\(\\frac{8}{100}\\)
\nP(B \u2229 C) = \\(\\frac{4}{100}\\)
\nP(C \u2229 A) = \\(\\frac{5}{100}\\)
\nP(A \u2229 B \u2229 C) = \\(\\frac{2}{100}\\)
\n\u2234 The percentage of the population who read atleast one newspaper,
\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= \\(\\frac{20}{100}+\\frac{16}{100}+\\frac{14}{100}-\\frac{8}{100}-\\frac{4}{100}-\\frac{5}{100}+\\frac{2}{100}\\)
\n= \\(\\frac{20+16+14-8-4-5+2}{100}=\\frac{35}{100}\\)
\n\u2234 35% readers read atleast one newspaper.<\/p>\n

\"TS<\/p>\n

Question 6.
\nIf two numbers are selected randomly from 20 consecutIve natural numbers. Find the probability that the sum of the two numbers is
\nI) an even number
\nII) an odd number. [March 2008]
\nSolution:
\nLet, S be the sample space.
\nNo. of ways of selecting 2 numbers from 20 numbers
\nn(S) = \\({ }^{20} C_2=\\frac{20 \\times 19}{1 \\times 2}\\) = 190<\/p>\n

I) Let, A be the event that the sum of the numbers is even. When two numbers are selected at random from 20 consecutive natural numbers.
\nSince, the sum of two numbers is even if the numbers are both even or both odd.
\n\u2234 n(A) = \\({ }^{10} C_2+{ }^{10} C_2\\)
\n= \\(\\frac{10 \\times 9}{2 \\times 1}+\\frac{10 \\times 9}{2 \\times 1}\\)
\n= 45 + 45 = 90<\/p>\n

II) The probability that the sum of 2 numbers is odd = probability of \\(\\overline{\\mathrm{A}}\\)
\n= P(\\(\\overline{\\mathrm{A}}\\)) = 1 – P(A)
\n= 1 – \\(\\frac{9}{19}\\)
\n= \\(\\frac{10}{19}\\) OR
\nThe sum of two numbers is odd if one is even and the other number is odd.
\nn(\\(\\overline{\\mathrm{A}}\\)) = \\({ }^{10} \\mathrm{C}_1 \\cdot{ }^{10} \\mathrm{C}_1\\) = 100
\nP(\\(\\overline{\\mathrm{A}}\\)) = \\(\\frac{n(\\bar{A})}{n(S)}=\\frac{100}{190}=\\frac{10}{19}\\)<\/p>\n

Question 7.
\nThe probability for a contractor to get a road contract is \\(\\frac{2}{3}\\) and to get a building contract is \\(\\frac{5}{9}\\). The probability to get atleast one contract is \\(\\frac{4}{5}\\). Find the probability that he get both the contracts. [TS – Mar. 2019; AP – Mar. 2016]
\nSolution:
\nLet A is the event of getting a road contract.
\nB is the event of getting a building contract.
\nGiven, the probability for a contractor to get a road contract is P(A) = \\(\\frac{2}{3}\\)
\nThe probability for a contractor to get a building contract is P(B) = \\(\\frac{5}{9}\\)
\nThe probability to get atleast one contract is P (A \u222a B)= \\(\\frac{4}{5}\\)
\nThe probabiLity that he gets both the contracts is P(A \u222a B) = ?
\nBy addition on probability,
\nP(A \u222a B) = P(A) P(B) – P(A \u2229 B)
\n\\(\\frac{4}{5}\\) = \\(\\frac{2}{3}\\) + \\(\\frac{5}{9}\\) – P(A \u2229 B)
\nP(A \u2229 B) = \\(\\frac{2}{3}+\\frac{5}{9}-\\frac{4}{5}=\\frac{11}{9}-\\frac{4}{5}\\)
\n= \\(\\frac{55-36}{45}=\\frac{19}{45}\\).<\/p>\n

\"TS<\/p>\n

Question 8.
\nIn a committee of 25 members, each member is proficient either in mathematics or in statistics or in both. If 19 of these are proficient in mathematics, 16 in Statistics, find the probability that a person selected from the committee is proficient in both. [TS – Mar. 2016; Mar. \u201894]
\nSolution:
\nLet, \u2018S be the sample space.
\nWhen a person is choosen at random from committee consisting of 25 members then
\nn(S) = \\({ }^{25} \\mathrm{C}_1\\) = 25
\nLet, A be the event that the person is proficient in mathematics n(A) = \\({ }^{19} \\mathrm{C}_1\\) = 19
\n\u2234 P(A) = \\(\\frac{n(A)}{n(S)}=\\frac{19}{25}\\)
\nLet, B be the event that the person is proficient in statistics n(B) = \\({ }^{16} \\mathrm{C}_1\\) = 16
\n\u2234 P(B) = \\(\\frac{n(B)}{n(S)}=\\frac{16}{25}\\)
\n[Since, 19 are proficient in mathematics and 16 are proficient in statistics]
\nSince, everyone is either proficient in mathematics or statistics or both then A \u222a B = S
\nP(A \u222a B) = P(S)
\nP(A \u222a B) = 1
\nP(A) + P(B) – P(A \u2229 B) = 1
\n\\(\\frac{19}{25}+\\frac{16}{25}\\) – P(A \u2229 B) = 1
\n\\(\\frac{35}{25}\\) – P (A \u2229 B) = 1
\n\\(\\frac{7}{5}\\) – P (A \u2229 B) = 1
\n\u2234 P (A \u2229 B) = \\(\\frac{7}{5}\\) – 1 = \\(\\frac{2}{5}\\).<\/p>\n

Question 9.
\nA, B, C are 3 horses in a race. The probability of A to win the race is twice that of B, and probability of B is twice that of C. What are the probabilities of A, B and C to win the race? [TS- Mar. \u201818 March 14, \u201813, \u201899; May \u201814, ’09, 07; AP – Mar. \u201919]
\nSolution:
\nLet A. B, C be the events that the horses A, B, C win the race respectively.
\nGiven that, the probability of A to win the race is twice that of B.
\n\u2234 P(A) = 2P(B)
\n\u21d2 P(B) = \\(\\frac{1}{2}\\) [P(A)]
\nThe probability of B to win the race is twice that of C.
\nP(B) = 2P(C)
\n\u21d2 P(C) = [P(B)]
\n= \\(\\frac{1}{2}\\) [\\(\\frac{1}{2}\\) [P(A)]]
\n= \\(\\frac{1}{4}\\) [P(A)]
\nSince, the horses A, B and C run the race A \u222a B \u222a C = S and A, B, C are mutually disjoint.
\nP(A \u222a B \u222a C) = P(S)
\nP(A \u222a B \u222a C) = 1
\nP(A) + P(B) + P(C) =1
\nP(A) + \\(\\frac{1}{2}\\) P(A) + \\(\\frac{1}{4}\\) P(A) = 1
\nP(A) [1 + \\(\\frac{1}{2}\\) + \\(\\frac{1}{4}\\)] = 1
\nP(A) [latex]\\frac{4+2+1}{4}[\/latex] = 1
\nP(A) \\(\\frac{7}{4}\\) = 1
\nP(A) = \\(\\frac{4}{7}\\)
\nP(A) = \\(\\frac{1}{2}\\) P(A)
\n= \\(\\frac{1}{2} \\times \\frac{4}{7}=\\frac{2}{7}\\)
\n\u2234 P(B) = \\(\\frac{2}{7}\\)
\nP(C) = \\(\\frac{1}{4}\\) P(A)
\n= \\(\\frac{1}{4} \\times \\frac{4}{7}=\\frac{1}{7}\\)
\nP(C) = \\(\\frac{1}{7}\\)
\n\u2234 P(A) = \\(\\frac{4}{7}\\), P(B) = \\(\\frac{2}{7}\\), P(C) = \\(\\frac{1}{7}\\).<\/p>\n

\"TS<\/p>\n

Question 10.
\nA bag contaIns 12 two rupee coins, 7 one rupee coins and 4 half rupee coins. If three coins are selected at random, then find the probability that
\nI) the sum of 3 coins is maximum
\nII) the sum of 3 coins is minimum
\nIII) each coin is of different value. [March \u201807]
\nSolution:
\nIn a bag, there are 12 two rupee coins, 7 one rupee coins and 4 half rupee coins.
\nTotal no. of coins = 12 + 7 + 4 = 23
\nLet, ‘S’ be the sample space.
\nTotal no. of ways of drawing 3 coins from 23 coins is n(S) = \\({ }^{23} \\mathrm{C}_3\\)
\nI) We get maximum amount with the coins of two rupee coins.
\n\u2234 No. of ways of drawing 3 two rupee coins = \\({ }^{12} \\mathrm{C}_3\\)
\n\u2234 n(A) = \\({ }^{12} \\mathrm{C}_3\\)
\n\u2234 P(A) = \\(\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{{ }^{12} \\mathrm{C}_3}{{ }^{23} \\mathrm{C}_3}\\)<\/p>\n

II) We get minimum amount if 3 coins are half rupee coins.
\n\u2234 No. of ways of drawing 3-half rupee coins = \\({ }^4 \\mathrm{C}_3\\)
\n\u2234 n(B) = \\({ }^4 \\mathrm{C}_3\\)
\n\u2234 P(B) = \\(\\frac{\\mathrm{n}(\\mathrm{B})}{\\mathrm{n}(\\mathrm{S})}=\\frac{{ }^4 \\mathrm{C}_3}{{ }^{23} \\mathrm{C}_3}\\)<\/p>\n

III) Each coin is of different value then we must draw one coin in each.
\nThis can be done in \\({ }^{12} \\mathrm{C}_1 \\cdot{ }^7 \\mathrm{C}_1 \\cdot{ }^4 \\mathrm{C}_1\\) ways
\n\u2234 n(C) = \\({ }^{12} \\mathrm{C}_1 \\cdot{ }^7 \\mathrm{C}_1 \\cdot{ }^4 \\mathrm{C}_1\\)
\n= 12 \u00d7 7 \u00d7 4
\n\u2234 P(C) = \\(\\frac{n(C)}{n(S)}=\\frac{12 \\times 7 \\times 4}{{ }^{23} C_3}\\)<\/p>\n

Question 11.
\nTwo dice are thrown. Find the probability of getting the same iiumber ou both the faces.
\nSolution:
\nLet, \u2018S\u2019 be the sample space.
\nThe total no.of ways of rolling two dice is n(S) = 62<\/sup> = 36
\nLet A be the event of getting the same number on both the faces of the two dice then A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
\n\u2234 n(A) = 6
\n\u2234 The probability of getting the same number on both the face
\nP(A) = \\(\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{6}{36}=\\frac{1}{6}\\).<\/p>\n

Question 12.
\nAn integer is picked from 1 to 20, both inclusive. Find the probability that is a prime.
\nSolution:
\nLet \u2018S\u2019 be the sample space.
\nTotal no.of ways of selecting one number from 20 numbers is n(S) = \\({ }^{20} \\mathrm{C}_1\\) = 20
\nLet, A be the event of getting a prim number then A = {2, 3, 5, 7, 11, 13, 17, 19}
\n\u2234 n(A) = 8
\n\u2234 The probability of a selected number be a prime is P(A) = \\(\\frac{n(A)}{n(S)}=\\frac{8}{20}=\\frac{2}{5}\\).<\/p>\n

\"TS<\/p>\n

Question 13.
\nA bag contains 4 red balls, 5 black balls and 6 blue balls. Find the probability that 2 balls drawn at random simultaneously from the black are a red and black ball.
\nSolution:
\nLet, S be the sample space.
\nTotal no.of balls = 15 (4 + 5 + 6)
\nTotal no.of ways of drawing 2 balls from 15 balls is
\nn(S) = \\({ }^{15} \\mathrm{C}_2\\)
\n= \\(\\frac{15.14}{2 \\times 1}\\)
\n= 15 \u00d7 7 = 105
\nLet, A be the event of getting a red and a black ball in a draw.
\nn(A) = \\({ }^4 \\mathrm{C}_1 \\cdot{ }^5 \\mathrm{C}_1\\)
\n= 4 \u00d7 5 = 20
\n\u2234 Required probability, P(A) = \\(\\frac{n(A)}{n(S)}=\\frac{20}{105}=\\frac{4}{21}\\).<\/p>\n

Question 14.
\nTen dice are thrown. Find the probability that none of the dice shows the number one.
\nSolution:
\nLet, S be the sample space that ten dice are rolled.
\nn(S) = 6.6 ………….. 6 (10 times) = 610<\/sup>
\nLet, A be the event of not getting a number 1 on the face of the dice.
\nn(A) = 5 . 5 …………………. 5 (10 times) = 510<\/sup>
\nRequired probability, P(A) = \\(\\frac{n(A)}{n(S)}=\\frac{5^{10}}{6^{10}}=\\left(\\frac{5}{6}\\right)^{10}\\)<\/p>\n

Question 15.
\nA number \u2018x\u2019 is drawn arbitrarly from the set {1, 2, ……………. 100}. Find the Probability that x + \\(\\frac{1}{x}\\) is greater than 29. [AP – May 2015]
\nSolution:
\nLet, S be the sample space.
\nTotal no. of ways of selecting one number from 100 numbers is
\nn(S) = \\({ }^{100} \\mathrm{C}_1\\) = 100
\nLet A be the event that x is selected at random from the set,
\nS = {1, 2, 3 , …………………, 100} has the property
\nx + \\(\\frac{100}{x}\\) > 29
\n\\(\\frac{x^2+100}{x}\\) > 29
\nx2<\/sup> + 100 > 29x
\nx2<\/sup> – 29x + 100 > 0
\nx2<\/sup> – 4x – 25x + 100 > 0
\nx (x – 4) – 25 (x – 4) > 0
\n(x – 4) (x – 25) > 0
\nx < 4 or x > 25
\n\u2234 A = {1, 2, 3, 26, 27, …………….., 100}
\nn(A) = 78
\n\u2234 Required probability, P(A) = \\(\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{78}{100}\\) = 0.78.<\/p>\n

\"TS<\/p>\n

Question 16.
\nA fair coin is tossed 200 times. Find the probability of getting a head an odd number of times.
\nSolution:
\nLet, S be the sample space.
\nThe total no.of ways a fair coin is tossed 200 times.
\nn(S) = 2200<\/sup>
\nLet, A be the event of getting a head an odd number of times then
\nn(A) = \\(200 c_1+200 c_3+200 c_5+\\ldots \\ldots \\ldots+200 c_{199}\\)
\n= 2100-1<\/sup> [2n-1<\/sup>]
\n= 2199<\/sup>
\n\u2234 Required probability, P(A) = \\(\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{2^{199}}{2^{200}}=\\frac{1}{2}\\).<\/p>\n

Question 17.
\nA and B are among 20 persons who sit at random along a round table. Find the probability that there are any 6 persons
\nbetween A and B.
\nSolution:
\nLet S be the sample space.
\nLet A occupy any seat at the round table then there are 19 seats left for B.
\n\u2234 n(S) = 19
\nLet E be the event if 6 persons are to be seated between A and B then B has only two ways to sit.
\n\u2234 Required probability, P(A) = \\(\\frac{n(A)}{n(S)}=\\frac{2}{19}\\)<\/p>\n

Question 18.
\nOut of 30 consecutive Integers, two integers are drawn at random. Find the probability that their sum is odd.
\nSolution:
\nLet S\u2019 be the sample space.
\nThe total no. of ways of choosing two integers out of 30 is n(S) = \\({ }^{30} \\mathrm{C}_2\\)
\nOut of the 30 numbers, 15 are even and 15 are odd.
\nIf the sum of the two numbers is to be odd, one should be even and other is odd.
\nHence, the no.of cases favourable to the required event is n(A) = \\({ }^{15} \\mathrm{C}_1 \\cdot{ }^{15} \\mathrm{C}_1\\)
\n\u2234 The required probability, P(A) = \\(\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{{ }^{15} \\mathrm{C}_1 \\cdot{ }^{15} \\mathrm{C}_1}{{ }^{30} \\mathrm{C}_2}=\\frac{\\frac{15.15}{30 \\times 29}}{2 \\times 1}=\\frac{15}{29}\\).<\/p>\n

\"TS<\/p>\n

Question 19.
\nFind the probability of throwing a total score of 7 with two dice. [May ’08]
\nSolution:
\nLet, S be the sample space.
\nThe total no. of ways of rolling two dice
\n\u2234 n(S) = 62<\/sup> = 36
\nLet, A\u2019 be the event of getting a total score of 7 with two dice then
\nA = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
\n\u2234 n(A) = 6
\n\u2234 Probability of throwing a total score of 7 with 2 dice is P(A) = \\(\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{6}{36}=\\frac{1}{6}\\).<\/p>\n

Question 20.
\nFind the probability of obtaining two tails and one head when three coins are tossed.
\nSolution:
\nLet, S be the sample space.
\nThe total no. of ways three coins are tossed,
\nS = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
\n\u2234 n(S) = 8
\nLet, A be the event of getting two tails and a head when three coins are tossed then
\nA = {HTT, THT, TTH}
\n\u2234 n(A) = 3
\n\u2234 The required probability, P(A) = \\(\\frac{n(A)}{n(S)}=\\frac{3}{8}\\).<\/p>\n

Question 21.
\nA page is opened at random from a book containing 200 pages. What is the prob ability that the number oil the page is a perfect square? [May ’12]
\nSolution:
\nLet, S be the sample space.
\nThe total no.of ways selecting a page from 200 pages is n(S) = \\({ }^{200} \\mathrm{C}_1\\) = 200
\nLet A be the event of drawing a page whose number is a perfect square.
\nA = {12<\/sup>, 22<\/sup>, 32<\/sup>, 42<\/sup>, 52<\/sup>, 62<\/sup>, 72<\/sup>, 82<\/sup>, 92<\/sup>, 102<\/sup>, 112<\/sup>, 122<\/sup>, 132<\/sup>, 142<\/sup>}
\n\u2234 n(A) = 14
\n\u2234 Required probability, P(A) = \\(\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{14}{200}=\\frac{7}{100}\\).<\/p>\n

\"TS<\/p>\n

Question 22.
\nFind the probability of drawing an ace or a spade from a well – shuffled pack of 52 playing cards. [TS – Mar. 2015]
\nSolution:
\nLet, S be the sample space.
\nTotal no. cards in the pack = 52
\n\u2234 n(S) = \\({ }^{52} \\mathrm{C}_1\\) = 52
\nIf A is the event of getting an ace card then
\nn(A) = \\({ }^4 \\mathrm{C}_1\\) = 4
\n\u2234 Probability of getting an ace card,
\nP(A) = \\(\\frac{\\mathrm{n}(\\mathrm{A})}{\\mathrm{n}(\\mathrm{S})}=\\frac{4}{52}=\\frac{1}{13}\\)
\nIf B is the event of getting a spade card then n(B) = \\({ }^{13} \\mathrm{C}_1\\) = 13
\n\u2234 Probability of getting a spade card,
\nP(B) = \\(\\frac{n(B)}{n(S)}=\\frac{13}{52}=\\frac{1}{4}\\)
\nA \u2229 B is common to the events A, B. There is a spade, ace.
\n\u2234 n(A \u2229 B) = \\({ }^1 C_1\\) = 1
\nprobability of getting A \u2229 B is P(A \u2229 B) = \\(\\frac{n(A \\cap B)}{n(S)}=\\frac{1}{52}\\)
\nBy addition theorem on probability,
\nP(A \u222a B) = P(A) + P(B) – P(A \u2229 B)
\n= \\(\\frac{1}{13}+\\frac{1}{4}-\\frac{1}{52}\\)
\n= \\(\\frac{4+13-1}{52}=\\frac{8}{52}=\\frac{4}{13}\\)<\/p>\n

Question 23.
\nA and B are events with P(A) 0.5, P(B) = 0.4 and P(A \u2229 B) = 0.3. Find the probability that
\nI) A does not occurs
\nII) Neither A nor B occurs [AP – Mar. \u201818, TS – Mar.\u201917 March \u201810]
\nSolution:
\nA and B are two events
\nGiven that, P(A) = 0.5.
\nP(B) = 0.4, P(A \u2229 B) = 0.3
\nI) The probability that A does not occur = P(\\(\\overline{\\mathrm{A}}\\))
\n= 1 – P(A)
\n= 1 – 0.5 = 0.5.<\/p>\n

II) The probability that neither A nor B occurs = \\(P(A \\cup B)^C\\)
\n= 1 – P(A \u222a B)
\n= 1 – [P(A) + P(B) – P(A \u2229 B)]
\n= 1 – [0.5 + 0.4 – 0.3]
\n= 1 – 0.6 = 0.4.<\/p>\n

Question 24.
\nIf A, B, C are three events, show that P(A \u222a B \u222a C) = P(A) + P(B) + P(C) – P(A \u2229 B) – P(B \u2229 C) – P(C \u2229 A) + P(A \u2229 B \u2229 C). [AP – Mar. 2015, AP – May 2016]
\nSolution:
\nA, B, C are three events
\nL.H.S = P(A \u222a B \u222a C)
\n= P(A) + P(B \u222a C) – P[A \u2229 (B \u222a C)]
\n= P(A) + P(B) + P(C) – P(B \u2229 C) – P[(A \u2229 B) \u222a (A \u2229 C)]
\n= P(A) + P(B) + P(C) – P(B \u2229 C) – [P(A \u2229 B) + P(A \u2229 C) – P[(A \u2229 B) \u2229 (A \u2229 C)]]
\n= P(A) + P(B) + P(C) – P(B \u2229 C) – P(A \u2229 B) – P(C \u2229 A) + P(A \u2229 B \u2229 C)
\n= P(A) + P(B) + P(C) – P(A \u2229 B) – P(B \u2229 C) – P(C \u2229 A) + P(A \u2229 B \u2229 C)
\n= RHS
\n\u2234 P(A \u222a B\u222a C) = P(A) + P(B) + P(C) – P(A \u2229 B) – P(B \u2229 C) – P(C \u2229 A) + P(A \u2229 B \u2229 C).<\/p>\n

\"TS<\/p>\n

Question 25.
\nThree screws are drawn at random from a lot of 50 screws, five of which are defective. Find the probability of the event that all 3 screws are nondefective. Assuming that the drawing is
\na) with replacement
\nb) without replacement.
\nSolution:
\nLet, \u2018S\u2019 be the sample space.
\nThe total no. of screws = 50
\nThe no.of defective screws = 5
\nThe no.of non-defective screws = 45
\nLet A be the event of getting 3 screws which are non-defective.
\na) With replacement:
\nP(A) = \\(\\frac{{ }^{45} \\mathrm{C}_1}{{ }^{50} \\mathrm{C}_1} \\cdot \\frac{{ }^{45} \\mathrm{C}_1}{{ }^{50} \\mathrm{C}_1} \\cdot \\frac{{ }^{45} \\mathrm{C}_1}{{ }^{50} \\mathrm{C}_1}\\)
\n= \\(\\frac{45}{50} \\cdot \\frac{45}{50} \\cdot \\frac{45}{50}=\\left(\\frac{9}{10}\\right)^3\\)
\n\u2234 P(A) = \\(\\left(\\frac{9}{10}\\right)^3\\)<\/p>\n

b) without replacement:
\nP(A) = \\(\\frac{{ }^{45} \\mathrm{C}_1}{{ }^{50} \\mathrm{C}_1} \\cdot \\frac{{ }^{44} \\mathrm{C}_1}{{ }^{49} \\mathrm{C}_1} \\cdot \\frac{{ }^{43} \\mathrm{C}_1}{{ }^{48} \\mathrm{C}_1}\\)
\n= \\(\\frac{45}{50} \\cdot \\frac{44}{49} \\cdot \\frac{43}{48}=\\frac{1419}{1960}\\)<\/p>\n

Question 26.
\nThere are 3 black and 4 white balls in one bag, 4 black and 3 whIte balls In the second bag A die is rolled and the first bag is selected if the die shows up 1 or 3, and the second bag for the rest. Find the probability of drawing a black ball from the bag thus selected. [March \u201809]
\nSolution:
\nProbability of selecting I bag = \\(\\frac{2}{6}=\\frac{1}{3}\\)
\nProbabthty of selecting II bag 1 – \\(\\frac{1}{3}\\) = \\(\\frac{2}{3}\\)
\nProbability of getting a black ball from I bag = \\(\\frac{{ }^3 \\mathrm{C}_1}{{ }^7 \\mathrm{C}_1}=\\frac{3}{7}\\)
\nProbability of getting a black ball from II bag = \\(\\frac{{ }^4 \\mathrm{C}_1}{{ }^7 \\mathrm{C}_1}=\\frac{4}{7}\\)
\n\u2234 The probability of drawing a black ball = \\(\\frac{1}{3} \\cdot \\frac{3}{7}+\\frac{2}{3} \\cdot \\frac{4}{7}\\)
\n= \\(\\frac{1}{7}+\\frac{8}{21}\\)
\n= \\(\\frac{3+8}{21}=\\frac{11}{21}\\).<\/p>\n

Question 27.
\nA, B, C are aiming to shoot a balloon. A will succeed 4 times out of 5 attempts. The chance of B to shoot the balloori is 3 out of 4 and that of C is 2 out of three. If the 3 aim the balloon simultaneously, then find the probability that atleast two of them hit the balloon.
\nSolution:
\nGiven that,
\nP(A) = \\(\\frac{4}{5}\\), P(B) = \\(\\frac{3}{4}\\), P(C) = \\(\\frac{2}{3}\\)
\nP(\\(\\overline{\\mathrm{A}}\\)) = 1 – P(A)
\n= 1 – \\(\\frac{4}{5}\\) = \\(\\frac{1}{5}\\)
\nP(\\(\\overline{\\mathrm{B}}\\)) = 1 – P(B)
\n= 1 – \\(\\frac{3}{4}\\) = \\(\\frac{1}{4}\\)
\nP(\\(\\overline{\\mathrm{C}}\\)) = 1 – P(C)
\n= 1 – \\(\\frac{2}{3}\\) = \\(\\frac{1}{3}\\)
\n\u2234 The probability that at least two of them hit balloon is equal to the probability of A, B hits the balloon then C will not hit or the probability of A, C hits the balloon then B will not hit or the probability of B. C hits the balloon then A will
\nnot hit or the probability of all the three will hit the balloon.
\n= P(A \u2229 B \u2229 \\(\\overline{\\mathrm{C}}\\)) + P(A \u2229 \\(\\overline{\\mathrm{B}}\\)\u2229 C) + P(\\(\\) \u2229 B \u2229 C) + P(A \u2229 B \u2229 C)
\n= P(A) . P(B). P(\\(\\overline{\\mathrm{C}}\\)) + P(A). P(\\(\\overline{\\mathrm{B}}\\)) . P(C) + P(\\(\\overline{\\mathrm{A}}\\)) . P(B) P(C) + P(A) + P(B) . P(C)
\n= \\(\\frac{4}{5} \\cdot \\frac{3}{4} \\cdot \\frac{1}{3}+\\frac{4}{5} \\cdot \\frac{1}{4} \\cdot \\frac{2}{3}+\\frac{1}{5} \\cdot \\frac{3}{4} \\cdot \\frac{2}{3}+\\frac{4}{5} \\cdot \\frac{3}{4} \\cdot \\frac{2}{3}\\)
\n= \\(\\frac{1}{5}+\\frac{2}{15}+\\frac{1}{10}+\\frac{2}{5}=\\frac{6+4+3+12}{30}=\\frac{25}{30}=\\frac{5}{6}\\).<\/p>\n

\"TS<\/p>\n

Question 28.
\nIf A, B are two events, then show that \\(\\mathbf{P}\\left(\\frac{A}{B}\\right) \\cdot P(B)+P\\left(\\frac{A}{B^C}\\right) \\mathbf{P}\\left(B^C\\right)=P(A)\\)
\nSolution:
\nA, B are two events.<\/p>\n

\"TS<\/p>\n

Question 29.
\nA pair of dice is rolled. What is the probability that their sum to 7 ? Give thatneither dice shows a 2.
\nSolution:
\nLet S be the sample space.
\nLet S does not shows 2 then S = 25
\nS = {(1, 1), (1, 3), (1, 4), (1, 5), (1, 6), (4, 1), (4, 3), (4, 4), (4, 5), (4, 6), (3, 1), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 3), (5, 4),
\n(5, 5), (5, 6), (6. 1), (6, 3), (6, 4), (6, 5), (6, 6)}
\n\u2234 n(S) = 25
\nLet A be the event that the sum to 7 one the pair of dice.
\nA = {(1, 6), (3, 4), (4, 3), (6, l)}
\n\u2234 n(A) = 4
\n\u2234 Required probability, P(A) = \\(\\frac{n(A)}{n(S)}=\\frac{4}{25}\\).<\/p>\n

Question 30.
\nA pair of dice is rolled. What is the probability that neither dice shows 2?
\nSolution:
\nGiven that S is the sum to 7.
\nLet S be the sample space.
\nLet S be the sum to 7.
\nWhen two dies are rolled,
\nS = {(l, 6), (2,5), (3,4), (4, 3), (5, 2), (6, 1)}
\n\u2234 n(S) = 6
\nLet A be the event that neither of dice shows 2.
\nA = {(1, 6), (3, 4), (4, 3), (6, 1)}
\n\u2234 n(A) = 4
\n\u2234 Required probability, P(A) = \\(\\frac{n(A)}{n(S)}=\\frac{4}{6}=\\frac{2}{3}\\)<\/p>\n

Question 31.
\nAn urn contains 12 red balls and 12 green balls. Suppose 2 balls are drawn one after another without replacement. Find the probability that the second ball drawn is green. The first ball drawn is red ball.
\nSolution:
\nLet, S be the sample space.
\nThe total no.of balls, n(S) = 24
\nLet, A be the event of drawing a red ball
\nn(A) = \\({ }^{12} \\mathrm{C}_1\\) = 12
\nThe probability of drawing a red in the first attempt
\nP(A) = \\(\\frac{n(A)}{n(S)}=\\frac{12}{24}=\\frac{1}{2}\\)
\nNow, there are 23 balls remaining.
\n\u2234 n(S) = 23
\nLet \\(\\frac{B}{A}\\) be the event of drawing a green in the second attempt \\(n\\left(\\frac{B}{A}\\right)={ }^{12} C_1\\) = 12
\n\u2234 Required probability, \\(P\\left(\\frac{B}{A}\\right)=\\frac{n\\left(\\frac{B}{A}\\right)}{n(S)}=\\frac{12}{23}\\).<\/p>\n

\"TS<\/p>\n

Question 32.
\nIf one card is drawn at random from a pack of cards then show that the event of getting an ace and getting a heart are independent events.
\nSolution:
\nLet, S be the sample space.
\nTotal no.of playing cards = 52
\nTotal no.of ways of selecting one card from a pack
\nn(S) = \\({ }^{52} \\mathrm{C}_1\\) = 52
\nLet A be the event of getting an ace n(A) = \\({ }^4 \\mathrm{C}_1\\) = 4
\n\u2234 P(A) = \\(\\frac{n(A)}{n(S)}=\\frac{4}{52}=\\frac{1}{13}\\)
\nLet B be the event of getting a heart
\nn(B) = \\({ }^{13} \\mathrm{C}_1\\) = 13
\n\u2234 P(B) = \\(\\frac{n(B)}{n(S)}=\\frac{13}{52}=\\frac{1}{4}\\)
\nA \u2229 B = event of getting an ace and a heart
\ni.e., heart ace
\n\u2234 n(A \u2229 B) = 1
\n\u2234 P(A \u2229 B) = \\(\\frac{\\mathrm{n}(\\mathrm{A} \\cap \\mathrm{B})}{\\mathrm{n}(\\mathrm{S})}=\\frac{1}{52}\\)
\n= \\(\\frac{1}{4} \\cdot \\frac{1}{13}\\) = P(A). P(B)
\n\u2234 P(A \u2229 B) = P(A). P(B)
\n\u2234 The events A, B are independent events.<\/p>\n

Question 33.
\nThe probability that a boy A will get a scholarship is 0.9 and that another boy B will get is 0.8. What is the probability that atleast one of them will get the scholar ship? [May 2003]
\nSolution:
\nLet, A be the event that A will get scholar ship.
\nB be the event that B will get scholarship.
\nGiven, P(A) = 0.9, P(B) = 0.8
\nA and B are independent events.
\nThe probability that atleast one of them will get a scholarship is
\nP(A \u222a B) = P(A) + P(B) – P(A r B)
\n= P(A) + P(B) – P(A) P(B)
\n= 0.9 + 0.8 – (0.9) (0.8)
\n= 1.7 – 0.72 = 0.98.<\/p>\n

Question 34.
\nIf A, B are two events with P(A \u222a B) = 0.65, P(AB) = 0.15 then find the value of \\(P\\left(\\mathbf{A}^C\\right)+P\\left(B^C\\right)\\). [MAr. ’13, ’05, May ’11, TS – Mar. May 2015]
\nSolution:
\nA, B are two events
\nGiven that P (A \u222a B) = 0.65,
\nP(A \u2229 B) = 0.15
\nFrom addition theorem on probability,
\nP(A \u222a B) = P(A) + P(B) – P(A \u2229 B)
\n0.65 = P(A) + P(B) – 0.15
\nP(A) + P(B) = 0.8
\n1 – \\(P\\left(A^C\\right)\\) + 1 – \\(P\\left(B^C\\right)\\) = 0.8
\n2 – 0.8 = \\(P\\left(A^C\\right)+P\\left(B^C\\right)\\)
\n\u2234 \\(P\\left(A^C\\right)+P\\left(B^C\\right)\\) = 1.2.<\/p>\n

\"TS<\/p>\n

Question 35.
\nA, B are two independent events such that the probability of both the events to occur is \\(\\frac{1}{6}\\) and the probability of both the events do not occur is \\(\\frac{1}{3}\\). Find P(A).
\nSolution:
\nA and B are independent events.
\nThen P(A \u2229 B) = P(A) P(B)
\nGiven that, the probability of both the events to occur is \\(\\frac{1}{6}\\)
\nP(A \u2229 B) = \\(\\frac{1}{6}\\)
\nP(A) . P(B) = \\(\\frac{1}{6}\\) ………………(1)
\nThe probability of both the events do not occur is \\(\\frac{1}{3}\\).
\n\\(\\mathrm{P}(\\overline{\\mathrm{A}} \\cap \\overline{\\mathrm{B}})=\\frac{1}{3}\\)
\n\\(\\mathrm{P}(\\overline{\\mathrm{A} \\cup \\mathrm{B}})=\\frac{1}{3}\\)
\n1 – P(A \u222a B) = \\(\\frac{1}{3}\\)
\n1 – [P(A) + P(B) – P(A \u2229 B)] = \\(\\frac{1}{3}\\)
\n1 – P(A) – P(B) + \\(\\frac{1}{6}\\) = \\(\\frac{1}{3}\\)
\n\\(\\frac{7}{6}\\) – P(A) + P(B) = \\(\\frac{1}{3}\\)
\nP(A) + P(B) = \\(\\frac{7}{6}\\) – \\(\\frac{1}{3}\\)
\nLet P(A) = x, P(B) = y then
\nfrom (1)
\n\u21d2 xy = \\(\\frac{1}{6}\\)
\nfrom (2) x + y = \\(\\frac{5}{6}\\)
\nWe know that,
\n(x – y)2<\/sup> = (x + y)2<\/sup> – 4xy
\n= \\(\\left(\\frac{5}{6}\\right)^2-4 \\cdot \\frac{1}{6}\\)
\n= \\(\\frac{25}{36}-\\frac{4}{6}=\\frac{1}{36}\\)
\n(x – y) = \u00b1 \\(\\frac{1}{6}\\)<\/p>\n

Case – 1:
\nx + y = \\(\\frac{5}{6}\\)
\nx – y = \\(\\frac{1}{6}\\)
\n2x = \\(\\frac{5}{6}\\) + \\(\\frac{1}{6}\\)
\n2x = 1
\nx = \\(\\frac{1}{2}\\)
\n\u21d2 P(A) = \\(\\frac{1}{2}\\)<\/p>\n

Case – 2:
\nx + y = \\(\\frac{5}{6}\\)
\nx – y = \\(\\frac{1}{6}\\)
\n2x = \\(\\frac{5}{6}\\) – \\(\\frac{1}{6}\\)
\n2x = \\(\\frac{4}{6}\\)
\n2x = \\(\\frac{2}{3}\\)
\nx = \\(\\frac{1}{3}\\)
\n\u21d2 P(A) = \\(\\frac{1}{3}\\)
\n\u2234 P(A) = \\(\\frac{1}{2}\\) or \\(\\frac{1}{3}\\) .<\/p>\n

Question 36.
\nA fair die is rolled. Consider the events A = {1, 3, 5}, B = {2, 3} and C = {2, 3, 4, 5}. Find
\nI) P(A \u2229 B), P(A \u222a B)
\nII) P(\\(\\left(\\frac{\\mathbf{A}}{\\mathbf{B}}\\right)\\)), \\(\\mathbf{P}\\left(\\frac{\\mathbf{B}}{\\mathbf{A}}\\right)\\)
\nIII) \\(\\mathbf{P}\\left(\\frac{\\mathbf{A}}{\\mathbf{C}}\\right), \\mathbf{P}\\left(\\frac{\\mathbf{C}}{\\mathbf{A}}\\right)\\)
\nIV) \\(\\mathbf{P}\\left(\\frac{\\mathbf{B}}{\\mathbf{C}}\\right), \\mathbf{P}\\left(\\frac{\\mathbf{C}}{\\mathbf{B}}\\right)\\)
\nSolution:
\nA fair die is rolled then n(S) = 6
\nn(A) = 3, n(B) = 2, n(C) = 4.
\n\\(P(A)=\\frac{n(A)}{n(S)}=\\frac{3}{6}=\\frac{1}{2}\\)
\n\\(\\mathrm{P}(\\mathrm{B})=\\frac{\\mathrm{n}(\\mathrm{B})}{\\mathrm{n}(\\mathrm{S})}=\\frac{2}{6}=\\frac{1}{3}\\)
\n\\(P(C)=\\frac{n(C)}{n(S)}=\\frac{4}{6}=\\frac{2}{3}\\)
\nGiven that, A = {1, 3, 5), B {2, 3}, C = {2, 3, 4, 5}
\nI) A = {1, 3, 5}, B = {2, 3)
\nA \u2229 B = (1, 3, 5} \u2229 {2, 3} = {3)
\n\u2234 n(A \u2229 B) = P
\n\u2234 P(A \u2229 B) = \\(\\frac{\\mathrm{n}(\\mathrm{A} \\cap \\mathrm{B})}{\\mathrm{n}(\\mathrm{S})}=\\frac{1}{6}\\)
\nA \u222a B = {1, 3, 5} \u222a {2, 3} = {1, 2, 3, 5}
\n\u2234 n(A \u222a B) = 4
\n\u2234 P(A \u222a B) = \\(\\frac{n(A \\cup B)}{n(S)}=\\frac{4}{6}=\\frac{2}{3}\\).<\/p>\n

\"TS<\/p>\n

II) \\(\\mathrm{P}\\left(\\frac{\\mathrm{A}}{\\mathrm{B}}\\right)=\\frac{\\mathrm{P}(\\mathrm{A} \\cap \\mathrm{B})}{\\mathrm{P}(\\mathrm{B})}=\\frac{\\frac{1}{6}}{\\frac{1}{3}}=\\frac{1}{2}\\)
\n\\(P\\left(\\frac{B}{A}\\right)=\\frac{P(A \\cap B)}{P(A)}=\\frac{\\frac{1}{6}}{\\frac{1}{2}}=\\frac{1}{3}\\)<\/p>\n

\"TS<\/p>\n

Question 37.
\nSuppose A and B are Independent events with P(A) = 0.6, P(B) = 0.7 then compute
\nI) P(A \u2229 B)
\nII) P(A \u222a B
\nIII) \\(\\mathbf{P}\\left(\\frac{\\mathbf{B}}{\\mathbf{A}}\\right)\\)
\nIV) \\(\\mathbf{P}\\left(\\mathbf{A}^{\\mathbf{C}} \\cap \\mathbf{B}^{\\mathbf{C}}\\right)\\) [May \u201814. Mar. \u201814] [AP – Mar.\u2019 17; TS – Mar. \u201818, May \u201816]
\nSolution:
\nA, B are independent events then
\nP(A \u2229 B) = P(A) . P(B)
\nGiven that, P(A) = 0.6, P(B) = 0.7
\nI) P(A \u2229 B) = P(A) . P(B)
\n= 0.6 xO.7 = 0.42.<\/p>\n

II) P(A \u222a B) = P(A) + P(B) – P(A \u2229 B)
\n= P(A) + P(B) – P(A). P(B)
\n= 0.6 + 0.7 – (0.6) (0.7)
\n= 1.3 – 0.42 = 0.88<\/p>\n

III) \\(P\\left(\\frac{B}{A}\\right)=\\frac{P(A \\cap B)}{P(A)}\\)
\n= \\(\\frac{P(A) \\cdot P(B)}{P(A)}\\) = P(B) = 0.7<\/p>\n

IV) \\(P\\left(A^C \\cap B^C\\right)=P(A \\cup B)^C\\)
\n= 1 – P(A \u222a B)
\n= 1 – 0.88 = 0.12.<\/p>\n

Question 38.
\nThe probability that Australia wins a match against india in a cricket game is given to be \\(\\frac{1}{3}\\). If India and Australia play 3 matches what Is the probability that
\nI) Australia will loose all the 3 matches
\nII) Australia will win atleast one match. [May ’12]
\nSolution:
\nSuppose, A is the event at Australia winning the match.
\nP(A) = \\(\\frac{1}{3}\\)
\nI) Probability that Australia will loose all the three matches
\n= \\(P(\\overline{\\mathrm{A}}) \\cdot P(\\overline{\\mathrm{A}}) \\cdot P(\\overline{\\mathrm{A}})\\)
\n= \\(\\frac{2}{3} \\cdot \\frac{2}{3} \\cdot \\frac{2}{3}=\\frac{8}{27}\\).<\/p>\n

II) Probability that Australia will win atleast one match
\n= 1 – \\(P(\\overline{\\mathrm{A}}) \\cdot P(\\overline{\\mathrm{A}}) \\cdot P(\\overline{\\mathrm{A}})\\)
\n= \\(1-\\frac{8}{27}=\\frac{19}{27}\\).<\/p>\n

\"TS<\/p>\n

Question 39.
\nIn a shooting test, the probability of A, B, C hitting the targets are \\(\\frac{1}{2}\\), \\(\\frac{2}{3}\\) and \\(\\frac{3}{4}\\) respectively. If all of them fire at the same target. Find the probability that
\nI) Only one of them hits the target
\nII) Atleast one of them hits the target
\nSolution:
\nGiven that,
\nP(A) = \\(\\frac{1}{2}\\)
\n\u21d2 P(\\(\\overline{\\mathrm{A}}\\)) = 1 – P(A)
\n= 1 – \\(\\frac{1}{2}\\) = \\(\\frac{1}{2}\\)
\nP(B) = \\(\\frac{2}{3}\\)
\n\u21d2 P(\\(\\overline{\\mathrm{B}}\\)) =1 – P(B)
\n= 1 – \\(\\frac{2}{3}\\) = \\(\\frac{1}{3}\\)
\nP(C) = \\(\\frac{3}{4}\\)
\n\u21d2 P(\\(\\overline{\\mathrm{C}}\\)) =1 – P(C)
\n= 1 – \\(\\frac{3}{4}\\) = \\(\\frac{1}{4}\\)<\/p>\n

I) The probability that one of them hits the target<\/p>\n

\"TS<\/p>\n

II) The probability that atleast one of them hits the target<\/p>\n

\"TS<\/p>\n

Question 40.
\nDefine conditional events conditional probability. [TS – May 2015: March \u201811, 06, May 11, ’07]
\nSolution:
\nConditional event:
\nIf A, B are two events in a sample space then the event of happening B after the event A happening is called conditional event. It is denoted by \\(\\frac{B}{A}\\).<\/p>\n

Conditional Probabillty :
\nIf A, B are two events in a sample space S and P(A) \u2260 0 then the probability of B after the event A has occurred \u00a1s called conditional probability of B given A. It is denoted by P(\\(\\frac{B}{A}\\)).
\nWe define P(\\(\\frac{B}{A}\\)) = \\(\\frac{P(B \\cap A)}{P(A)}\\), P(A) \u2260 0.<\/p>\n

Question 41.
\nA pair of dice is thrown. Find the probability that either of the dice shows 2 when their sum is 6.
\nSolution:
\nLet S be the sample space.
\nLet S be the sum to 6, when two dice are rolled.
\nS = {1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
\n\u2234 n(S) = 5
\nLet A be the event that either of the dice shows 2
\n\u2234 A = {(2, 4), (4, 2))
\n\u2234 n(A)= 2
\n\u2234 Required probability, P(A) = \\(\\frac{n(A)}{n(S)}=\\frac{2}{5}\\).<\/p>\n

\"TS<\/p>\n

Question 42.
\nAn urn contains 7 red and 3 black balls. 2 balls are drawn without replacement. What is the propability that the 2\u2019 ball is red \u00a1fit is known that the 1 ball drawn is red? [Mar. ’05]
\nSolution:
\nTotal number of balls in urn = 10
\nRed balls = 7
\nBlack balls = 3
\nLet R1<\/sub> be the event of drawing the first ball red and R2<\/sub> be the event of drawing the second ball red.
\nThe probability that the first ball drawn is red is P(R1<\/sub>) = \\(\\frac{{ }^7 \\mathrm{C}_1}{{ }^{10} \\mathrm{C}_1}=\\frac{7}{10}\\)
\nNow, after one red ball is drawn out. 6 red balls and three black balls remain in the urn.
\nThe probability that the second ball is red it is known that the first ball drawn is red.
\n\\(\\mathrm{P}\\left(\\frac{\\mathrm{R}_2}{\\mathrm{R}_1}\\right)=\\frac{{ }^6 \\mathrm{C}_1}{{ }^9 \\mathrm{C}_1}=\\frac{6}{9}=\\frac{2}{3}\\).<\/p>\n

Question 43.
\nLet A and B be independent events with P(A) = 0.2, P(B) = 0.5. Find
\nI) \\(\\mathrm{P}\\left(\\frac{\\mathrm{A}}{\\mathrm{B}}\\right)\\)
\nII) \\(\\mathrm{P}\\left(\\frac{\\mathrm{B}}{\\mathrm{A}}\\right)\\)
\nIII) P(A \u2229 B)
\nIV) P(A \u222a B) [Mar. \u201812, \u201809, \u201806, May \u201811]
\nSolution:
\nSince, A and B are independent events then P(A \u2229 B) = P(A) . P(B)
\nGiven that,
\nP(A) = 0.2, P(B) = 0.5
\nI) \\(\\mathrm{P}\\left(\\frac{\\mathrm{A}}{\\mathrm{B}}\\right)\\) = P(A) = 0.2
\nII) \\(\\mathrm{P}\\left(\\frac{\\mathrm{B}}{\\mathrm{A}}\\right)\\) = P(B) = 0.5
\nIII) P(A \u2229 B) = P(A) . P(B)
\n= 0.2 \u00d7 0.5 = 0.01
\nIV) P(A \u222a B) = P(A) + P(B) – P(A \u2229 B)
\n= P(A) + P(B) – P(A). P(B)
\n= 0.2 + 0.5 – 0.1 = 0.6.<\/p>\n

Question 44.
\nBag B1<\/sub> contains 4 white and 2 black balls. Bag B2<\/sub> contains 3 whIte and 4 black balls. A bag is drawn at random and a ball is choosen at random from it. What is the probability that the ball drawn is white? [March \u201911, May ’11]
\nSolution:
\nProbability of selecting first bag = \\(\\frac{1}{2}\\)
\nProbability of selecting second bag = \\(\\frac{1}{2}\\)
\nProbability of getting a white ball from first bag = \\(\\frac{{ }^4 C_1}{{ }^6 C_1}=\\frac{4}{6}=\\frac{2}{3}\\)
\nProbability of getting a white ball from second bag = \\(\\frac{{ }^3 \\mathrm{C}_1}{{ }^7 \\mathrm{C}_1}=\\frac{3}{7}\\)
\nProbability of drawing a white ball = \\(\\frac{1}{2} \\cdot \\frac{2}{3}+\\frac{1}{2} \\cdot \\frac{3}{7}=\\frac{1}{3}+\\frac{3}{14}=\\frac{23}{42}\\).<\/p>\n

\"TS<\/p>\n

Question 45.
\nDefine mutually exclusive events and give an example. [May 2013]
\nSolution:
\nMutually Exclusive Events:
\nTwo events A, B in a sample space, S are said to be disjoint or mutually exclusive if A \u2229 B = \u03a6
\nE.g.:
\nThe events {1, 2}, {3, 5) are disjoint in the sample space, S = {1, 2, 3, 4, 5, 6}<\/p>\n

Question 46.
\nSuppose there are 12 boys and 4 girls in a class if we choose 3 children one after another in succession at random, find the probability that all the 3 are boys.
\nSolution:
\nTotal no. of children = 16
\nBoys = 12
\nGirls = 4
\nThe probability ol selecting first child is boy is \\(\\frac{{ }^{12} \\mathrm{C}_1}{{ }^{16} \\mathrm{C}_1}=\\frac{12}{16}=\\frac{3}{4}\\)
\nIf the first child selected is boy, 15 children are left and 11 out of them are boys.
\nThe probability that the second child selected is boy is \\(\\frac{{ }^{11} \\mathrm{C}_1}{{ }^{15} \\mathrm{C}_1}=\\frac{11}{15}\\)
\nIf the first two children selected are boys, 14 children are left and 10 out of them are boys.
\nThe probability that the third child selected is boy is \\(\\frac{{ }^{10} \\mathrm{C}_1}{{ }^{14} \\mathrm{C}_1}=\\frac{10}{14}=\\frac{5}{7}\\)
\nHence, by the multiplication theorem the required probability = \\(\\frac{3}{4} \\cdot \\frac{11}{15} \\cdot \\frac{5}{7}=\\frac{11}{28}\\).<\/p>\n

Question 47.
\n\u2018A\u2019 speaks truth In 75% of the cases and \u2018B\u2019 in 80% cases. What is the probability that their statements about an incident do not match? [AP & TS – Mar. 2016; May ’10, AP – Mar. 2019]
\nSolution:
\nLet A be the event that ‘A\u2019 speaks truth about an incident.
\nP(A) = \\(\\frac{75}{100}=\\frac{3}{4}\\)
\n\\(P\\left(A^c\\right)\\) = 1 – P(A)
\n= 1 – \\(\\frac{3}{4}\\) = \\(\\frac{1}{4}\\)
\nLet B be the event that ‘B’ speaks truth about an incident
\nP(B) = \\(\\frac{80}{100}=\\frac{4}{5}\\)
\n\\(P\\left(B^c\\right)\\) = 1 – P(B)
\n= 1 – \\(\\frac{4}{5}\\) = \\(\\frac{1}{5}\\)
\nThe probability that their statements about an incident do not match = \\(P\\left(A \\cap B^C\\right)+P\\left(A^C \\cap B\\right)\\)
\nSince A, B are independent events = \\(P(A) \\cdot P\\left(B^C\\right)+P\\left(A^C\\right) \\cdot P(B)\\)
\n= \\(\\frac{3}{4} \\cdot \\frac{1}{5}+\\frac{1}{4} \\cdot \\frac{4}{5}\\)
\n= \\(\\frac{3}{20}+\\frac{1}{5}=\\frac{3+4}{20}=\\frac{7}{20}\\).<\/p>\n

\"TS<\/p>\n

Question 48.
\nA problem in Calculus is given to two students A and B whose chances of solving it are \\(\\frac{1}{3}\\) and \\(\\frac{1}{4}\\) respectively. Find the probability of the problem being solved if both of them try independently. [AP – Mar. \u201818, \u201815, May \u201815; Mar.\u201905, Board Paper]
\nSolution:
\nLet A be the event that the problem is solved by \u2018A\u2019.
\nP(A) = \\(\\frac{1}{3}\\)
\nLet B be the event that the problem is solved by B\u2019
\nP(B) = \\(\\frac{1}{4}\\)
\nThe probability of the problem being solved is
\nP(A \u222a B) = P(A) + P(B) – P(A \u2229 B)
\nSince A, B are independent events = P(A) + P(B) – P(A) . P(B)
\n= \\(\\frac{1}{3}+\\frac{1}{4}-\\frac{1}{3} \\cdot \\frac{1}{4}\\)
\n= \\(\\frac{1}{3}+\\frac{1}{4}-\\frac{1}{12}\\)
\n= \\(\\frac{4+3-1}{12}=\\frac{6}{12}=\\frac{1}{2}\\).<\/p>\n

Question 49.
\nA and B toss a coin 50 times each simultaneously. Find the probability that both of them will not get tails at the same toss.
\nSolution:
\nLet E be the event that A and B both will not get tails at the same toss.
\nIn each toss, we have the following 4 choices:
\n1) A : Head B : Head
\n2) A : Head B : Tall
\n3) A : Tail B : Head
\n4) A : Tail B : Tail
\nL.et S be the sample space.
\nThere are 50 Trials.
\nThe total no. of choices, n(S) = 4.4.4 …………………….4 (50 times) = 450<\/sup>
\nLet E be the event that A and B both will not get tails at same toss.
\nn(E) = 3.3.3. 3 (50 times) = 350<\/sup>
\nThe probability that both of them will not get tails at the same toss.
\nP(E) = \\(\\frac{n(E)}{n(S)}=\\frac{3^{50}}{4^{50}}=\\left(\\frac{3}{4}\\right)^{50}\\)<\/p>\n

\"TS<\/p>\n

Question 50.
\nIf A and B are independent events of a random experiment then show that A and BC are also independent. [Oct. ’99], [TS – May 2016]
\nSolution:
\nSince, A and B are independent events then
\nP(A \u2229 B) = P(A) . P(B)
\nNow, \\(P\\left(A^C \\cap B^C\\right)=P(A \\cup B)^C=P(\\overline{A \\cup B})\\)
\n= 1 – P (A \u222a B)
\n= 1 – [P(A) + P(B) – P(A \u2229 B)]
\n= 1 – P(A) – P(B) . P(A \u2229 B)
\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))
\n= \\(P\\left(A^C\\right) \\cdot P\\left(B^C\\right)\\)
\n\u2234 \\(A^C\\) and \\(B^C\\) are independent.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type to help strengthen their preparations for exams. TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type Question 1. If four fair coins are tossed simultaneously then find the probability that two heads … 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 Short Answer Type - TS Board 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:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type - TS Board Solutions\" \/>\n<meta property=\"og:description\" content=\"Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type to help strengthen their preparations for exams. TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type Question 1. If four fair coins are tossed simultaneously then find the probability that two heads ... Read more\" \/>\n<meta property=\"og:url\" content=\"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/\" \/>\n<meta property=\"og:site_name\" content=\"TS Board Solutions\" \/>\n<meta property=\"article:published_time\" content=\"2022-12-07T06:41:25+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2022-12-07T10:25:31+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/TS-Board-Solutions.png\" \/>\n<meta name=\"author\" content=\"Mahesh\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Mahesh\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"34 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/\"},\"author\":{\"name\":\"Mahesh\",\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/person\/32b03c2fe28184bd0170db96f7ed6aae\"},\"headline\":\"TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type\",\"datePublished\":\"2022-12-07T06:41:25+00:00\",\"dateModified\":\"2022-12-07T10:25:31+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/\"},\"wordCount\":6790,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/tsboardsolutions.com\/#organization\"},\"articleSection\":[\"TS Inter 2nd Year\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/\",\"url\":\"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/\",\"name\":\"TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type - TS Board Solutions\",\"isPartOf\":{\"@id\":\"https:\/\/tsboardsolutions.com\/#website\"},\"datePublished\":\"2022-12-07T06:41:25+00:00\",\"dateModified\":\"2022-12-07T10:25:31+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/tsboardsolutions.com\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type\"}]},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/tsboardsolutions.com\/#website\",\"url\":\"https:\/\/tsboardsolutions.com\/\",\"name\":\"TS Board Solutions\",\"description\":\"Telangana TS Board Textbook Solutions for Class 6th, 7th, 8th, 9th, 10th, Inter 1st & 2nd Year\",\"publisher\":{\"@id\":\"https:\/\/tsboardsolutions.com\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/tsboardsolutions.com\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"Organization\",\"@id\":\"https:\/\/tsboardsolutions.com\/#organization\",\"name\":\"TS Board Solutions\",\"url\":\"https:\/\/tsboardsolutions.com\/\",\"logo\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/logo\/image\/\",\"url\":\"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/cropped-TS-Board-Solutions-1.png\",\"contentUrl\":\"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/cropped-TS-Board-Solutions-1.png\",\"width\":488,\"height\":40,\"caption\":\"TS Board Solutions\"},\"image\":{\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/logo\/image\/\"}},{\"@type\":\"Person\",\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/person\/32b03c2fe28184bd0170db96f7ed6aae\",\"name\":\"Mahesh\",\"image\":{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/tsboardsolutions.com\/#\/schema\/person\/image\/\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/460dd5c928927786394035e387cc998d?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/460dd5c928927786394035e387cc998d?s=96&d=mm&r=g\",\"caption\":\"Mahesh\"},\"url\":\"https:\/\/tsboardsolutions.com\/author\/mahesh\/\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type - TS Board Solutions","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/","og_locale":"en_US","og_type":"article","og_title":"TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type - TS Board Solutions","og_description":"Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type to help strengthen their preparations for exams. TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type Question 1. If four fair coins are tossed simultaneously then find the probability that two heads ... Read more","og_url":"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/","og_site_name":"TS Board Solutions","article_published_time":"2022-12-07T06:41:25+00:00","article_modified_time":"2022-12-07T10:25:31+00:00","og_image":[{"url":"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/TS-Board-Solutions.png"}],"author":"Mahesh","twitter_card":"summary_large_image","twitter_misc":{"Written by":"Mahesh","Est. reading time":"34 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/#article","isPartOf":{"@id":"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/"},"author":{"name":"Mahesh","@id":"https:\/\/tsboardsolutions.com\/#\/schema\/person\/32b03c2fe28184bd0170db96f7ed6aae"},"headline":"TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type","datePublished":"2022-12-07T06:41:25+00:00","dateModified":"2022-12-07T10:25:31+00:00","mainEntityOfPage":{"@id":"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/"},"wordCount":6790,"commentCount":0,"publisher":{"@id":"https:\/\/tsboardsolutions.com\/#organization"},"articleSection":["TS Inter 2nd Year"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/","url":"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/","name":"TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type - TS Board Solutions","isPartOf":{"@id":"https:\/\/tsboardsolutions.com\/#website"},"datePublished":"2022-12-07T06:41:25+00:00","dateModified":"2022-12-07T10:25:31+00:00","breadcrumb":{"@id":"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/tsboardsolutions.com\/maths-2a-probability-important-questions-short-answer-type\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/tsboardsolutions.com\/"},{"@type":"ListItem","position":2,"name":"TS Inter Second Year Maths 2A Probability Important Questions Short Answer Type"}]},{"@type":"WebSite","@id":"https:\/\/tsboardsolutions.com\/#website","url":"https:\/\/tsboardsolutions.com\/","name":"TS Board Solutions","description":"Telangana TS Board Textbook Solutions for Class 6th, 7th, 8th, 9th, 10th, Inter 1st & 2nd Year","publisher":{"@id":"https:\/\/tsboardsolutions.com\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/tsboardsolutions.com\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/tsboardsolutions.com\/#organization","name":"TS Board Solutions","url":"https:\/\/tsboardsolutions.com\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/tsboardsolutions.com\/#\/schema\/logo\/image\/","url":"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/cropped-TS-Board-Solutions-1.png","contentUrl":"https:\/\/tsboardsolutions.com\/wp-content\/uploads\/2022\/10\/cropped-TS-Board-Solutions-1.png","width":488,"height":40,"caption":"TS Board Solutions"},"image":{"@id":"https:\/\/tsboardsolutions.com\/#\/schema\/logo\/image\/"}},{"@type":"Person","@id":"https:\/\/tsboardsolutions.com\/#\/schema\/person\/32b03c2fe28184bd0170db96f7ed6aae","name":"Mahesh","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/tsboardsolutions.com\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/460dd5c928927786394035e387cc998d?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/460dd5c928927786394035e387cc998d?s=96&d=mm&r=g","caption":"Mahesh"},"url":"https:\/\/tsboardsolutions.com\/author\/mahesh\/"}]}},"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/posts\/39697"}],"collection":[{"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/comments?post=39697"}],"version-history":[{"count":4,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/posts\/39697\/revisions"}],"predecessor-version":[{"id":39738,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/posts\/39697\/revisions\/39738"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/media?parent=39697"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/categories?post=39697"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.com\/wp-json\/wp\/v2\/tags?post=39697"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}