{"id":32789,"date":"2022-03-30T11:30:33","date_gmt":"2022-03-30T06:00:33","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=32789"},"modified":"2022-03-30T11:36:47","modified_gmt":"2022-03-30T06:06:47","slug":"ncert-solutions-for-class-12-maths-chapter-13-ex-13-4","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-13-ex-13-4\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 13 Probability Ex 13.4 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-13-ex-13-4\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.4<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nState which of the following are not the probability distributions of a random variable. Give reasons for your answer.
\n\"NCERT
\nSolution:
\nP (0) + P (1) + P (2) = 0.4 + 0.4 + 0.2 = 1
\nIt is a probability distribution.<\/p>\n

(ii) P (3) = – 0.1 which is not possible.
\nThus it is not a probability distribution.<\/p>\n

(iii) P(-1)+P(0)+P(1) = 0.6 + 0.1 + 0.2 = 0.9 \u2260 1
\nThus it is not a probability distribution.<\/p>\n

(iv) P (3) + P (2) + P (1) + P (0) + P (-1)
\n= 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05 \u2260 1
\nHence it is not a probability distribution.<\/p>\n

Question 2.
\nAn urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X ?. Is X a random variable?
\nSolution:
\nThese two balls may be selected as RR, RB, BR, BB, where R represents red and B represents black ball, variable X has the value 0,1,2, i.e., there may be no black balls, may be one black ball, or both the balls are.black. Yes , X is a random variable.<\/p>\n

Question 3.
\nLet X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?
\nSolution:
\nLet \u2018w\u2019 denote the number of heads and \u2018n\u2019 the number of tails when a coin is tossesd six times
\nX is the difference between m and n
\n\u2234 X = |m – n|
\n\"NCERT
\n\u2234 The possible values of X are 0, 2, 4, 6.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nFind the probability distribution of
\n(a) number of heads in two tosses of a coin.
\n(b) number of tails in the simultaneous tosses of three coins.
\n(c) number of heads in four tosses of a coin.
\nSolution:
\ni. When a coin is tossed twice, the sample
\nspace S = {HH, HT, TH, TT}
\nLet X denote the number of heads.
\nThen X takes the values 0, 1, 2.
\nP(X = 0) = P(two tails) = P{TT} = \\(\\frac { 1 }{ 4 }\\)
\nP(X = 1) = P(one head & one tail)
\n= P{HT,TH} = \\(\\frac { 2 }{ 4 }\\) = \\(\\frac { 1 }{ 2 }\\)
\nP(X = 2) = P(two heads) = P{HH} = \\(\\frac { 1 }{ 4 }\\)
\n\u2234 The probability distribution of X is
\n\"NCERT<\/p>\n

ii. When a coin is tossed 3 times, the sample space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
\nLet X denote the number of tails.
\nThen X take values 0, 1, 2, 3.
\nP(X = 0) = P(no tail) = P{HHH} = \\(\\frac { 1 }{ 8 }\\)
\nP(X = 1) = P(one tail & two heads)
\n= P{HHT, THH, HTH} = \\(\\frac { 3 }{ 8 }\\)
\nP(X = 2) = P(two tails & one head)
\n= P{HTT, THT, TTH} = \\(\\frac { 3 }{ 8 }\\)
\nP(X = 3) = P(three tails) = P(TTT) = \\(\\frac { 1 }{ 8 }\\)
\nThe probability distribution of X is
\n\"NCERT<\/p>\n

iii. When four coins are tossed, sample space, S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
\nLet X denote the number of heads in the four tosses of a coin.
\nThen X can take the values 0, 1,2, 3 and 4
\nP(H) = \\(\\frac { 1 }{ 2 }\\) ,P(T) = \\(\\frac { 1 }{ 2 }\\)
\nP(X = 0) = P(TTTT) = \\(\\frac { 1 }{ 2 }\\) x \\(\\frac { 1 }{ 2 }\\) x \\(\\frac { 1 }{ 2 }\\) x \\(\\frac { 1 }{ 2 }\\) = \\(\\frac { 1 }{ 16 }\\)
\nP(X = 1) = P{HTTT, THTT, TTHT, TTTH}
\n\"NCERT<\/p>\n

Question 5.
\nFind the probability distribution of the number of successes in two tosses of a die, where a success is defined as
\ni. number greater than 4
\nii. six appears on atleast one die
\nSolution:
\ni. When a die is tossed the sample space S = {1, 2, 3, 4, 5, 6}.
\nLet A denote the success
\nA: getting a number greater than 4
\nA = {5, 6}
\nP(A) = \\(\\frac { 2 }{ 6 }\\) = \\(\\frac { 1 }{ 3 }\\)
\nP(A’) = 1 – P(A)= 1 – \\(\\frac { 1 }{ 3 }\\) = \\(\\frac { 2 }{ 3 }\\)
\nLet X denote the number of successes in \u2018two tosses of a die.
\nThen X takes the values 0, 1, 2
\nP(X = 0) = P(A’A’) = \\(\\frac { 2 }{ 3 }\\) x \\(\\frac { 2 }{ 3 }\\) = \\(\\frac { 4 }{ 9 }\\)
\nP(X = 1) = P(A A’ or A’ A)
\n= \\(\\frac { 1 }{ 3 }\\) x \\(\\frac { 2 }{ 3 }\\) + \\(\\frac { 2 }{ 3 }\\) x \\(\\frac { 1 }{ 3 }\\) = \\(\\frac { 4 }{ 9 }\\)
\nP(X = 2) = P(AA) = \\(\\frac { 1 }{ 3 }\\) x \\(\\frac { 1 }{ 3 }\\) = \\(\\frac { 1 }{ 9 }\\)
\nThe Probability distribution of X is
\n\"NCERT<\/p>\n

ii. Let B denote the event of getting 6 on atleast one die
\n\u2234 B = {(1, 6), (2, 6), (3, 6), (4, 6), 6), (6, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
\nLet X denote the number of success
\nThen X takes the values 0 and 1
\nP(X = 0) = P(B’)= 1 – P(B)
\n= 1 – \\(\\frac { 11 }{ 36 }\\) = \\(\\frac { 25 }{ 36 }\\)
\nP(X = 1) = P(B) = \\(\\frac { 11 }{ 36 }\\)
\n\u2234 The probability distribution of X is
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 6.
\nFrom a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
\nSolution:
\nLet D denote a defective bulb
\n\u2234 P(D) = \\(\\frac { 6 }{ 30 }\\)
\n\u2234 P(D’) = 1 – P(D) = 1 – \\(\\frac { 1 }{ 5 }\\) = \\(\\frac { 4 }{ 5 }\\)
\nLet X denote the number of defective bulbs in the sample of 4 bulbs.
\nThen X takes the values 0, 1, 2, 3, 4
\n\"NCERT<\/p>\n

Another Method:
\nThere are 6 defective bulbs and 24 non defective bulbs
\nP(getting a defective bulb) = \\(\\frac { 6 }{ 30 }\\) = \\(\\frac { 1 }{ 5 }\\)
\nP(getting a non defective bulb) = \\(\\frac { 24 }{ 30 }\\) = \\(\\frac { 4 }{ 5 }\\)
\nLet X denote the number of defective bulbs.
\nThen X takes the values 0, 1, 2, 3, 4
\nP(X = 0) = P(no defective bulb)
\n\"NCERT<\/p>\n

Question 7.
\nA coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tads.
\nSolution:
\n\"NCERT<\/p>\n

Question 8
\nA random variable X has the following probability distribution:
\n\"NCERT
\nDetermine
\n(i) k
\n(ii) P(X < 3) (iii) P(X > 6)
\n(iv) P(0 < X < 3)
\nSolution:
\n(i) k
\n\"NCERT<\/p>\n

Question 9.
\nThe random variable X has a probability distribution P (X) of the following form, where k is some number
\n\"NCERT
\n(a) Determine the value of k
\n(b) Find P(X < 2), P (X \u2264 2), P(X \u2265 2)
\nSolution:
\n\"NCERT<\/p>\n

Question 10.
\nFind the mean number of heads in three tosses of a fair coin.
\nSolution:
\nWhen 3 coins are tossed, the sample space S = {HHH, HHT, FITH, HTT, THH, THT, TTH, TTT}
\nLet X denote the number of heads
\nThen X takes the values 0, 1, 2, 3
\nP(X = 0) = P(no head) = \\(\\frac { 1 }{ 8 }\\)
\nP(X = 1) = P(one head) = \\(\\frac { 3 }{ 8 }\\)
\nP(X = 2) = P(two heads) = \\(\\frac { 3 }{ 8 }\\)
\nP(X = 3) = P(three heads) = \\(\\frac { 1 }{ 8 }\\)
\nThe probability distribution of X is
\n\"NCERT<\/p>\n

Question 11.
\nTwo dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
\nSolution:
\nLet X denote the number of 6\u2019s when two dice are thrown.
\nThen X takes the values 0, 1, 2
\nP(X = 0) = P(no six on both dice)
\n\"NCERT<\/p>\n

Question 12.
\nTwo numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E(X).
\nSolution:
\nThere are six numbers 1, 2, 3, 4, 5, 6 one of them is selected in 6 ways
\nWhen one of the numbers has been selected, 5 numbers are left, one number out of 5 may be select in 5 ways
\n\u2234 No. of ways of selecting two numbers without replacement out of 6 positive integers = 6 x 5 = 30
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 13.
\nLet X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.
\nSolution:
\nWhen two dice are rolled, then the sample space S has 36 simple events.
\nLet X denote the sum of numbers on the two dice Then X takes the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
\nP(X = 2) = P{(1,1)} = \\(\\frac { 1 }{ 36 }\\)
\nP(X = 3) = P{(1,2), (2,1)} = \\(\\frac { 2 }{ 36 }\\)
\nP(X = 4) = P{(1,3),(2,2),(3,1)} = \\(\\frac { 3 }{ 36 }\\)
\nP(X = 5) = P{(1,4),(2,3),(3,2),(4,1)} = \\(\\frac { 4 }{ 36 }\\)
\nP(X = 6) = P{(1,5), (2,4), (3,3), (4,2), (5,1)} = \\(\\frac { 5 }{ 36 }\\)
\nP(X = 7) = P{(1,6), (2,5), (3,4), (4,3), (5,2), (6, 1)} = \\(\\frac { 6 }{ 36 }\\)
\nP(X = 8) = P{(2,6), (3, 5), (4,4), (5, 3), (6,2)} = \\(\\frac { 5 }{ 36 }\\)
\nP(X = 9) = P{(3,6), (4,5), (5,4), (6,3)} = \\(\\frac { 4 }{ 36 }\\)
\nP(X = 10) = P{(4, 6), (5, 5), (6,4)} = \\(\\frac { 3 }{ 36 }\\)
\nP(X=11) = P{(5,6),(6, 5)} = \\(\\frac { 52}{ 36 }\\)
\nP(X =12) = P{(6, 6)} = \\(\\frac { 1 }{ 36 }\\)
\nThe probability distribution of X is
\n\"NCERT<\/p>\n

Question 14.
\nA class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded.What is the probability distribution of the random variable X ? Find mean, variance and standard deviation of X?
\nSolution:
\nX denotes the age of the student selected.
\n\u2234 X takes the values 14, 15, 16, 17, 18, 19, 20, 21
\nThe data can be summarised into the following table
\n\"NCERT<\/p>\n

Question 15.
\nIn a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0, if he opposed, and X = 1, if he is in favour Find E (X) and Var (X).
\nSolution:
\nX takes the values 0 and 1
\n\"NCERT<\/p>\n

Question 16.
\nThe mean of the number obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
\n(a) 1
\n(b) 2
\n(c) 5
\n(d) \\(\\frac { 8 }{ 3 }\\)
\nSolution:
\nLet X denote the number written on the face of the die.
\nThen X takes the values 1, 2, 5
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 17.
\nSuppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. What is the value of E (X)?
\n(a) \\(\\frac { 37 }{ 221 }\\)
\n(b) \\(\\frac { 5 }{ 13 }\\)
\n(c) \\(\\frac { 1 }{ 13 }\\)
\n(d) \\(\\frac { 2 }{ 13 }\\)
\nSolution:
\nLet X denote the number of aces
\nThen X can take the values 0, 1, 2
\nP(X = 0) = P(no ace)
\n\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-13-ex-13-4\/ NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.4 Question 1. State which of the following are not the probability distributions of a random variable. Give reasons for your …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 - MCQ Questions<\/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:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-13-ex-13-4\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-13-ex-13-4\/ NCERT Solutions for Class 12 Maths Chapter 13 Probability Exercise 13.4 Question 1. 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State which of the following are not the probability distributions of a random variable. 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