{"id":161863,"date":"2023-10-13T17:17:08","date_gmt":"2023-10-13T11:47:08","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=161863"},"modified":"2023-10-13T17:17:08","modified_gmt":"2023-10-13T11:47:08","slug":"ml-aggarwal-class-12-maths-solutions-section-a-chapter-10-chapter-test","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-12-maths-solutions-section-a-chapter-10-chapter-test\/","title":{"rendered":"ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Chapter Test"},"content":{"rendered":"

The availability of step-by-step ML Aggarwal Class 12 ISC Solutions<\/a> Chapter 10 Probability Chapter Test can make challenging problems more manageable.<\/p>\n

ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Chapter Test<\/h2>\n

Question 1.
\nTwo coins are tossed together. Find P(E | F) if E : no tails appear and F : no heads appear.
\nAnswer:
\nWhen two coins are tossed together S = {HH, HT, TH, TT}
\nE : no tails appear = {HH}
\nF : no heads appears = {TT}
\n\u2234 E \u2229 F = \u03a6 \u2234 P (E \u2229 F) = 0
\n\u2234 P (E\/F) = \\(\\frac{P(E \\cap F)}{P(F)}\\) = 0<\/p>\n

Question 2.
\nGiven that the numbers appearing on rolling two dice together are different, what is the probability that the sum of the numbers appearing on the dice is 6 ?
\nAnswer:
\nWhen two dice rolled together
\nThen total no. of outcomes = 62<\/sup> = 36
\nE : numbers appearing on rolling two dice together are different
\nF : Sum of the numbers appearing on dice is 6.
\nOut of 36 outcomes, the outcomes {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} are leaving out.
\n\u2234 n (E) = 36 – 6 = 30
\nHere E \u2229 F = {(1, 5), (5, 1), (2, 4), (4, 2)}
\n\u2234 n (E \u2229 F) = 4
\nSince (3, 3) lies in F but not in E.
\nn(E \u2229 F) = 4
\nThus P (F\/E) = \\(\\frac{n(\\mathrm{E} \\cap \\mathrm{F})}{n(\\mathrm{E})}=\\frac{4}{30}=\\frac{2}{15}\\)<\/p>\n

Question 3.
\nIf A and B are independent events, and P(A) = \\(\\frac{1}{2}\\), P(B) = \\(\\frac{1}{3}\\) then find P (Ac<\/sup> \u2229 Bc<\/sup>).
\nAnswer:
\nGiven P (A) = \\(\\frac{1}{2}\\) P (B) = \\(\\frac{1}{3}\\)
\nP (Ac<\/sup> \u2229 Bc<\/sup>) = P ((A \u222a B)c<\/sup>)
\n= l- 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[v A and B are independent events Then P (A \u2229 B) = P (A) P (B)]
\n= (1 – P(A))(1 – P(B))
\n= (1 – \\(\\frac{1}{2}\\)) (1 – \\(\\frac{1}{3}\\))
\n= \\(\\frac{1}{2} \\times \\frac{2}{3}=\\frac{1}{3}\\)<\/p>\n

\"ML<\/p>\n

Question 3(Old).
\nChoose the correct answer in each of the following :
\n(i) If A and B are two events such that P (A) \u2260 0 and P (B | A) = 1 then
\n(a) A \u2282 B
\n(b) B \u2282 A
\n(c) B = \u03a6
\n(d) A = \u03a6
\nAnswer:
\nSince P (B\/A) = 1 \u21d2 \\(\\frac{P(A \\cap B)}{P(A)}\\) = 1
\nP (A \u2229 B) = P (A)
\nThus A \u2282 B, \u21d2 A \u2229 B = A
\nP (A \u2229 B) = P (A)<\/p>\n

(ii) If P (A | B) > P (A) then
\n(a) P (B | A) < P (B)
\n(b) P (A \u2229 B) < P (A) . P(B) (c) P (B | A) > P (B)
\n(d) P (B | A) = P (B)
\nAnswer:
\nGiven P (A\/B) > P (A)
\n\u21d2 \\(\\frac{P(A \\cap B)}{P(B)}\\) >P(A)
\nP (A \u2229 B) > P (A) P (B)
\nNow P(B\/A) = \\(\\frac{P(A \\cap B)}{P(A)}>\\frac{P(A) P(B)}{P(A)}\\) = P(B)
\n[using (1)]<\/p>\n

(iii) If A and B are any two events such that P (A) + P (B) – P (A and B) = P (A) then
\n(a) P (B | A) = 1
\n(b) P (A | B) = 1
\n(c) P (B | A) = 0
\n(d) P (A | B) = 0 (NCERT)
\nAnswer:
\nGiven P (A) + P (B) – P (A \u2229 B) = P (A)
\n\u21d2 P (B) = P (A \u2229 B)
\n\u21d2 1 = \\(\\frac{P(A \\cap B)}{P(B)}\\) = P(A\/B)<\/p>\n

Question 4.
\nThe probabilities that a husband and wife will be alive 20 years from now are 0-7 M and 0-8. Find that in 20 years, husband ‘ will be a widower.
\nAnswer:
\nLet A: event that husband will alive 20 years from now
\nB : event that wife will alive 20 years from now.
\n\u2234 P (A) = 0.7 ; P (B) = 0.8
\nP(A\u0304) = 1 – P (A) = 1 – 0.7 = 0.3
\nP(B\u0304) = 1 – P(B) = 1 – 0.8 = 0.2
\n\u2234 required probability = P(A \u2229 B\u0304)
\n= P(A)P(B\u0304) = 0.7 \u00d7 0.2 = 0.14
\n[\u2235 A and B are independent events]<\/p>\n

Question 5.
\nAn urn contains 4 square pieces, 5 round pieces and 3 triangular pieces. Two pieces are drawn. Find the probability of drawing two different shapes.
\nAnswer:
\nTotal no. of square pieces = 4
\nTotal no. of round pieces = 5
\nTotal no. of triangular pieces = 3
\n\u2234 Total no. of pieces = 4 + 5 + 3 = 12
\nrequired probability = P (drawing one square and one round piece) + P (drawing one square and one triangular) + P (drawing one round and one triangular piece)
\n= \\(\\frac{{ }^4 \\mathrm{C}_1 \\times{ }^5 \\mathrm{C}_1}{{ }^{12} \\mathrm{C}_2}+\\frac{{ }^4 \\mathrm{C}_1 \\times{ }^3 \\mathrm{C}_1}{{ }^{12} \\mathrm{C}_2}+\\frac{{ }^5 \\mathrm{C}_1 \\times{ }^3 \\mathrm{C}_1}{{ }^{12} \\mathrm{C}_2}\\)
\n= \\(\\frac{4 \\times 5+4 \\times 3+5 \\times 3}{\\frac{12 \\times 11}{2}}=\\frac{47 \\times 2}{12 \\times 11}=\\frac{47}{66}\\)<\/p>\n

Question 6.
\nThree faces of a fair die are yellow, two faces red and one blue. The die is tossed A three times. Find the probability that colours yellow, red and blue appear in the first, second and third toss respectively.
\nAnswer:
\nGiven, no. of yellow faces of a fair die = 3
\nno. of red faces of a fair die = 2
\nno. of blue faces of a fair die = 1
\nTotal no. of faces in a fair die = 3 + 2 + 1 = 6
\nP (getting yellow face) = \\(\\frac{3}{6}=\\frac{1}{2}\\)<\/p>\n

P (getting red face) = \\(\\frac{2}{6}=\\frac{1}{3}\\)<\/p>\n

P (getting blue face) = \\(\\frac{1}{6}\\)<\/p>\n

\u2234 required probability = \\(\\frac{1}{2} \\times \\frac{1}{3} \\times \\frac{1}{6}=\\frac{1}{36}\\)<\/p>\n

Question 7.
\nFind the probability that a given three letter word (in English) has all letters repeated.
\nAnswer:
\nFor three letter word, first place can be filled 26 26 26
\nbe 26 alphabets similarly 2nd and 3rd place can be filled by 26 alphabets each.
\nThus the total no. of possible 3 letter words
\ncan be formed when letters may repeated = 26 \u00d7 26 \u00d7 26
\nSince 3 letters of the word can be filled by any of the 26 English alphabets and all letters are same e.g. aaa, bbb, ………………. zzz.
\nSo total no. of favourable outcomes = 26
\n\u2234 required probability = \\(\\frac{\\text { favourable outcomes }}{\\text { Total outcomes }}\\)
\n= \\(\\frac{26}{26 \\times 26 \\times 26}=\\frac{1}{676}\\)<\/p>\n

Question 8.
\nIs it possible to construct an unfair die such that probability of getting a six is \\(\\frac{1}{2}\\) that of getting any other number is \\(\\frac{1}{6}\\)?
\nAnswer:
\nIf prob. of getting a six = \\(\\frac{1}{2}\\)
\nProb. of getting 1,2,3,4 and 5 each is equal t0 \\(\\frac{1}{6}\\)
\nHere sum of probabilities of all events
\n= \\(\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{2}\\)
\n= \\(\\frac{5}{6}+\\frac{1}{2}=\\frac{5+3}{6}=\\frac{8}{6}=\\frac{4}{3}\\) \u2260 1
\nSo construction of such on unfair die is not possible.<\/p>\n

\"ML<\/p>\n

Question 9.
\nFrom each of the four married couples, one of the partners is selected at random. Find the probability that the selected are of the same sex.
\nAnswer:
\nSince the prob. of getting male from each married couple = \\(\\frac{1}{2}\\)
\nProb. of selecting female from each married couple = \\(\\frac{1}{2}\\)
\nrequired probability = P (selecting male partner from each married couple) + P (selecting female partner from each married couple)
\n= \\(\\left(\\frac{1}{2}\\right)^4+\\left(\\frac{1}{2}\\right)^4=\\frac{1}{8}\\)
\n[since there are 4 married couples]<\/p>\n

Question 10.
\nWhat is the probability that a couple\u2019s ^ second child will be a boy, given that the first child is a girl ?
\nAnswer:
\nHere sample space with the experiment be {GB, GG}
\nTotal no. of outcomes = 2
\nno. of favourable outcomes =1 {GB}
\nrequired probability = \\(\\frac{1}{2}\\)<\/p>\n

Question 11.
\nFind the probability that when a hand of 7 cards is drawn from a well-shuffled pack of 52 cards, it contains
\n(i) all kings
\n(ii) exactly 3 kings
\n(iii) atleast 3 kings.
\nAnswer:
\nTotal number of ways of selecting 7 cards out of 52 cards = 52<\/sup>C7<\/sub><\/p>\n

(i) required prob. of containing all 4 kings
\n\"ML
\n(ii) required prob. of containing atleast 3 kings
\n\"ML
\n(iii) required prob. of containing atleast 3 kings
\n\"ML<\/p>\n

Question 12.
\nFour married couples have gathered in a room. Two persons are selected at i\u2019^-Trandom from amongst them. Find the probability that the selected persons are
\n(i) a husband and his wife
\n(ii) a gentleman and a lady but not a couple.
\nAnswer:
\nGiven no. of married couples = 4
\n\u2234 Total no. of persons = 8<\/p>\n

(i) Total no. of ways of selecting 2 persons out of 8 persons
\nrequired probability = \\(\\frac{{ }^4 \\mathrm{C}_1}{{ }^8 \\mathrm{C}_2}=\\frac{4 \\times 2}{8 \\times 7}=\\frac{1}{7}\\)
\n[since there are 4 married couples]<\/p>\n

(ii) P (of selecting a gentleman and a lady)
\n= \\(\\frac{{ }^4 \\mathrm{C}_1 \\times{ }^4 \\mathrm{C}_1}{{ }^8 \\mathrm{C}_2}=\\frac{4 \\times 4 \\times 2}{8 \\times 7}=\\frac{4}{7}\\)
\n\u2234 required probability = P (of selecting a gentleman and a lady) – P (selecting a couple)
\n= \\(\\frac{4}{7}-\\frac{1}{7}=\\frac{3}{7}\\)<\/p>\n

Question 13.
\nA bag contains 5 red, 4 blue and m green balls. Two balls are drawn at random from the bag. If the probability of both being green is \\(\\frac{1}{7}\\) then find m.
\nAnswer:
\nGiven no. of red balls = 5
\nno. of blue balls = 4
\nand no. of green balls = m
\n\u2234 Total no. of balls = 5 + 4 + m = 9 + m
\n\u2234 prob. of drawing 2 green balls = \\(\\frac{{ }^m \\mathrm{C}_2}{{ }^{9+m} \\mathrm{C}_2}\\)<\/p>\n

also given, prob. of drawing both green balls = \\(\\frac{1}{7}\\)
\n\u21d2 \\(\\frac{{ }^m \\mathrm{C}_2}{{ }^{m+9} \\mathrm{C}_2}=\\frac{1}{7}\\)
\n\u21d2 \\(\\frac{m !}{\\frac{(m-2) ! 2 !}{\\frac{(m+9) !}{(m+7) ! 2 !}}}=\\frac{1}{7}\\)
\n\u21d2 \\(\\frac{m(m-1)}{(m+9)(m+8)}=\\frac{1}{7}\\)
\n\u21d2 7m (m – 1) = m2<\/sup> + 17m + 72
\n\u21d2 6m2<\/sup> – 24m – 72 = 0
\n\u21d2 m2<\/sup> – 4m – 12 = 0
\n\u21d2 (m – 6) (m + 2) = 0
\n\u21d2 m = 6, – 2
\nSince number of balls cannot be negative m = 6
\nHence the required no. of green balls = 6<\/p>\n

Question 14.
\nThe odds that a book will be reviewed favourably by three independent critics are 5 : 2, 4 : 3 and 3 : 4 respectively. What is the probability that of the three reviews, a majority will be favourable ?
\nAnswer:
\nThe odds that a book will be reviewed favourably by three independent critics be 5 : 2, 4 : 3 and 3 :4.
\n\"ML<\/p>\n

Question 15.
\nFour dice are thrown. What is the probability that the sum of the numbers appearing on j^the dice is 23 ?
\nAnswer:
\nWhen four dice are thrown.
\nTotal no. of possible outcomes = 64<\/sup> = 1296
\nHere favourable outcomes are
\n{(6, 6, 6, 5), (6, 6, 5, 6), (6. 5, 6, 6). (5. 6, 6. 6)}
\n\u2234 no. of favourable outcomes = 4
\nrequired probability = \\(\\frac{\\text { No. of favourable outcomes }}{\\text { Total no. of outcomes }}=\\frac{4}{1296}=\\frac{1}{324}\\)<\/p>\n

Question 16.
\n(i) Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is neither divisible by 3 nor by 4 ?
\n(ii) Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is divisible by 3 or 4 ?
\nAnswer:
\n(i) When two dice are thrown together
\nTotal no. of possible outcomes = 62<\/sup> = 36
\nLet A : event the sum of numbers on two faces is divisible by 3.
\nB : event the sum of numbers on two faces is divisible by 4.
\nA = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)}
\nB = {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}
\nA \u2229 B = {6, 6}
\nP(A) = \\(\\frac{12}{36}=\\frac{1}{3}\\); P(B) = \\(\\frac{9}{36}=\\frac{1}{4}\\)
\nP(A \u2229 B) = \\(\\frac{1}{36}\\)
\n\u2234 required probability = P(A\u0304 \u2229B\u0304) = P\\((\\overline{A \\cup B})\\)
\n= 1 – P(A \u222a B)
\n= 1 – {P(A) + P(B) – P(A \u2229B)
\n= 1 – \\(\\frac{1}{3}-\\frac{1}{4}+\\frac{1}{36}\\)
\n= \\(\\frac{36-12-9+1}{36}=\\frac{16}{36}=\\frac{4}{9}\\)<\/p>\n

(ii) required probability = P(A \u222a B)
\n= P(A) + P(B) – P(A \u2229 B)
\n= \\(\\frac{1}{3}+\\frac{1}{4}-\\frac{1}{36}=\\frac{12+9-1}{36}\\)
\n= \\(\\frac{20}{36}=\\frac{5}{9}\\)<\/p>\n

Question 18.
\nIn an essay competition, the odds in favour of the four competitors A, B, C and D are 1 : 2, 1 : 3, 1 : 4 and 1 : 5 respectively. Find the probability that one of them wins the competition.
\nAnswer:
\nGiven odds in favour of four competitors A, B, C and D are 1 : 2, 1 : 3, 1 :4 and 1 : 5 respectively.
\n\u2234 P(A) = \\(\\frac{1}{3}\\)
\nP(B) = \\(\\frac{1}{4}\\)
\nP(C) = \\(\\frac{1}{5}\\)
\nP(D) = \\(\\frac{1}{6}\\)<\/p>\n

P(A\u0304) = 1 – P(A)
\n= 1 – \\(\\frac{1}{3}=\\frac{2}{3}\\)<\/p>\n

P(B\u0304) = 1 – P(B)
\n= 1 – \\(\\frac{1}{4}=\\frac{3}{4}\\)<\/p>\n

P(C\u0304) = 1 – P(C)
\n= 1 – \\(\\frac{1}{5}=\\frac{4}{5}\\)<\/p>\n

P(D\u0304) = 1 – P(D)
\n= 1 – \\(\\frac{1}{6}=\\frac{5}{6}\\)<\/p>\n

required probability that one of them will win the competition
\n= P (A) + P (B) + P (C) + P (D)
\n= \\(\\frac{1}{3}+\\frac{1}{4}+\\frac{1}{5}+\\frac{1}{6}=\\frac{20+15+12+10}{60}\\)
\n= \\(\\frac{57}{60}=\\frac{19}{20}\\)<\/p>\n

\"ML<\/p>\n

Question 19.
\nA bag contains 3 white, 3 black and 2 red bails. Three balls are drawn one by one without replacement. Find the probability that the third ball is red.
\nAnswer:
\nGiven bag contains 3 white, 3 black and 2 red balls Total no. of balls = 3 + 3 + 2 = 8
\n\u2234 required probability = P (WBR) + P (BWR) + P (WWR) + P (BBR) + P (BRR) + P (RBR) + P (WRR) + P (RWR)
\n\"ML<\/p>\n

Question 20.
\nThere are three urns A, B and C. A contains 4 white balls and 5 blue balls. Urn B contains 4 white balls and 3 blue balls. Urn C contains 3 w hite balls and 6 blue balls. One ball is drawn from each of these urns. What is the probability that out of these three balls drawn, two are white and one is blue ?
\nAnswer:
\nurn A contains 4 white and 5 blue balls
\nurn B contains 4 white and 3 blue balls
\nurn C contains 3 white and 6 blue balls
\nrequired probability = P (W1<\/sub>W2<\/sub>B3<\/sub>) + P (W1<\/sub>B2<\/sub>W3<\/sub>) + P (B1<\/sub>W2<\/sub>W3<\/sub>)
\n= P (W1<\/sub>) P (W2<\/sub>) P (B3<\/sub>) + P (W1<\/sub>) P (B2<\/sub>) P (W3<\/sub>) + P (B1<\/sub>) P (W2<\/sub>) P (W3<\/sub>)
\n= \\(\\frac{4}{9} \\times \\frac{4}{7} \\times \\frac{6}{9}+\\frac{4}{9} \\times \\frac{3}{7} \\times \\frac{3}{9}+\\frac{5}{9} \\times \\frac{4}{7} \\times \\frac{3}{9}\\)
\n= \\(\\frac{96+36+60}{9 \\times 7 \\times 9}\\)
\n= \\(\\frac{64}{189}\\)<\/p>\n

Question 21
\nIf each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive ? (NCERT)
\nAnswer:
\nLet the second order determinant A = \\(\\left|\\begin{array}{ll}
\na & b \\\\
\nc & d
\n\\end{array}\\right|\\)
\nWhere each entry a, b, c or d is either 0 or 1.
\nThus for each entry we have 2 possibilities
\nTotal no. of possible cases = 24 = 16
\nHere favourable outcomes are those determinants whose value is positive and these are
\n\\(\\left|\\begin{array}{ll}
\n1 & 0 \\\\
\n0 & 1
\n\\end{array}\\right| ;\\left|\\begin{array}{ll}
\n1 & 1 \\\\
\n0 & 1
\n\\end{array}\\right| ;\\left|\\begin{array}{ll}
\n1 & 0 \\\\
\n1 & 1
\n\\end{array}\\right|\\)
\nIn each case the value of \u0394 = 1 > 0.
\nrequired probability = \\(\\frac{3}{16}\\)<\/p>\n

Question 22.
\nA bag contains (n + 1) coins. It is known that one of these coins shows head on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that the toss results in head is \\(\\frac{7}{12}\\) then find the value of n.
\nAnswer:
\nTotal no. of coins in bag = n + 1
\nHere total no. of fair coins = n
\nand no. of biased coin = 1
\nLet us define the events E1<\/sub>, E2<\/sub> and A as follows :
\nE1<\/sub> : drawing a fair coin from bag
\nE2<\/sub> : drawing an unfair coin from bag
\nA : toss results in head.
\nThen P (E1<\/sub>) = \\(\\frac{n}{n+1}\\) ; P (E2<\/sub>) = \\(\\frac{1}{n+1}\\)
\nHere E1<\/sub> and E2<\/sub> both mutually exclusive and exhaustive events.
\nThen P(A\/E1<\/sub>) = \\(\\frac{1}{2}\\); P(A\/E2<\/sub>) = \\(\\frac{2}{2}\\)
\nThus by law of total probability; we have
\nP(A) = P(E1<\/sub>) P(A\/E1<\/sub>) + P(E2<\/sub>)P(A\/E2<\/sub>)
\n\u21d2 \\(\\frac{7}{12}=\\frac{n}{n+1} \\cdot \\frac{1}{2}+\\frac{1}{n+1}\\) . 1
\n\u21d2 \\(\\frac{7}{12}=\\frac{n+2}{2(n+1)}\\)
\n\u21d2 7(n + 1) = 6(n + 2)
\n\u21d2 7n – 6n = 12 – 7
\n\u21d2 n = 5<\/p>\n

Question 23.
\nA bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag. (NCERT)
\nAnswer:
\nLet us define the events E1<\/sub>, E2<\/sub> and A as follows:
\nE1<\/sub>: bag I is chosen
\nE2<\/sub>: bag I is chosen
\nA: drawn ball will be red
\nThen P(E1<\/sub>) = P(E2<\/sub>) = \\(\\frac{1}{2}\\)
\n\u2234 both events E1<\/sub> and E2<\/sub> are mutually exclusive and exhaustive events.
\nP (A\/E1<\/sub>) = prob. of drawing red ball from bag I
\n= \\(\\frac{4}{4+4}=\\frac{1}{2}\\)<\/p>\n

P (A\/E2<\/sub>) = probability of drawing red ball from bag II
\n=\\(\\frac{2}{2+6}=\\frac{2}{8}=\\frac{1}{4}\\)<\/p>\n

we want to find P (E1<\/sub>\/A)
\nThen by Baye\u2019s Theorem, we have
\n\"ML<\/p>\n

\"ML<\/p>\n

Question 24.
\nA company launches a new product and estimated that a person who comes across its advertisement will buy the product with a probability of 0.7, and who does \u2019 not see the advertisement will buy the product with a probability of 0.3. If 70% of the people come across the advertisement, then what is the probability that a person who buys the product had come across the advertisement ?
\nAnswer:
\nLet us define the events E1<\/sub>, E2<\/sub> and A are as follows:
\nE1<\/sub> : people came across the advertisement
\nE2<\/sub> : people does not came across the advertisement
\nA : person will buy the product
\nThen P (E1<\/sub>) = 70% = \\(\\frac{70}{100}\\) ;
\nP(E2<\/sub>) = 30% = \\(\\frac{30}{100}\\)
\nE1<\/sub> and E2<\/sub> are mutually exclusive and exhaustive events.
\nP(A\/E1<\/sub>) = probability of that person will buy the product and he does not came across the advertisement = 0.7
\nP(A\/E2<\/sub>) = probability that person will buy the product and he already came across the advertisement = 1 – 0.7 = 0.3
\nWe want to find P (E1<\/sub>\/A)
\nThen by Baye\u2019s Theorem, we have
\n\"ML<\/p>\n

Question 25.
\nProbability that A speaks truth is \\(\\frac{4}{5}\\). A coin is tossed. A reports that head appears. What is the probability that actually it was head ? (NCERT)
\nAnswer:
\nLet us define the events E1<\/sub>, E2<\/sub> and A are as follows :
\nE1<\/sub> : Coin shows head
\nE2<\/sub> : Coin does not shows head
\nA : A reports that head appears
\nThenP(E1<\/sub>) = \\(\\frac{1}{2}\\) = P (E2<\/sub>)
\nE1<\/sub> and E2<\/sub> are mutually exclusive and exhaustive events.
\nP (A\/E1<\/sub>) = probability that A reports that head appears and given that head has occured.
\nprobability that man speaks truth = \\(\\frac{4}{5}\\)<\/p>\n

P (A\/E2<\/sub>) = prob. that A reports that head appears and given that head has not occured.
\n= probability that A does not speaks truth
\n= 1 –\\(\\frac{4}{5}=\\frac{1}{5}\\)<\/p>\n

Then by Baye\u2019s Theorem, we have
\nP(E1<\/sub>\/A) = \\(\\frac{P\\left(A \/ E_1\\right) P\\left(E_1\\right)}{P\\left(A \/ E_1\\right) P\\left(E_1\\right)+P\\left(A \/ E_2\\right) P\\left(E_2\\right)}\\)
\n= \\(\\frac{\\frac{4}{5} \\times \\frac{1}{2}}{\\frac{4}{5} \\times \\frac{1}{2}+\\frac{1}{5} \\times \\frac{1}{2}}=\\frac{\\frac{4}{5}}{\\frac{4}{5}+\\frac{1}{5}}\\)
\n= \\(\\frac{4}{5}\\)<\/p>\n

Question 26.
\nAn insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver ?
\nAnswer:
\nLet E1<\/sub>, E2<\/sub>, E3<\/sub> and A are the events defined as follows :
\nE1<\/sub> : A scooter driver is insured
\nE2<\/sub> : A car driver is insured
\nE3<\/sub> : A truck driver is insured
\n\"ML
\nLet A : event that insured person met an accident
\nP (A\/E1<\/sub>) = probability that insured scooter driver will meet an accident = 0.01
\nP (A\/E2<\/sub>) = probability that insured car driver will meet an accident = 0.03
\nP (A\/E3<\/sub>) = probability that the person will meet an accident given that he is a truck driver = 0.15
\nWe want to find P (E1<\/sub>\/A).
\nThen by Baye\u2019s Theorem, we have
\n\"ML<\/p>\n

Question 27.
\nThree bags contain balls as shown in the table below :
\n\"ML
\nA bag is chosen at random and two balls are drawn from it. They happened to be white and red. What is the probability that they came from bag III ?
\nAnswer:
\nLet E1<\/sub>, E2<\/sub>, E3<\/sub> and A are the events as follows :
\nE1<\/sub> : bag-I is chosen;
\nE2<\/sub> : bag-11 is chosen ;
\nE3<\/sub> : bag III is chosen
\nA : a white and red ball is drawn
\nThen P(E1<\/sub>) = P(E2<\/sub>) = P(E3<\/sub>) = \\(\\frac{1}{3}\\)
\nHere, E1<\/sub>, E2<\/sub>, E3<\/sub> are mutually exclusive and exhaustive events.
\nP(A\/E1<\/sub>) = Prob. of drawing one white and one red ball from bag I
\n= \\(\\frac{{ }^1 \\mathrm{C}_1 \\times{ }^3 \\mathrm{C}_1}{{ }^6 \\mathrm{C}_2}=\\frac{3 \\times 2}{6 \\times 5}=\\frac{1}{5}\\)<\/p>\n

P(A\/E2<\/sub>) = Prob. of drawing one white and one red ball from bag II
\n= \\(\\frac{{ }^2 \\mathrm{C}_1 \\times{ }^1 \\mathrm{C}_1}{{ }^4 \\mathrm{C}_2}=\\frac{2}{\\frac{4 \\times 3}{2}}=\\frac{4}{4 \\times 3}=\\frac{1}{3}\\)<\/p>\n

P(A\/E3<\/sub>) = Prob. of drawing one white and one red ball from bag III
\n= \\(\\frac{{ }^4 \\mathrm{C}_1 \\times{ }^2 \\mathrm{C}_1}{{ }^9 \\mathrm{C}_2}=\\frac{4 \\times 2 \\times 2}{9 \\times 8}=\\frac{2}{9}\\)<\/p>\n

We want to find P (E3<\/sub>\/A)
\nThen by Baye\u2019s theorem ; we have
\n\"ML<\/p>\n

Question 27(Old).
\nA manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A ? (NCERT)
\nAnswer:
\nLet E1<\/sub>, E2<\/sub>, E3<\/sub> and E are the events defined as follows :
\nE1<\/sub> : Machine A is producing items
\nE2<\/sub> : Machine B is producing items
\nE3<\/sub> : Machine C is producing items<\/p>\n

E : A defective item is produced
\nThen P(E1<\/sub>) = \\(\\frac{50}{100}=\\frac{1}{2}\\)
\nP(E2<\/sub>) = \\(\\frac{30}{100}=\\frac{3}{10}\\)
\nP(E3<\/sub>) = \\(\\frac{20}{100}=\\frac{1}{5}\\)<\/p>\n

E1<\/sub>, E2<\/sub> and E3<\/sub> are mutually exclusive and exhaustive events.
\nP (E\/E1<\/sub>) = P (getting a defective item produced by machine A)
\n= 1% = \\(\\frac{1}{100}\\)<\/p>\n

P (E\/E2<\/sub>) = P (defective item is produced by machine B)
\n= 5% = \\(\\frac{5}{100}\\)<\/p>\n

P (E\/E3<\/sub>) = P (defective item is produced by machine C)
\n= 7% = \\(\\frac{7}{100}\\)<\/p>\n

We want to find P (E1<\/sub>\/E)
\nThen by Baye\u2019s Theorem, we have
\n\"ML<\/p>\n

Question 28.
\nThe probability function of a random variable X is given by
\n\"ML
\nwhere p is a constant. Find the value of p. Calculate P (0 \u2264 X < 3) and P (X > 1).
\nAnswer:
\nGiven
\n\"ML
\nSince \u03a3 P (x) = 1 \u21d2 P (1) + P (2) + P (3) + …. = 1
\n\u21d2 2p + p + 4p + 0 …. + 0 …….. = 1
\n\u21d2 7p = 1
\n\u21d2p = \\(\\frac{1}{7}\\)
\nP (0 \u2264 X < 3) = P (X = 0) + P (X = 1) + P (X = 2) = 0 + 2p + p = 3p = \\(\\frac{3}{7}\\) P(X > 1) = 1 – P(X \u2264 1)= 1 – P(X = 0) – P(X = 1)
\n= 1 – 0 – 2p
\n= 1 – \\(\\frac{2}{7}=\\frac{5}{7}\\)<\/p>\n

\"ML<\/p>\n

Question 29.
\nFind the probability distribution of Y in two throws of two dice, where Y represents the number of times a total of 9 appears.
\nAnswer:
\nWhen two dice thrown, all the 36 outcomes are equally likely so it is case of binomial distribution.
\nLet p = probability of getting a total of 9 = \\(\\)
\nSince favourable outcomes are {(4, 5), (5, 4), (3, 6), (6, 3)}
\nq = 1 – p = 1 – \\(\\frac{1}{9}=\\frac{8}{9}\\) Here n = 2
\nGiven random variable Y represents the number of times a total of 9 appears. Y takes values 0, 1,2.
\nNow P (X = r) = n<\/sup>Cr<\/sub> pr<\/sup> qn – r<\/sup>
\n\"ML<\/p>\n

Thus the probability of distribution of Y is given below:
\n\"ML<\/p>\n

Question 30.
\nTwo cards were drawn without replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of cards of honour (i.e. Jack, Queen, King and Ace).
\nAnswer:
\nSince there are 16 honour cards i.e. 4 kings, 4 queens, 4 jacks and 4 aces.
\nClearly the cards are drawn without replacement so trials or events are not independent,
\nit is not a problem of binomial distribution.
\nLet X be the random variable denotes the number of honour cards and X takes values 0, 1,2.
\nP(X = 0) = Prob. of drawing no honour card
\n= Prob. of getting two 2 cards other than honour cards
\n= \\(\\frac{{ }^{36} \\mathrm{C}_2}{{ }^{52} \\mathrm{C}_2}=\\frac{36 \\times 35}{52 \\times 51}=\\frac{105}{221}\\)<\/p>\n

P (X = 1) = prob. of drawing one honour card and one non-honour card
\n= \\(\\frac{{ }^{16} \\mathrm{C}_1 \\times{ }^{36} \\mathrm{C}_1}{{ }^{52} \\mathrm{C}_2}=\\frac{16 \\times 36 \\times 2}{52 \\times 51}=\\frac{96}{221}\\)<\/p>\n

P (X = 2) = prob. of drawing both honour cards
\n= \\(\\frac{{ }^{16} \\mathrm{C}_2}{{ }^{52} \\mathrm{C}_2}=\\frac{16 \\times 15}{52 \\times 51}=\\frac{20}{221}\\)<\/p>\n

Thus the required probability distribution of X is given below :
\n\"ML<\/p>\n

Question 31.
\nFind the probability distribution of the number of boys in families with 3 children, assuming ^ equal probabilities for boys and girls.
\nAnswer:
\nSince probabilities of both boys and girls are same,
\ni.e. \\(\\frac{1}{2}\\) i.e. trials are independent and hence it is a problem of binomial distribution.
\np = probability of getting boy = \\(\\frac{1}{2}\\)
\nq = 1 -p = 1 – \\(\\frac{1}{2}=\\frac{1}{2}\\) and here n = 3<\/p>\n

Let X be the binomial variate and be the number of boys in families and X can takes values 0, 1, 2, 3.
\n\"ML
\nThus, the required probability distribution of X is given by
\n\"ML<\/p>\n

Question 32.
\nFind the probability distribution of the number of white balls drawn in a random draw of 3 balls without replacement from a bag containing 4 white and 6 red balls. Also find the mean and variance of the distribution.
\nAnswer:
\nSince balls are drawn one by one without replacement so it is not a case of binomial distribution.
\nLet X be the random variable denotes the no. of white balls drawn in a random sample of 3 draws.
\nSo X can takes values 0, 1, 2, 3.
\nP(X = 0) = prob. (of drawing no white ball)
\n= P (drawing 3 red ball one by one without replacement)
\n= \\(\\frac{6}{10} \\times \\frac{5}{9} \\times \\frac{4}{8}=\\frac{120}{720}=\\frac{1}{6}\\)<\/p>\n

P (X = 1) = P (of drawing one white ball)
\n= P (drawing one white and 2 red balls)
\n= P (WRR, RWR, RRW)
\n= P (WRR) + P (RWR) + P (RRW)
\n= \\(\\frac{4}{10} \\times \\frac{6}{9} \\times \\frac{5}{8}+\\frac{6}{10} \\times \\frac{4}{9} \\times \\frac{5}{8}+\\frac{6}{10} \\times \\frac{5}{9} \\times \\frac{4}{8}\\)
\n= \\(\\frac{360}{720}=\\frac{1}{2}\\)<\/p>\n

P (X = 2) = P (drawing two white balls and one red ball)
\n= P (WWR, WRW, RWW)
\n= P (WWR) + P (WRW) + P (RWW)
\n= \\(\\frac{4}{10} \\times \\frac{3}{9} \\times \\frac{6}{8}+\\frac{4}{10} \\times \\frac{6}{9} \\times \\frac{3}{8}+\\frac{6}{10} \\times \\frac{4}{9} \\times \\frac{3}{8}\\)
\n= \\(\\frac{216}{720}=\\frac{3}{10}\\)<\/p>\n

P (X = 3) = P (drawing 3 white balls one by one without replacement)
\n= \\(\\frac{4}{10} \\times \\frac{3}{9} \\times \\frac{2}{8}=\\frac{24}{720}=\\frac{1}{30}\\)<\/p>\n

The probability distribution of X is given below ;
\n\"ML
\n\"ML<\/p>\n

Question 36.
\nThe probability distribution of a discrete random variable X is given as under
\n\"ML
\nCalculate :
\n(i) the value of A if E (X) = 2 -94
\n(ii) variance of X.
\nAnswer:
\nGiven probability distribution of discrete random variable given below
\n\"ML
\nMean = E (X) = \u03a3X P (X) = 1 \u00d7 \\(\\frac{1}{2}\\) + 2 \u00d7 \\(\\frac{1}{5}+\\frac{12}{25}+\\frac{2 \\mathrm{~A}}{10}+\\frac{3 \\mathrm{~A}}{25}+\\frac{5 \\mathrm{~A}}{25}\\)
\n\u21d2 2.94 = \\(\\frac{1}{2}+\\frac{2}{5}+\\frac{12}{25}+\\frac{\\mathrm{A}}{5}+\\frac{3 \\mathrm{~A}}{25}+\\frac{\\mathrm{A}}{5}\\)
\n\u21d2 2.94 = \\(\\frac{25+20+24+10 \\mathrm{~A}+6 \\mathrm{~A}+10 \\mathrm{~A}}{50}\\)
\n\u21d2 147 = 69 + 26A
\n\u21d2 78 = 26A
\n\u21d2 A = \\(\\frac{78}{26}\\) = 3<\/p>\n

probability distribution of X reduces to :
\n\"ML
\n\u2234 Var (X) = \u03a3X2<\/sup> P(X) – \u00b52<\/sup>
\n\"ML<\/p>\n

\"ML<\/p>\n

Question 34.
\nA and B play a game in which A\u2019s chance of w inning the game is \\(\\frac{3}{5}\\). In a series of 6 games, find the probability that A will win atleast 4 games.
\nAnswer:
\nLet p = probability of A\u2019s winning the game = \\(\\frac{3}{5}\\)
\nq = 1 – p = 1 – \\(\\frac{3}{5}=\\frac{2}{5}\\)
\nHere n = 6
\nIt is case of binomial distribution
\n\"ML<\/p>\n

Question 35.
\nSuppose that 90% of the people are right handed. What is the probability that atmost 6 people of a random sample of 10 people are right handed ? (NCERT)
\nAnswer:
\nLet p probability that people are right handed = 90% = \\(\\frac{90}{100}=\\frac{9}{10}\\)
\nq = 1 – p = 1 – \\(\\frac{9}{10}=\\frac{1}{10}\\), Here n = 10
\nThen by binomial distribution, we have
\nP (X = r) = n<\/sup>Cr<\/sub> pr<\/sup> qn – r<\/sup> = 10<\/sup>Cr<\/sub> \\(\\left(\\frac{19}{10}\\right)^r\\left(\\frac{1}{10}\\right)^{10-r}\\)
\nrequired probability = P (X \u2264 6) = 1 – P (X \u2265 7)
\n= 1 – [P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)]
\n= 1 – \\(\\sum_{r=7}^{10}{ }^{10} \\mathrm{C}_r\\left(\\frac{9}{10}\\right)^{\\mathrm{r}}\\left(\\frac{1}{10}\\right)^{10-r}\\)<\/p>\n

Question 35(Old).
\nLet X denote the num ber of colleges w here you w iUapply after your results and P (X = x) denotes your probability of getting admission in x number of colleges. It is given that
\n\"ML
\nwhere A is a positive constant
\n(i) Find the value of k.
\n(ii) What is the probability that you will get admission in exactly two colleges ?
\n(iii) Find the mean and variance of the probability distribution.
\nAnswer:
\n(i) Given P (X = x)
\n\"ML
\nSince P (X = x) denotes your probability of getting admission in x number of colleges.
\n\u2234 \u03a3pi<\/sub> = 1
\n\u21d2 P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1
\n\u21d2 k . 0 + k . 1 + 2k (2) + k (5 – 3) + k (5 – 4) = 1
\n\u21d2 0 + k + 4k + 2k + k = 1
\n\u21d2 8k = 1
\n\u21d2 k = \\(\\frac{1}{8}\\)<\/p>\n

(ii) required probability = P (X = 2) = 2k \u00d7 2 = 4k = \\(\\frac{4}{8}=\\frac{1}{2}\\)
\n(iii)
\n\"ML
\n\u2234 Mean = \u03bc = \u03a3X P (X)
\n= 02<\/sup> \u00d7 0 + 1 2<\/sup> \u00d7 k + 22<\/sup> \u00d7 4k + 32<\/sup> \u00d7 2k + 42<\/sup> \u00d7 k – \\(\\)
\n= k + 8k + 6k + 4k – \\(\\left(\\frac{19}{8}\\right)^2\\)
\n= 51k – \\(\\frac{361}{64}=\\frac{51}{8}-\\frac{361}{64}\\)
\n= \\(\\frac{408-361}{54}=\\frac{47}{64}\\)<\/p>\n

Question 36.
\nIf 20% of the bolts produced by a machine are defective, determine the probability that out of 4 bolts chosen at random
\n(i) 1
\n(ii) 0
\n(iii) less than 2 bolts will be defective.
\nAnswer:
\nLet p probability of getting a detective bolt produced by machine
\n= 20% = \\(\\frac{20}{100}=\\frac{1}{5}\\)
\nq = 1 – p = 1 – \\(\\frac{1}{5}=\\frac{4}{5}\\) and here n = 4
\nThen by binomial distribution, we have
\n\"ML<\/p>\n

\"ML<\/p>\n

Question 37.
\nA die is thrown 6 times. What is the probability that there will be (i) no ace (ii) not more than one ace (iii) not more than 4 aces ? (Note that ace means a number 1, or one dot, on , the upper face of the die.)
\nAnswer:
\nHere p = probability of getting ace in single throw of dice = \\(\\frac{1}{6}\\)
\nq = 1 – p = 1 – \\(\\frac{1}{6}=\\frac{5}{6}\\) and here n = 6
\nSince all the outcomes are equally likely. So it is a problem of binomial distribution.
\nThen by binomial distribution, we have
\nP(X = r) = n<\/sup>Cr<\/sub> pr<\/sup> qn – r<\/sup> = 6<\/sup>Cr<\/sub><\/p>\n

(i) Prob. of getting no ace = P (X = 0)
\n= 6<\/sup>C0<\/sub>\\(\\left(\\frac{1}{6}\\right)^0\\left(\\frac{5}{6}\\right)^6=\\left(\\frac{5}{6}\\right)^6\\)<\/p>\n

(ii) Probability of getting not more than one ace
\n= P (X < 1) = P (X = 0) + P (X = 1)
\n= 6<\/sup>C0<\/sub>\\(\\left(\\frac{1}{6}\\right)^0\\left(\\frac{5}{6}\\right)^6\\) + 6<\/sup>C1<\/sub>\\(\\left(\\frac{1}{6}\\right)^1\\left(\\frac{5}{6}\\right)^5\\)
\n= \\(\\left(\\frac{5}{6}\\right)^5\\left[\\frac{5}{6}+\\frac{6}{6}\\right]=\\frac{11}{6}\\left(\\frac{5}{6}\\right)^5\\)<\/p>\n

(iii) Probability of getting not more than 4 aces
\n= P (X < 4) = 1 – P (X > 4) = 1 – P (X = 5) – P (X = 6)
\n= 1 – 6<\/sup>C5<\/sub>\\(\\left(\\frac{1}{6}\\right)^5\\left(\\frac{5}{6}\\right)\\) – 6<\/sup>C5<\/sub>\\(\\left(\\frac{1}{6}\\right)^6\\left(\\frac{5}{6}\\right)^0\\)
\n= 1 – \\(\\frac{6 \\times 5}{6^6}-\\frac{1}{6^6}\\)
\n= 1 – \\(\\frac{31}{6^6}\\)
\n= 1 – \\(\\left(\\frac{1}{6}\\right)^6\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

The availability of step-by-step ML Aggarwal Class 12 ISC Solutions Chapter 10 Probability Chapter Test can make challenging problems more manageable. ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Chapter Test Question 1. Two coins are tossed together. Find P(E | F) if E : no tails appear and F : no …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/161863"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=161863"}],"version-history":[{"count":7,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/161863\/revisions"}],"predecessor-version":[{"id":161945,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/161863\/revisions\/161945"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=161863"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=161863"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=161863"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}