Vectors and Matrices Assignment

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Vectors and Matrices
TABLE OF CONTENTTASK 1............................................................................................................................................11.1 Probabilities of familiar events.............................................................................................11.2 Relative frequencies as probability measure........................................................................41.3 Combined probabilities.........................................................................................................61. 4 Binomial, Poisson and Gaussian distribution......................................................................71.5 Simple problem solutions using nCr, nPr and n!..................................................................9TASK 2............................................................................................................................................92.1 Estimates of modalism, skew and standard deviation..........................................................9...................................................................................................................................................102.2 Confidence intervals and standard error..............................................................................112.3 Implications of central limit theorem (CLT).......................................................................122.4 Probability of statistics and interpretations of the results...................................................12
TASK 11.1 Probabilities of familiar eventsMutually exclusive:These are defined as the events which cannot occur simultaneously. For instance theevent A is defined as occurrence of head and event B is known as occurrence of tail whiletossing a coin. When event A occurs then, it is not possible that event B will occur. Such types ofevents in which A∩B = 0 then A and B are called mutually exclusive events.Conditional probability:The probability of an event A is called conditional probability when the occurrence of adepends upon the knowledge of other event B which has already occurred. The conditionalprobability of A is represented by P (A, B) and is given by:P (A|B) = P (A and B) / P (A)Dependent and independent:The two events are known as dependent events when the outcomes of first event affectthe occurrence or probability of other event. For example if two events A and B are dependentthen their probability can be derived by using formula of conditional probability.Similarly, the events whose occurrence does not depend upon the occurrence of anotherevent are known as independent events. When two events A and B are independent then theysatisfy the following relation:P (A and B) = P (A). P (B)1.Two fair spinners with faces numbered 1-4Probability of spinner indicating sum of 5 = ?SolutionPossible outcomes of both the spinners= (1, 1) (1, 2) (1, 3) (1, 4)(2, 1) (2, 2) (2, 3) (2, 4)(3, 1) (3, 2) (3, 3) (3, 4)(4, 1) (4, 2) (4, 3) (4, 4)Total number of outcomes = 16Favourable number of outcomes with sum equals to 5 and number 3 on at least one spinner = 21
[(3, 2) (2, 3)]Required probability = 2/16 = 1/82.)P(C) = 0.2P (D) = 0.6P (C| D) = 0.3SolutionFrom the formula of conditional probability we have:P (C| D) = P (C ∩ D) / P(C)On putting values from above we get:P (C ∩ D) = 0.3 * 0.2 = 0.06P (D| C) = P (C ∩ D) / P(D) = 0.06 / 0.6 = 0.1P (C U D) = P (C) + P (D) - P (C ∩ D)P (C U D) = 0.2 + 0.6 – 0.06 = 0.74Now using the formula: P (C' ∩ D') = 1 - P (C U D)P (C' ∩ D') = 1 – 0.74 = 0.26Using the formula: P (C' ∩ D) = P (D) - (C ∩ D)= 0.6 – 0.06 = 0.54P (D| C) = 0.1P (C' ∩ D') = 0.26P (C' ∩ D) = 0.543.)Total number of cards = 52Probability of card to be an ace, given that it is diamond = ?SolutionLet A be the event that card is an ace and B is that event that card is diamondRequired probability = P (A|B)2
P (B) = 13/52 = 0.25P (A and B) = p (A ∩ B) = 1/52 = 0.019P (A|B) = p (A ∩ B) / P (B) = 0.076Calculate probabilities1.)A and B are mutually exclusive events.P (A) = 0.2P (B) = 0.4Solutiona.) P (A U B)For the mutually exclusive events: P (A U B) = P (A) + P (B)P (A U B) = 0.2 +0.4 = 0.6b.) P (A ∩ B')P (AB') = P (A) - (AB)For mutually exclusive events (AB)= 0 soP (A ∩ B') = P (A) – 0 = 0.2c.) P (A' ∩ B')P (A' ∩ B') = 1 - P (A U B) = 1 – 0.6 = 0.42.)A and B are independent eventsP (A) = ¼ = 0.25P (B) = 1/5 = 0.2Solutiona.)P (A ∩ B)For independent events P (A ∩ B) = P (A). P (B)P (A ∩ B)= 0.25 * 0.2 = 0.05b.) P (A ∩ B')P (A ∩ B') = P (A) - P (A ∩ B) = 0.25 – 0.05 = 0.23
P (A ∩ B') = 0.2c.) P (A' ∩ B')P (A U B) =P (A) + P (B) - P (A ∩ B)= 0.25 + 0.2 – 0.05 = 0.4P (A' ∩ B') = 1 - P (A U B) = 1- 0.4P (A' ∩ B') = 0.63.)Number of red beads = 3Number of blue beads = 5Total beads in bag = 8Solutiona.) Probability that both beads are blueRequired probability = (5/8) * (5/8) = 25/64 = 0.39P (Both beads are blue) = 0.39b.) Probability that second bead is blueRequired probability = P (both are blue) + P (First is red and second is blue)= 0.39 + [(3/8) * (5/8)] = 0.62P (second bead is blue) = 0.621.2 Relative frequencies as probability measureNo. of meals a dayFrequency16223356414a.) Relative frequency of choosing a person eating 3 meals a daySolution:4
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