{"id":35218,"date":"2022-08-25T11:22:07","date_gmt":"2022-08-25T05:52:07","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35218"},"modified":"2022-08-25T11:22:07","modified_gmt":"2022-08-25T05:52:07","slug":"inter-2nd-year-maths-2a-random-variables-and-probability-distributions-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2a-random-variables-and-probability-distributions-important-questions\/","title":{"rendered":"Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions"},"content":{"rendered":"

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

Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions<\/h2>\n

Question 1.
\nA Poisson variable satisfies P(X = 1) = P(X = 2). Find P(X = 5) (Mar.14; May ’13,\u201906).
\nSolution:
\n\"Inter<\/p>\n

Question 2.
\nThe probability that a person chosen at random is left handed (in hand writing) is 0.1. What is the probability that in a group of ten people there is one, who is left handed? (TS Mar.\u201916; AP Mar. ’17 ’15)
\nSolution:
\nHere n = 10
\np = 0.1
\nq = 1 – p = 1 – 0.1 = 0.9
\nP(X = 1) = 10<\/sup>C1<\/sub> (0.1)1<\/sup> (0.9)10 – 1<\/sup>
\n= 10 \u00d7 0.1 \u00d7 (0.9)
\n= 1 \u00d7 (0.9)
\n= (0.9)<\/p>\n

\"Inter<\/p>\n

Question 3.
\nA Poisson variable satisfies P(X = 1) = P(X = 2). Find P(X = 5) (Mar.14; May ’13, ’06)
\nSolution:
\nGiven P(X = 1) = P(X = 2)
\n\"Inter<\/p>\n

Question 4.
\n\"Inter
\nis the probability distribution of a random variable X. Find the value of K and the variance of X. ( March 2006) (TS Mar. 17)
\nSolution:
\nSum of the probabilities = 1
\n0.1 + k + 0.2 + 2k + 0.3 + k = 1
\n4k + 0.6 = 1
\n4k = 1 – 0.6 = 0.4
\nk = \\(\\frac{0.4}{4}\\) = 0.1
\nMean = (-2) (0.1) + (-1) k + 0(0.2) + 1(2k) + 2(0.3) + 3k
\n= -0.2 – k + 0 + 2k + 0.6 + 3k
\n= 4k + 0.4
\n=4(0.1) + 0.4
\n= 0.4 + 0.4
\n= 0.8
\n\u03bc = 0.8
\n\"Inter
\n\u2234 Variance = 4(0.1) + 1(k) + 0(0.2) + 1(2k) + 4(0.3) + 9k – \u03bc2<\/sup>
\n= 0.4 + k + 0 + 2k + 4(0.3) + 9k – \u03bc2<\/sup>
\n= 12k + 0.4 + 1.2 – (0.8)2<\/sup>
\n= 12(0.1) + 1.6 – 0.64
\n= 1.2 + 1.6 – 0.64 .
\n\u2234 \u03c32<\/sup> = 2.8 – 0.64 = 2.16<\/p>\n

Question 5.
\nA random variable X has the following probability distribution. (TS & AP Mar. \u201816)
\n\"Inter
\nFind
\ni) k
\nii) the mean and
\niii) P(0 < X < 5).
\nSolution:
\nSum of the probabilities = 1
\n\u21d2 0 + k + 2k + 2k + 3k + k2<\/sup> + 2k2<\/sup> + 7k2<\/sup> + k = 1
\n\u21d2 10k2<\/sup> + 9k = 1
\n\u21d2 10k2<\/sup> + 9k – 1 = 0
\n\u21d2 10k(k + 1) – 1(k + 1) = 0
\n\u21d2 (10k – 1) (k + 1) = 0
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

iii) P(0 < x < 5)
\nP(0 < x < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
\n= k + 2k + 2k + 3k
\n= 8k
\n= 8\\(\\frac{1}{10}\\)
\n= \\(\\frac{8}{10}\\) = \\(\\frac{4}{5}\\)<\/p>\n

Question 6.
\nThe range of a random variable X is {0, 1, 2}. Given that P(X = 0) = 3c3<\/sup>, P(X = 1) = 4c – 10c2<\/sup>, P(X = 2) = 5c – 1
\ni) Find the value of c
\nii) P(X < 1),P(1 < X \u2264 2) and P (0 < X \u2264 3) (AP & TS Mar. 15, 13, \u201811, 07, 05; May \u201811\u2019)
\nSolution:
\nP(X = 0) + P(X = 1) + P(X = 2) = 1
\n3c3<\/sup> + 4c – 10c2<\/sup> + 5c – 1 = 1
\n3c3<\/sup> – 10c2<\/sup> + 9c – 2 = 0
\nc = 1 satisfy this equation
\nc = 1 \u21d2 P(X = 0) = 3 which is not possible
\nDividing with c – 1, we get
\n3c2<\/sup> – 7c + 2 = 0
\n(c – 2) (3c – 1) = 0
\n\"Inter<\/p>\n

ii) P(1 < X \u2264 2) = P(X = 2) = 5c – 1
\n= \\(\\frac{5}{3}\\) – 1 = \\(\\frac{2}{3}\\)<\/p>\n

iii)P(0 < X \u2264 3) = P(X = 1) + P(X = 2)
\n= 4c – 10c2<\/sup> + 5c – 1
\n= 9c – 10c2<\/sup> – 1
\n= 9.\\(\\frac{1}{3}\\) – 10.\\(\\frac{1}{9}\\) – 1
\n= 3 – \\(\\frac{10}{9}\\) – 1 = 2 – \\(\\frac{10}{9}\\) = \\(\\frac{8}{9}\\)<\/p>\n

Question 7.
\nOne in 9 ships is likely to be wrecked, when they are set on sail, when 6 ships are on sail, find the probability for
\ni) Atleast one will arrive safely
\nii) Exactly, 3 will arrive safely. (Mar. 2008)
\nSolution:
\np = probability of ship to be wrecked = \\(\\frac{1}{9}\\)
\n\"Inter<\/p>\n

Question 8.
\nIf the mean and variance of a binomial variable X are 2.4 and 1.44 respectively, find P(1 < X \u2264 4). (May ’06)
\nSolution:
\nMean = np = 2.4 …… (1)
\nVariance = npq = 1.44 …… (2)
\nDividing (2) by (1),
\n\\(\\frac{\\mathrm{npq}}{\\mathrm{np}}\\) = \\(\\frac{1.44}{2.4}\\)
\nq = 0.6 = \\(\\frac{3}{5}\\)
\n2
\np = 1 – q = 1 – 0.6 = 0.4 = \\(\\frac{2}{5}\\)
\nSubstituting in (1)
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 9.
\nThe probability distribution of a random variable X is given below. (AP Mar. \u201817\u2019) (Mar. \u201814; May \u201813)
\n\"Inter
\nFind the value of K, and the mean and variance of X.
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 10.
\nThe mean and variance of a binomial distribution are 4 and 3 respectively. Fix the distribution and find P(X \u2265 1)
\n(AP Mar. \u201816 TS Mar. 17 \u201815, ’08)
\nSolution:
\nGiven distribution \u00a1s Binomial distribution with mean = np = 4
\nvariance = npq = 3
\n\u2234 \\(\\frac{n p q}{n p}\\) = \\(\\frac{3}{4}\\)
\n\u21d2 q = \\(\\frac{3}{4}\\)
\nso that p = 1 – q
\n= 1 – \\(\\frac{3}{4}\\) = \\(\\frac{1}{4}\\)
\n\u2234 np = 4
\nn\\(\\frac{1}{4}\\) = 4
\n\u21d2 n = 16
\n\"Inter<\/p>\n

Question 11.
\nA cubical die is thrown. Find the mean and variance of X, giving the number on the face that shows up.
\nSolution:
\nLet S be the sample space and X be the random variable associated with S, where P(X) is given by the following table
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 12.
\nThe probability distribution of a random variable X is given below.
\n\"Inter
\nFind the value of k, and the mean and variance of X
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 13.
\nIf x is a random variable with probability distribution. P(X = k) = \\(\\frac{(k+1) c}{2^{k}}\\), k = 0, 1, 2 then find c.
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 14.
\nLet X be a random variable such that P(X = -2) =P(X = -1) = P(X = 2) = P(X = 1) = \\(\\frac{1}{6}\\) and P(X = 0) = \\(\\frac{1}{3}\\). Find the mean and variance of X.
\nSolution:
\n\"Inter
\n\"Inter<\/p>\n

Question 15.
\nTwo dice are rolled at random. Find the probability distribution of the sum of the numbers on them. Find the mean of the random variable.
\nSolution:
\nWhen two dice are rolled, the sample space
\nS contains 6 \u00d7 6 = 36 sample points.
\nS = {(1, 1), (1, 2) (1, 6), (2, 1), (2, 2) (6, 6)}
\nLet X denote the sum of the numbers on the tw0 dice.
\nThen the range of X = {2, 3, 4, ……… 12}
\nThe Prob. distribution of X is given by the following table.
\n\"Inter<\/p>\n

Question 16.
\n8 coins are tossed simultaneously. Find the probability of getting atleast 6 heads.
\nSolution:
\np = Probability of getting head = \\(\\frac{1}{2}\\)
\nq = 1 – p = 1 – \\(\\frac{1}{2}\\) = \\(\\frac{1}{2}\\) ; n = 8
\nP(X \u2265 6) = P(X = 6) + P(X = 7) + P(X = 8)
\n\"Inter<\/p>\n

Question 17.
\nThe mean and variance of a binomial distribution are 4 and 3 respectively. Fix the distribution and find P(X \u2265 1)
\n(A.P. Mar. ’16, T.S. Mar. ’15, ’08)
\nSolution:
\nGiven distribution is Binomial distribution with mean = np = 4
\nvariance = npq = 3
\n\u2234 \\(\\frac{n p q}{n p}\\) = \\(\\frac{3}{4}\\)
\n\u21d2 q = \\(\\frac{3}{4}\\)
\nso that p = 1 – q
\n= 1 – \\(\\frac{3}{4}\\) = \\(\\frac{1}{4}\\)
\n\u2234 np = 4
\nn\\(\\frac{1}{4}\\) = 4
\n\u21d2 n = 16
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 18.
\nThe probability that a person chosen at random is left handed (in hand writing) is 0.1. What is the probability that in a group of ten people there is one, who is left handed ? (Mar. 16, AP. Mar. ’15)
\nSolution:
\nHere n = 10
\np = 0.1
\nq = 1 – p = 1 – 0.1 = 0.9
\nP(X = 1) = 10<\/sup>C1<\/sub> (0.1)1<\/sup> (0.9)10 – 1<\/sup>
\n= 10 \u00d7 0.1 \u00d7 (0.9)
\n= 1 \u00d7 (0.9)
\n= (0.9)<\/p>\n

Question 19.
\nIn a b\u00f5ok of 450 pages, there are 400 typographical errors. Assumi\u00f1g that the number of err\u00f6rs per page follow the
\npoisson law, find the probability that a random sample of 5 pages will contain no typographical error.
\nSolution:
\nThe average number of errors per page in the book is
\n\"Inter
\nThe required probability that a random sample of 5 pages will contain no error is
\n[P(X = 0)]5<\/sup> = \\(\\left(e^{-8 \/ 9}\\right)^{5}\\)<\/p>\n

Question 20.
\nDeficiency of red cells in the blood cells is determined by examining a specimen of blood under a microscope. Suppose a small fixed volume contains on an average 20 red cells for normal persons. Using the poisson distribution, find the probability that a specimen of blood taken from a normal person will contain less than 15 red cells.
\nSolution:
\n\"Inter<\/p>\n

Question 21.
\nA Poisson variable satisfies P(X = 1) = P(X = 2). Find P(X = 5) (Mar. \u201814; May \u201806, \u201813)
\nSolution:
\nGiven P(X = 1) = P(X = 2)
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 2A Important Questions Inter 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions which are most likely to be asked in the exam. Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Important Questions Question 1. A Poisson variable satisfies P(X = 1) = P(X = 2). Find …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35218"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35218"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35218\/revisions"}],"predecessor-version":[{"id":35219,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35218\/revisions\/35219"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35218"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35218"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35218"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}