{"id":22103,"date":"2024-01-10T16:28:17","date_gmt":"2024-01-10T10:58:17","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22103"},"modified":"2024-01-11T12:39:43","modified_gmt":"2024-01-11T07:09:43","slug":"inter-2nd-year-maths-2a-probability-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-2nd-year-maths-2a-probability-important-questions\/","title":{"rendered":"Inter 2nd Year Maths 2A Probability Important Questions"},"content":{"rendered":"

Students get through Maths 2A Important Questions<\/a> Inter 2nd Year Maths 2A Probability Important Questions which are most likely to be asked in the exam.<\/p>\n

Intermediate 2nd Year Maths 2A Probability Important Questions<\/h2>\n

Question 1.
\nIf 4 fair coins are tossed simultaneously, then find the probability that 2 heads and 2 tails appear. (Mar. ’08)
\nSolution:
\n4 coins are tossed simultaneously.
\nTotal number of ways = 24<\/sup> = 16
\nn(S) = 16
\nFrom 4 heads we must get 2 heads.
\nA number of ways of getting 2 heads.
\n= 4<\/sup>C2<\/sub> = \\(\\frac{4.3}{1.2}\\) = 6
\n\u2234 n(E) = 6
\nP(E) = \\(\\frac{n(E)}{n(S)}\\) = \\(\\frac{6}{16}\\) = \\(\\frac{3}{8}\\)
\n\u2234 Probability of getting 2 heads and 2 tails = \\(\\frac{3}{8}\\)<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the probability that a non – leap year contains
\ni) 53 Sundays
\nii) 52 Sundays only. (Mar. ’07, May ’06)
\nSolution:
\nA non – leap year contains 35 days 52 weeks and 1 day more.
\ni) We get 53 sundays when the remaining day is Sunday.
\nNumber of days in the week = 7
\n\u2234 n(S) = 7
\nNumber of ways getting 53 Sundays.
\nn(E) = 1
\n\u2234 P(E) = \\(\\frac{n(E)}{n(S)}\\) = \\(\\frac{1}{7}\\)
\n\u2234 Probability of getting 53 Sundays = \\(\\frac{1}{7}\\)<\/p>\n

ii) Probability of getting 52 Sundays
\nP(E) = 1 – P(E)
\n= 1 – \\(\\frac{1}{7}\\) = \\(\\frac{6}{7}\\)<\/p>\n

Question 3.
\nIf one ticket is randomly selected from tickets numbered 1 to 30. Then find the probability that the number on the ticket is a multiple of 3 or 5 (Mar. \u201808)
\nSolution:
\nSuppose A is the event of getting a multiple of 3 and B is the event of getting a multiple of 5.
\nA = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30}
\nB = {5, 10, 15, 20, 25, 30}
\nA \u2229 B = {15, 30}
\n\"Inter<\/p>\n

Question 4.
\nIf two numbers are selected randomly from 20 consecutive natural numbers, find the probability that the sum of the two numbers is
\n(i) an even number
\n(ii) an odd number. (Mar. \u201808)
\nSolution:
\ni) Let A be the event that the sum of the numbers is even when two numbers are selected out of 20 consecutive natural numbers.
\nIn 20 consecutive natural numbers, we have 10 odd and 10 even natural numbers.
\nThe sum of two odd natural numbers is an even number and the sum of two even natural numbers is also an even number
\n\"Inter<\/p>\n

ii) Probability that the sum of two numbers is an odd number
\nP(\\(\\bar{A}\\)) = 1 – P(A) = 1 – \\(\\frac{9}{19}\\) = \\(\\frac{10}{19}\\)<\/p>\n

\"Inter<\/p>\n

Question 5.
\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 on contract is \\(\\frac{4}{5}\\). Find the probability to get both the contracts. (AP Mar. \u201816)
\nSolution:
\nSuppose A is the event of getting a road contract.
\nB is the event of getting a building contract
\n\"Inter<\/p>\n

Question 6.
\nIf one card is drawn at random from a pack of cards then show that event of getting an ace and getting heart are independent events. (Mar. 13)
\nSolution:
\nSuppose A is the event of getting an ace and B is the event of getting a heart.
\n\u2234 P(A) = \\(\\frac{4}{52}\\) = \\(\\frac{1}{13}\\)
\nP(B) = \\(\\frac{13}{52}\\) = \\(\\frac{1}{14}\\)
\nA \u2229 B is the event of getting a Hearts ace
\nP(A \u2229 B) = \\(\\frac{1}{52}\\) = \\(\\frac{1}{13}\\).\\(\\frac{1}{4}\\) = P(A).P(B)
\n\u2234 A and B are independent events.<\/p>\n

13 If A. B, are two events with P(A \u222a B) = 0.65 and P(A \u2229 B) = 0.15, then find the value of p(AC<\/sup>) + P (BC<\/sup>). (TS Mar. ’15, ’13, ’05; May ’11)
\nSolution:
\nBy addition theorem on probability
\nP(A \u222a B) = P(A) + P(B) – P(A \u2229 B)
\nP(A) + P(B) = P(A \u222a B) + P(A \u2229 B)
\n= 0.65 + 0.15
\n= 0.8 —— (1)
\nP(AC<\/sup>) + P(BC<\/sup>) = 1 – P(A) + 1 – P(B)
\n= 2 – [P(A) + P(B)]
\n= 2 – [P(A) + P(B)]
\n= 2 – 0.8 by (1)
\n= 1.<\/p>\n

Question 7.
\nSuppose A and B are independent events with P(A) = 0.6, P(B) = 0.7 then compute
\n(i) P(A \u2229 B)
\n(ii) P(A \u222a B)
\n(iii) \\(\\mathbf{P}\\left(\\frac{\\mathbf{B}}{\\mathrm{A}}\\right)\\)
\n(iv) P(AC<\/sup> \u2229 BC<\/sup>). (AP Mar. ’17; Mar. \u201814)
\nSolution:
\nGiver A, B are independent events and
\nP(A) 0.6, P(B) = 0.7<\/p>\n

i) P(A \u2229 B) = P(A) P(B) = 0.6 \u00d7 0.7 = 0.42<\/p>\n

ii) P(A \u222a B) = 1(A) + P(B) – P(A \u2229 B)
\n= 0.6 + 0.7 – 0.42
\n= 1.3 – 0.42 = 0.88<\/p>\n

iii) \\(\\mathrm{P}\\left(\\frac{\\mathrm{B}}{\\mathrm{A}}\\right)\\) = P(B) = 0.7<\/p>\n

iv) P(AC<\/sup> \u2229 BC<\/sup>) = P(AC<\/sup>). P(BC<\/sup>)
\n(AC<\/sup> & BC<\/sup> are also independent events)
\n= [1 – P(A) [1- P(B)]
\n= (1 – 0.6)(1 – 0.7)
\n= 0.4 \u00d7 0.3 = 0.12<\/p>\n

\"Inter<\/p>\n

Question 8.
\nFind the probability of drawing an ace or a spade from a well shuffled pack of 52 cards? (TS Mar. ’17, ’15)
\nSolution:
\nHint : A pack of cards means of pack containing 52 cards, 26 of them are red and 26 of them are black coloured. These 52 cards are divided into 4 sets namely hearts, Spades, Diamonds and Clubs. Each set contains of 13 cards names, A, 2, 3, 4, 5, 6, 7, 8, 9, 10, K, Q, J.
\nLet E1<\/sub> be the event of drawing a spade and E2<\/sub> be the event of drawing an ace. E1<\/sub>, E2<\/sub> are not mutually exclusive.
\nn(A) = 13, n(B) = 4, n(A \u2229 B) = 1
\n= \\(\\frac{13}{52}\\) + \\(\\frac{4}{52}\\) – \\(\\frac{1}{52}\\) = \\(\\frac{16}{52}\\) = \\(\\frac{4}{13}\\)<\/p>\n

Question 9.
\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) +
\nP(A \u2229 B \u2229 C). (AP Mar. ’15)
\nSolution:
\nWrite B \u222a C = D then P(A \u222a B \u222a C) = P(A \u222a D)
\n\u2234 P(A \u222a D) = P(A) + P(D) – P(A \u2229 D)
\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 \u222a C) – [P(A \u2229 B) + P(A \u2229 C) – P(A \u2229 B \u2229 D \u2229 C]
\nP(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

Question 10.
\nA speaks the truth in 75% of the cases; B in 80% \u00e7ases. What is the probability that their statements about an incident do not match? (TS & AP Mar.\u201916)
\nSolution:
\nLet E1<\/sub>, E2<\/sub> be the events that A and B respectively speak truth about an incident.
\n\"Inter
\nLet E be the event that their statements do not match about the incident. Then this happens in two mutually exclusive ways.
\ni) A speaks truth, B tells lie
\nii) A tells lie, B speaks truth. These two events are represented by E1<\/sub> \u2229 \\(E_{2}^{c}\\), \\(E_{1}^{c}\\), \u2229 E2<\/sub>
\n\"Inter<\/p>\n

Question 11.
\nA problem in Calculus is given to two students A and B whose chances of solving it are 1\/3 and 1\/4. What is the probability that the problem will be solved if both of them try independently ? (Mar.\u201915, ’05)
\nSolution:
\nLet E1<\/sub> and E2<\/sub> denote the events that the problem is solved by A and B respectively.
\nGiven that
\nP(E1<\/sub>) = \\(\\frac{1}{3}\\) and P(E2<\/sub>) = \\(\\frac{1}{4}\\)
\nNote that these two are independent events. Therefore the required probability
\nP(E1<\/sub> \u222a E2<\/sub>) = P(E1<\/sub>) + P(E2<\/sub>) – P(E1<\/sub> \u2229 E2<\/sub>)
\n= P(E1<\/sub>) + P(E2<\/sub>) – P(E1<\/sub>) P(E2<\/sub>)
\n(\u2235 E1<\/sub>, E2<\/sub> are independent)
\n= \\(\\frac{1}{3}\\) + \\(\\frac{1}{4}\\) – \\(\\frac{1}{12}\\) = \\(\\frac{1}{2}\\)<\/p>\n

\"Inter<\/p>\n

Question 12.
\nSuppose A 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 occur.
\nii) neither A nor B occurs. (TS Mar.\u201917)
\nSolution:
\nWe have AC<\/sup> = the event \u201cA does not occur\u201d
\n(A \u222a B)C<\/sup> = neither A nor B occurs.
\n\u2234 P(AC<\/sup>) = 1 – P(A) = 1 – 0.5 = 0.5
\nSince P(A \u222a B) = P(A) + P(B) – P(A \u2229 B)
\n= 0.5 + 0.4 – 0.3
\n= 0.6
\nP[(A \u03bd BC<\/sup>)] = 1 – P (A \u222a B)
\n= 1 – 0.6 = 0.4<\/p>\n

Question 13.
\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. 16)
\nSolution:
\nWhen a person is chosen at random from the academy consistin9 of 25 members, let A be the event that the person is proficient in Mathematics, B be the event that the person is proficient in Statistics and S be the sample space. Since 19 members are proficient in Mathematics and 16 members ar\u00e9 proficient in Statistics.
\nP(A) = \\(\\frac{19}{25}\\), P(B) = \\(\\frac{16}{25}\\)
\nSince every one is either proficient in Mathematics or Statistics or in both
\n\"Inter<\/p>\n

Question 14.
\nA, B, C are three 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? (Mar. \u201814, ’03)
\nSolution:
\nLet A, B, C be the events that the horses A, B, C win the race respectively,
\nGiven P(A) = 2P(B), P(B) = 2P(C)
\n\u2234 P(A) = 2P(B) = 2[2P(C)] = 4P(C)
\nSince the horses A, B and C run the race,
\nA \u222a B \u222a C = S and A, B, C are mutually disjoint.
\n\"Inter<\/p>\n

Question 15.
\nIf A, B, C are three independent events of an experiment such that P(A \u2229 BC<\/sup> \u2229 CC<\/sup>) = \\(\\frac{1}{4}\\), P(AC<\/sup> \u2229 BC<\/sup> \u2229 CC<\/sup>) = \\(\\frac{1}{4}\\) then find P(A), P(B) and P(C). (TS Mar. ’15)
\nSolution:
\nSince A, B, C are independent events
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 16.
\nAddition theorem on probability: Statement: If E1<\/sub>, E2<\/sub> an any two events of a random experiment and P is a probability function, then P(E1<\/sub> \u222a E2<\/sub>) = P(E1<\/sub>) + P(E2<\/sub>) – P(E1<\/sub> \u2229 E2<\/sub>)
\nSolution:
\nProof: Case (i) E1<\/sub> \u2229 E2<\/sub> = \u03c6
\nThen P(E1<\/sub> \u2229 E2<\/sub>) = 0
\n\u2234 P(E1<\/sub> \u222a E2<\/sub>) = P(E1<\/sub>) + P(E2<\/sub>)
\n= P(E1<\/sub>) + P(E2<\/sub>) – 0
\n= P(E1<\/sub>) + P(E2<\/sub>) – P(E1<\/sub> \u2229 E2<\/sub>)<\/p>\n

Case (ii) : Suppose E1<\/sub> \u2229 E2<\/sub> \u2260 \u03c6
\nThen E1<\/sub> \u222a E2<\/sub> = E1<\/sub> \u222a (E2<\/sub> – E1<\/sub>) and E1<\/sub> \u2229 (E2<\/sub> – E1<\/sub>) = 4
\n\u2234 P(E1<\/sub> \u222a E2<\/sub>) = P[E1<\/sub> \u222a (E2<\/sub> – E1<\/sub>)]
\n= P(E1<\/sub>) + P(E2<\/sub> – E1<\/sub>)
\n= P(E1<\/sub>) + P[E2<\/sub> – (E1<\/sub> \u2229 E2<\/sub>)
\nSince E2<\/sub> \u2229 (E1<\/sub> \u2229 E2<\/sub>) = \u03c6.
\n= P(E1<\/sub>) + P(E2<\/sub>) – P(E1<\/sub> \u2229 E2<\/sub>)
\nHence P(E1<\/sub> \u222a E2<\/sub>) = P(E1<\/sub>) + (E2<\/sub>) – P(E1<\/sub> \u2229 E2<\/sub>).<\/p>\n

Question 17.
\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 balck
\nOne of the Urn is selected at random and a ball is drawn. Pt turns out to be white. Find the probability that it come from Urn III. (AP. Mar. ’17)
\nSolution:
\nLet Ei<\/sub> be the event of Choosing the Urn i = 1, 2, 3 and P(Ei<\/sub>) be the probability of
\nchoosing the Urn i = 1, 2, 3. Then P(E1<\/sub>) = P(E2<\/sub>) = P(E3<\/sub>) = \\(\\frac{1}{3}\\).
\nHaving choose b the Urn i, the probability of drawing a white ball, P(W\/Ei<\/sub>), is given by
\nP(W\/E1<\/sub>) = \\(\\frac{1}{3}\\),
\nP(W\/E2<\/sub>) = \\(\\frac{2}{3}\\)
\nP(W\/E3<\/sub>) = \\(\\frac{2}{4}\\)
\nWe have to find the probability P(E3<\/sub>\/W) by Baye’s theorem.
\nP(E3<\/sub>\/W) =
\n\"Inter<\/p>\n

Question 18.
\nFind the Probability, of getting the same number on both the dice when two dice are thrown.
\nSolution:
\nLet E be the event of getting the same number on both the dice when two dice are thrown and S be the sample space.
\n\u2234 n(S) = 62<\/sup> = 36
\nE = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
\nn(E) = 6
\nP(E) = \\(\\frac{n(E)}{n(S)}\\)
\n= \\(\\frac{6}{36}\\)
\n= \\(\\frac{1}{6}\\)<\/p>\n

Question 19.
\nAn integer in picked from 1 to 20, both inclusive. Find the probability that it is a prime.
\nSolution:
\nLet E be the event that the number picked from 1 to 20 is a prime and S be the sample space.
\n\u2234 n(S) = 20<\/sup>C1<\/sub> = 20
\nE = {2, 3, 5, 7, 11, 13, 17, 19}
\nn(E) = 8
\nP(E) = \\(\\frac{n(E)}{n(S)}\\) = \\(\\frac{8}{20}\\) = \\(\\frac{2}{5}\\)<\/p>\n

\"Inter<\/p>\n

Question 20.
\nA bag contains 4 red, 5 black and 6 blue balls. Find the probability that two balls drawn at random simultaneously from the bag are a red and a black ball.
\nSolution:
\nLet E be the event that getting a red and black ball when two balls are drawn at random from a bag containing 4 red, 5 black, 6 blue balls and S be the sample space. Total no.of Balls = 4 + 5 + 6 = 15
\nn(S) = 15<\/sup>C2<\/sub>
\nn(E) = 4<\/sup>C1<\/sub>. 5<\/sup>C1<\/sub>
\n\u2234 P(E) = \\(\\frac{{ }^{4} C_{1} \\cdot{ }^{5} C_{1}}{{ }^{15} C_{2}}\\)
\n= \\(\\frac{4.5}{105}\\)
\n= \\(\\frac{4}{21}\\)<\/p>\n

Question 21.
\nTen dice are thrown. Find the probability that none of the dice shows the number 1.
\nSolution:
\nLet A be the event that none of the dice shows the numbers 1 when the dice thrown.
\nn(S) = 610<\/sup>
\nn(A) = 510<\/sup>
\nP(A) = \\(\\frac{n(A)}{n(S)}\\)
\n= \\(\\frac{5^{10}}{6^{10}}\\)
\n= \\(\\left(\\frac{5}{6}\\right)^{10}\\)<\/p>\n

Question 22.
\nA number x is drawn arbitrarily from the set {1, 2, 3, ……. 100}. What is the probability that (x + \\(\\frac{100}{x}\\)) > 29
\nSolution:
\nHere the total number of cases is 100.
\nLet A be the event that x selected from the set {1, 2, 3, ……. 100} has the property
\nx + \\(\\frac{100}{x}\\) > 29
\nNow x + \\(\\frac{100}{x}\\) > 29
\n\u21d4 x2<\/sup> – 29x + 100 > 0
\n\u21d4 (x – 4) (x – 25) > 0
\n\u21d4 x > 25 or x < 4
\n\u21d4 x \u2208 {1, 2, 3, 26. 27, … 100) = A (say),
\nso that the number of cases favourable to A is 78
\n\u2234 The required probability P(A) = \\(\\frac{78}{100}\\)<\/p>\n

Question 23.
\nTwo squares are choosen at random on a chess board. Show that the probability that they have a side in common is \\(\\frac{1}{18}\\).
\nSolution:
\nThe number of ways of choosing the first square is 64 and that of the second is 63
\n\u2234 The number of ways of choosing the first and second squares = 64 \u00d7 63
\n\u2234 n(S) = 64 \u00d7 63
\nLet E be the event that these squares have a side in common.<\/p>\n

If the first square happens to be one of the squares in the four corners of the chess board, the second square (with common side) can be choosen in 2 ways.<\/p>\n

If the first square happens to be any one of the remaining 24 squares along the four sides of the chess board other than the corner, the second squ\u00e4re can be choosen, in 3 ways.<\/p>\n

If the first square happens to be any one of the remaining 36 inner squares, then the second square can be choosen in 4 ways.<\/p>\n

Hence the number of cases favourable to E is (4 \u00d7 2) + (24 \u00d7 3) + (36 \u00d7 4) = 224
\n\u2234 The required probability =
\n\\(\\frac{n(E)}{n(S)}\\) = \\(\\frac{224}{64 \\times 63}\\) = \\(\\frac{1}{18}\\)<\/p>\n

\"Inter<\/p>\n

Question 24.
\nA fair coin is tossed 200 times. Find the probability of getting a head an odd number of times.
\nSolution:
\nThe total number of cases is 2200<\/sup>
\nThe number of favourable cases is
\n= 200<\/sup>C1<\/sub> + 200<\/sup>C3<\/sub> + 200<\/sup>C5<\/sub> + ….. + 200<\/sup>C199<\/sub>
\n= \\(\\frac{2^{200}}{2}\\)
\n= 2199<\/sup>
\n\u2234 Probability = \\(\\frac{2^{191}}{2^{200}}\\) = \\(\\frac{1}{2} .\\)<\/p>\n

Question 25.
\nA and B are among 20 persons sit at random along a round table. Find the probability that there are any 6 persons
\nbetween A and B.
\nSolution:
\nLet \u2018A\u2019 occupy any seat around the table. Then ther\u00e9 are 19 seats available for B. But if there are to be six persons between A and B, then B has only two ways to sit.
\n\u2234 Probability = [\/latex] = \\(\\frac{2}{19}\\)<\/p>\n

Question 26.
\nOut of 30 consecutive integers two are drawn at random. Then what is the probability that their sum is odd.
\nSolution:
\nThe total number of ways of choosing 2 out of 30 numbers = 30<\/sup>C2<\/sub>
\nOut of these 30 numbers, 15 are even and 15 are odd.
\nFor the sum of the choosen two numbers to be odd, one should be odd and the other even.
\n\u2234 The number of cases favourable
\n\"Inter<\/p>\n

Question 27.
\nOut of 1,00,000 new born babies 77,181 survived till the age of 20. Find the probability that a new baby survives till 20 years of age.
\nSolution:
\nHere m = 77,181
\nn = 1,00,000
\nRequired probability = \\(\\frac{\\mathrm{m}}{\\mathrm{n}}\\)
\n= \\(\\frac{77,181}{1,00,000}\\)
\n= 0.77181.<\/p>\n

Question 28.
\nAddition theorem on probability: Statement: If E1<\/sub>, E2<\/sub> an any two events of a random experiment and P is a probability function, then P(E1<\/sub> \u222a E2<\/sub>) = P(E1<\/sub>) + P(E2<\/sub>) – P(E1<\/sub> \u2229 E2<\/sub>) (Mar. \u201814, \u201813)
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 29.
\nFind is the probability of throwing a total score of 7 with two dice.
\nSolution:
\nLet S be the sample space and A be the event of getting a total score of 7 when two dice are thrown
\nS = {(1, 1) (1, 2)…. (1, 6), (2, 1) …. (2, 6) … (6, 1), (6, 2) …. (6, 6)}
\nn(S) = 36
\nHint: 62<\/sup> = 36
\nA = {(1, 6) (6, 1), (2, 5) (5,2) (3,4) (4, 3)}
\nn(A) = 6
\n\u2234 P(A) = \\(\\frac{n(A)}{n(S)}\\) = \\(\\frac{6}{36}\\) = \\(\\frac{1}{6}\\)<\/p>\n

\"Inter<\/p>\n

Question 30.
\nFind the probability of obtaining two tails and one head when three coins are tossed.
\nSolution:
\nL\u00e8t S be the sample space and A be the event of getting two tails and one head when three coins are tossed.
\nn(S) = 23<\/sup> = 8
\nA = [H T T, T H T, T T H]
\nn(A) = 3
\nP(A) = \\(\\frac{3}{8}\\)<\/p>\n

Question 31.
\nA page is opened at random from a book containing 200 pages. What is the probability that the number of the page is a perfect square?
\nSolution:
\nLet S be the sample space. Let A be the event of getting on the page is perfect square.
\nn(S) = 200
\nLet A be the event of drawing a page whose number is perfect square.
\nA = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196}
\nn(A) = 14
\nP(A) = \\(\\frac{14}{200}\\) = \\(\\frac{7}{100}\\) = 0.07<\/p>\n

Question 32.
\nFind the probability of drawing an ace or a spade from a well shuffled pack of 52 cards? (T.S. Mar. \u201815)
\nSolution:
\nHint : A pack of cards means of pack containing 52 cards, 26 of them are red and 26 of them are black coloured. These 52 cards are divided into 4 sets namely Hearts, Spades, Diamonds and Clubs. Each set contains of 13 cards names, A, 2, 3, 4, 5, 6, 7, 8, 9, 10, K, Q, J.
\nLet E1<\/sub> be the event of drawing a spade and E2<\/sub> be the event of drawing an ace. E1<\/sub>, E2<\/sub> are not mutually exclusive.
\n\u2234 P(A \u222a B) = P(A) + P(B) – P(A \u2229 B)
\n= \\(\\frac{13}{52}\\) + \\(\\frac{4}{52}\\) – \\(\\frac{1}{52}\\) = \\(\\frac{16}{52}\\) = \\(\\frac{4}{13}\\)<\/p>\n

If A and B are two events, show that
\ni) P(A \u2229 BC<\/sup>) = P(A) – P(A \u2229 B) and
\nii) the probability that one of them occurs is given by
\nP(A) + P(B) – 2P(A \u2229 B)
\nSolution:
\ni) We have A = (A \u2229 B) \u222a (A \u2229 BC<\/sup>)
\nand (A \u2229 B) \u2229 (A \u2229 BC<\/sup>) = \u03c6
\n\u2234 P(A) = (A \u2229 B) + P (A \u2229 BC<\/sup>)
\n\u2234 P(A \u2229 BC<\/sup>) = P(A) – P(A \u2229 B) ———(1)<\/p>\n

ii) Let E be the event that exactly one of them, (i.e.,) either A or B occws. Given
\nE = (A – B) \u222a (B – A)
\n= (A \u2229 BC<\/sup>) \u222a (B \u2229 AC<\/sup>)
\n\u2235 So P(E) = P(A \u2229 BC<\/sup>) + P(B \u2229 AC<\/sup>)
\n(A n BC) n (B n AC)
\n\u2235 P(E) = P(A) – P(A \u2229 B) + P(B) – P(A \u2229 B)
\nby (1)
\n\u2234 P(E) = P(A) + P(B) – 2 P(A \u2229 B)<\/p>\n

\"Inter<\/p>\n

Question 33.
\nSuppose A 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 occur,
\nii) neither A nor B occurs.
\nSolution:
\nWe have AC<\/sup> = the event A does not occurs.
\n(A \u222a B)C<\/sup> = neither A nor B occurs.
\n\u2234 p(AC<\/sub>) = 1 – P(A) = 1 – 0.5 = 0.5
\nSince P(A \u222a B) = P(A) + P(B) – P(A \u2229 B)
\n= 0.5 + 0.4 – 0.3
\n= 0.6
\nP[(A \u03bd BC<\/sup>)] = 1 – P (A \u222a B)
\n= 1 – 0.6 = 0.4<\/p>\n

Question 34.
\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. ’15)
\nSolution:
\nWrite B \u222a C = D then P(A \u222a B \u222a C) = P(A \u222a D)
\n\u2234 P(A \u222a D) = P(A) + P(D) – P(A \u2229 D)
\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 D \u2229 C]
\nP(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

Question 35.
\nMULTIPLICATION THEOREM OF PROBABILITY Statement: If A and B are two events of a random experiment and P(A) > 0 and P(B) > 0 then
\nP(A \u2229 B) = P(A)P\\(\\left(\\frac{\\mathbf{B}}{\\mathbf{A}}\\right)\\) = P(B)\\(\\left(\\frac{\\mathbf{A}}{\\mathbf{B}}\\right)\\).
\nSolution:
\nProof: Let S be the sample space associated with the random. Experiment and A, B be two events of S such that P(A) > 0 and P(B) > 0.
\nBy the definition of conditional probability, we have
\n\"Inter
\nInterchanging A, B we have
\n\"Inter<\/p>\n

Question 36.
\nA Pair of dice of thrown. Find probability that either of the dice shows 2 when their sum is 6.
\nSolution:
\nLet A be the event that 2 appears on either of dice and B be the event that the sum of the two number on the event that the sum of the two numbers on the dice is 6 when two dice is thrown.
\nA = {(2, 1), (2, 2), (2,3), (2, 4), (2,4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}
\nn(A) = 11
\n\"Inter<\/p>\n

Question 37.
\nA box contains 4 defective and 6 good bulbs. Two bulbs are drawn at random without replacement. Find the probability that both the bulbs drawn are good.
\nSolution:
\nLet a denote the event that drawing a good bulb is the first draw and B denote the event that the second draw is also good when two bulbs are drawn at random without replacement and S be the sample space.
\n\u2234 P(A) = \\(\\frac{6}{10}\\) and P\\(\\left(\\frac{B}{A}\\right)\\) = \\(\\frac{5}{9}\\)
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 38.
\nSuppose there are 12 boys and 4 girls in a class. If we choose three children one after another in succession, what is the probability that all the three are boys?
\nSolution:
\nLet E, be the event of choosing a boy child in ith<\/sup> trial (i = 1, 2, 3). We have to find
\n\"Inter<\/p>\n

Question 39.
\nA speaks the truth in 75% of the cases; B in 80% cases. What is the probability that their statements about an incident do not match? (T.S.& AP. Mar. \u201816)
\nSolution:
\nLet E1<\/sub>, E2<\/sub> be the events that A and, B respectively speak truth about an incident.
\nThen P(E1<\/sub>) = \\(\\frac{75}{100}\\) = \\(\\frac{3}{4}\\), P(E2<\/sub>) = \\(\\frac{80}{100}\\) = \\(\\frac{4}{5}\\)
\ns0 that P\\(\\left(\\mathrm{E}_{1}{ }^{c}\\right)\\) = \\(\\frac{1}{4}\\), P\\(\\left(\\mathrm{E}_{2}{ }^{c}\\right)\\) = \\(\\frac{1}{5}\\)
\nLet E be the event that their statements do not match about the incident. Then this happens in two mutually exclusive ways.
\ni) A speaks truth, B tells lie
\nii) A tells lie, B speaks truth. These two events are represented by E1<\/sub> \u2229 \\(\\left(\\mathrm{E}_{2}{ }^{c}\\right)\\), \\(\\left(\\mathrm{E}_{1}{ }^{c}\\right)\\) \u2229 E2<\/sub>
\n\"Inter<\/p>\n

Question 40.
\nA problem in Calculus is given to two students A and B whose chances of solving it are 1\/3 and 1\/4. What is the
\nprobability that the problem will be solved if both of them try independently? (A.P. Mar. \u201815, \u201805).
\nSolution:
\nLet E1<\/sub> and E2<\/sub> denote the events that the problem is solved by A and B respectively.
\nGiven that
\nP(E1<\/sub>) = \\(\\frac{1}{3}\\) and P(E2<\/sub>) = \\(\\frac{1}{4}\\)
\nNote that these two are independent events. Therefore the required probability
\nP(E1<\/sub> \u222a E2<\/sub>) = P(E1<\/sub>) + P(E2<\/sub>) – P(E1<\/sub> \u2229 E2<\/sub>)
\n= P(E1<\/sub>) + P(E2<\/sub>) – P(E1<\/sub>)P(E2<\/sub>)
\n(\u2235 E1<\/sub>, E2<\/sub> are independent)
\n= \\(\\frac{1}{3}\\) + \\(\\frac{1}{4}\\) – \\(\\frac{1}{12}\\) = \\(\\frac{1}{2}\\)<\/p>\n

Question 41.
\nA and B toss a fair coin 50 times each simultaneously. Then find the probability that both of them will not get tails at the same toss
\nSolution:
\nIn each toss there are four choices
\ni) A gets H, B gets H
\nii) A gets T, B gets H
\niii) A gets H, B gets T
\niv) A gets T, B gets T .
\nTherefore the total number of choices 450<\/sup> Out of the four cases listed above, (i), (ii) and (iii) are favourable.
\n(iv) is not favourable to the occurrence of the required event, say E.
\n\u2234 P(E) = \\(\\frac{3^{50}}{4^{50}}\\) = \\(\\left(\\frac{3}{4}\\right)^{50}\\)<\/p>\n

\"Inter<\/p>\n

Question 42.
\nIf A and B are independent events of a random experiment sh\u00f6w that AC<\/sup> and BC<\/sup> are also independent.
\nSolution:
\nIf A and B are independent then
\nP(A \u2229 B) = P(A) P(B)
\nNow P(AC<\/sup> \u2229 BC<\/sup>) = P[(A \u222a B)C<\/sup>]
\n= 1 – P(A \u222a 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)] [1 – P(B)] = P(AC<\/sup>)P(BC<\/sup>).
\n\u2234 AC<\/sup>, BC<\/sup> are independent.<\/p>\n

Question 43.
\nA bag contains 10 identical balls, of which 4 are blue and 6 are red. Three balls are taken out at random from the bag one after the other. Find the probability that all the three balls drawn are red without replacement.
\nSolution:
\nThe probability that the first ball drawn to be red is \\(\\frac{6}{10}\\)
\nThe probability that the second ball drawn to be red is \\(\\frac{5}{9}\\)
\nThe probability that the third ball drawn to be red is \\(\\frac{4}{8}\\)
\nBy the multiplication theorem Required probability = \\(\\frac{6}{10}\\).\\(\\frac{5}{9}\\).\\(\\frac{4}{8}\\)
\n= \\(\\frac{1}{6}\\).<\/p>\n

Question 44.
\nAn urn contains 7 red and 3 black balls. Two balls are drawn without replacement. What is the probability that the second ball is red it is known that the first ball drawn is red.
\nSolution:
\nLet R1<\/sub> be the event of drawing the first ball is red and R2<\/sub> be the event of drawing the second ball also red.
\n\"Inter<\/p>\n

Question 45.
\nLet A and B be independent events with P(A) = 0.2, P(B) = 0.5.
\nFind
\n(i) \\(P\\left(\\frac{A}{B}\\right)\\)
\n(ii) \\(\\mathbf{P}\\left(\\frac{\\mathbf{B}}{\\mathbf{A}}\\right)\\)
\n(iii) P(A \u2229 B)
\n(iv) P(A \u222a B)
\nSolution:
\nGiven P(A) = 0.2, P(B) = 0.5 and A, B are independent events.
\n\"Inter<\/p>\n

ii) \\(P\\left(\\frac{B}{A}\\right)\\) = P(B) = 0.5<\/p>\n

iii) P(A \u2229 B) = P(A). P(B)
\n= (0.2) . (0.5)
\n= 0.1<\/p>\n

iv) P(A \u222a B) = P(A) + P(B) – P(A \u2229 B)
\n= 0.2 + 0.5 – 0.1
\n= 0.6<\/p>\n

\"Inter<\/p>\n

Question 46.
\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 chosen at random from it. What is the probability that the ball drawn is white.
\nSolution:
\nLet E1<\/sub>, E2<\/sub> denote the events or choosing bags B1<\/sub> and B2<\/sub>
\n\u2234 P(E1<\/sub>) = P(E2<\/sub>) = \\(\\frac{1}{2}\\)
\nLet w be the even that the ball chosen from the selected bag is white.
\n\"Inter<\/p>\n

Question 47.
\nBAYES THEOREM : (TS, Mar. 16, A.P. Mar. 15)
\nSolution:
\nStatement : Let E1<\/sub>, E2<\/sub>, …… En<\/sub> are mutually exclusive and exhaustive events or a random experiment with P(E1<\/sub>) \u2260 0 for
\ni = 1, 2, 3 , n. Then for any event A or the random experiment with P(A) \u2260 0
\n\"Inter
\nBy multiplication theorem
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 48.
\nThree boxes B1<\/sub>, B2<\/sub> and B3<\/sub> contain balls detailed below.
\n\"Inter
\nA die is thrown. B1<\/sub>, is chosen if either 1 or 2 turns up; B2<\/sub> is chosen if 3 or 4 turns up
\nand B3<\/sub> is chosen if 5 or 6 turns up. Having chosen a box in this way, a ball is chosen at random from this box. If the ball drawn is of red colour, what is the probability that it comes from box B2<\/sub>?
\nSolution:
\nLet P(Ei<\/sub>) be the probability of choosing the box Bi<\/sub> (i = 1, 2, 3).
\nThen P(Ei<\/sub>) = \\(\\frac{2}{6}\\) = \\(\\frac{1}{3}\\) ; for i = 1, 2, 3
\nHaving chosen the box B., the probability of drawing a red ball, say, P(R\/Ei<\/sub>) is given by
\nP(R\/E1<\/sub>) = \\(\\frac{2}{5}\\). P(R\/E2<\/sub>) = \\(\\frac{4}{9}\\) and P(R\/E3<\/sub>) = \\(\\frac{2}{9}\\)
\nWe have to find the probability P(E2<\/sub>\/R). By Bayeras theorem, we get
\n\"Inter<\/p>\n

Question 49.
\nAn urn contain w white balls and b black balls. Two players Q and R alternately draw a with replacement from the urn. The player that draws a white ball first wins the game. If Q begins the game, find the probability that Q wins the game.
\nSolution:
\nLet W denote the event of drawing a white ball at any draw and B that of a black ball.
\nThen
\nP(W) = \\(\\frac{w}{w+b}\\), P(B) = \\(\\frac{b}{w+b}\\)
\nLet E be the event that Q wins the game.
\n= P(W P(E) BBW P(E) BBBBW P(E) …)
\n= P(W) + P(BBW) + P(BBBBW) + ….
\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> + ….)
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Probability Important Questions which are most likely to be asked in the exam. Intermediate 2nd Year Maths 2A Probability Important Questions Question 1. If 4 fair coins are tossed simultaneously, then find the probability that 2 heads and 2 tails appear. (Mar. ’08) … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22103"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=22103"}],"version-history":[{"count":2,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22103\/revisions"}],"predecessor-version":[{"id":22105,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22103\/revisions\/22105"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22103"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22103"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22103"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}